Nama : Aditya Isnugraha
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER 12
- REVIEW QUESTIONS
11. What is the message protocol of an object?
Answer :
12. From where are Smalltalk objects allocated?
Answer :
13. Explain how Smalltalk messages are bound to methods. When does this take place?
Answer :
14. What type checking is done in Smalltalk? When does it take place?
Answer :
15. What kind of inheritance, single or multiple, does Smalltalk support?
Answer :
- PROBLEM SET
11. Explain the advantages and disadvantages of having all values in a language be objects.
Answer :
12. What exactly does it mean for a subclass to have an is-a relationship with its parent class?
Answer :
13. Describe the issue of how closely the parameters of an overriding method must match those of the method it overrides.
Answer :
14. Explain type checking in Smalltalk.
Answer :
15. The designers of Java obviously thought it was not worth the additional efficiency of allowing any method to be statically bound, as is the case with C++. What are the arguments for and against the Java design?
Answer :
Rabu, 14 Januari 2015
CHAPTER 11
Nama : Aditya Isnugraha
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER 11
- REVIEW QUESTIONS
11. What is the use of the Ada use clause?
Answer :
12. What is the fundamental difference between a C++ class and an Ada package?
Answer :
13. From where are C++ objects allocated?
Answer :
14. In what different places can the definition of a C++ member function appear?
Answer :
15. What is the purpose of a C++ constructor?
Answer :
- PROBLEM SET
11. What are the arguments for and against the Objective-C design that method access cannot be restricted?
Answer :
12. Why are destructors rarely used in Java but essential in C++?
Answer :
13. What are the arguments for and against the C++ policy on inlining of methods?
Answer :
14. Describe a situation where a C# struct is preferable to a C# class.
Answer :
15. Explain why naming encapsulations are important for developing large programs.
Answer :
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER 11
- REVIEW QUESTIONS
11. What is the use of the Ada use clause?
Answer :
12. What is the fundamental difference between a C++ class and an Ada package?
Answer :
13. From where are C++ objects allocated?
Answer :
14. In what different places can the definition of a C++ member function appear?
Answer :
15. What is the purpose of a C++ constructor?
Answer :
- PROBLEM SET
11. What are the arguments for and against the Objective-C design that method access cannot be restricted?
Answer :
12. Why are destructors rarely used in Java but essential in C++?
Answer :
13. What are the arguments for and against the C++ policy on inlining of methods?
Answer :
14. Describe a situation where a C# struct is preferable to a C# class.
Answer :
15. Explain why naming encapsulations are important for developing large programs.
Answer :
CHAPTER 10
Nama : Aditya Isnugraha
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER10
- REVIEW QUESTIONS
11. What is an EP, and what is its purpose?
Answer :
12. How are references to variables represented in the static-chain method?
Answer :
It is represented by static depth.
13. Name three widely used programming languages that do not allow nested subprograms.
Answer :
14. What are the two potential problems with the static-chain method
Answer :
1. It is difficult for a programmer working on a time-critical program to estimate the costs of nonlocal references, because the cost of each reference depends on the depth of nesting between the reference and the scope of declaration.
2. Subsequent code modifications may change nesting depths, thereby changing the timing of some references, both in the changed code and possibly in code far from the changes.
15. Explain the two methods of implementing blocks.
Answer :
- Blocks can be implemented by using the static-chain process for implementing nested subprograms. Blocks are treated as parameterless subprograms that are always called from the same place in the program. Therefore, every block has an activation record. An instance of its activation record is created every time the block is executed.
- Blocks can also be implemented in a different and somewhat simpler and more efficient way. The maximum amount of storage required for block variables at any time during the execution of a program can be statically determined, because blocks are entered and exited in strictly textual order. This amount of space can be allocated after the local variables in the activation record. Offsets for all block variables can be statically computed, so block variables can addressed exactly as if they were local variables.
- PROBLEM SET
8. Pascal allows gotos with nonlocal targets. How could such statements be handled if static chains were used for nonlocal variable access? Hint: Consider the way the correct activation record instance of the static par- ent of a newly enacted procedure is found (see Section 10.4.2).
Answer : Following the hint stated with the question, the target of every goto in a program could be represented as an address and a nesting_depth, where the nesting_depth is the difference between the nesting level of the procedure that contains the goto and that of the procedure containing the target. Then, when a goto is executed, the static chain is followed by the number of links indicated in the nesting_depth of the goto target. The stack top pointer is reset to the top of the activation record at the end of the chain.
9. The static-chain method could be expanded slightly by using two static links in each activation record instance where the second points to the static grandparent activation record instance. How would this approach affect the time required for subprogram linkage and nonlocal references?
Answer : Including two static links would reduce the access time to nonlocals that are defined in scopes two steps away to be equal to that for nonlocals that are one step away. Overall, because most nonlocal references are relatively close, this could significantly increase the execution efficiency of many programs.
10. Design a skeletal program and a calling sequence that results in an acti- vation record instance in which the static and dynamic links point to dif- ferent activation-recorded instances in the run-time stack.
Answer :
>\verb+ + X : Integer;\\
\verb+ +procedure Bigsub is\\
\verb+ +\verb+ + A, B, C : Integer;\\
\verb+ +\verb+ + procedure Sub1 is\\
\verb+ +\verb+ +\verb+ + A, D : Integer;\\
\verb+ +\verb+ +\verb+ + begin — of Sub1\\
\verb+ +\verb+ +\verb+ + A := B + C; $\longleftarrow$ 1\\
\verb+ +\verb+ +\verb+ + …\\
\verb+ + end; — of Sub1\\
\verb+ + procedure Sub2(X : Integer) is\\
\verb+ +\verb+ + B, E : Integer;\\
\verb+ +\verb+ + procedure Sub3 is\\
\verb+ +\verb+ +\verb+ + C, E : Integer;\\
\verb+ +\verb+ +\verb+ + begin — of Sub3\\
\verb+ +\verb+ +\verb+ + …\\
\verb+ +\verb+ +\verb+ + Sub1;\\
\verb+ +\verb+ +\verb+ + …\\
\verb+ +\verb+ +\verb+ + E := B + A; $\longleftarrow$ 2\\
\verb+ +\verb+ + end; — of Sub3\\
\verb+ +\verb+ + begin — of Sub2\\
\verb+ +\verb+ + …\\
\verb+ +\verb+ + Sub3;\\
\verb+ +\verb+ + …\\
\verb+ +\verb+ + A := D + E; $\longleftarrow$ 3\\
\verb+ + end; — of Sub2\\
\verb+ + begin — of Bigsub\\
\verb+ +\verb+ + …\\
\verb+ +\verb+ + Sub2(7);\\
\verb+ +\verb+ + …\\
\verb+ + end; — of Bigsub\\
begin — of Main\_2\\
\verb+ + …\\
\verb+ + Bigsub;\\
\verb+ + …\\
end; — of Main\_2\\
\\
The sequence of procedure calls is:\\
Main\_2 calls Bigsub\\
Bigsub calls Sub2\\
Sub2 calls Sub3\\
Sub3 calls Sub1\\
\\
The activation records with static and dynamic links is as follows:\\
\begin{figure}
\centering
\includegraphics[scale=0.5]{ari}
\end{figure}
11. If a compiler uses the static chain approach to implementing blocks, which of the entries in the activation records for subprograms are needed in the activation records for blocks?
Answer : There are two options for implementing blocks as parameterless subprograms: One way is to use the same activation record as a subprogram that has no parameters. This is the most simple way, because accesses to block variables will be exactly like accesses to local variables. Of course, the space for the static and dynamic links and the return address will be wasted. The alternative is to leave out the static and dynamic links and the return address, which saves space but makes accesses to block variables different from subprogram locals.
12. Examine the subprogram call instructions of three different architec- tures, including at least one CISC machine and one RISC machine, and write a short comparison of their capabilities. (The design of these instructions usually determines at least part of the compiler writer’s design of subprogram linkage.)
Answer :
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER10
- REVIEW QUESTIONS
11. What is an EP, and what is its purpose?
Answer :
12. How are references to variables represented in the static-chain method?
Answer :
It is represented by static depth.
13. Name three widely used programming languages that do not allow nested subprograms.
Answer :
14. What are the two potential problems with the static-chain method
Answer :
1. It is difficult for a programmer working on a time-critical program to estimate the costs of nonlocal references, because the cost of each reference depends on the depth of nesting between the reference and the scope of declaration.
2. Subsequent code modifications may change nesting depths, thereby changing the timing of some references, both in the changed code and possibly in code far from the changes.
15. Explain the two methods of implementing blocks.
Answer :
- Blocks can be implemented by using the static-chain process for implementing nested subprograms. Blocks are treated as parameterless subprograms that are always called from the same place in the program. Therefore, every block has an activation record. An instance of its activation record is created every time the block is executed.
- Blocks can also be implemented in a different and somewhat simpler and more efficient way. The maximum amount of storage required for block variables at any time during the execution of a program can be statically determined, because blocks are entered and exited in strictly textual order. This amount of space can be allocated after the local variables in the activation record. Offsets for all block variables can be statically computed, so block variables can addressed exactly as if they were local variables.
- PROBLEM SET
8. Pascal allows gotos with nonlocal targets. How could such statements be handled if static chains were used for nonlocal variable access? Hint: Consider the way the correct activation record instance of the static par- ent of a newly enacted procedure is found (see Section 10.4.2).
Answer : Following the hint stated with the question, the target of every goto in a program could be represented as an address and a nesting_depth, where the nesting_depth is the difference between the nesting level of the procedure that contains the goto and that of the procedure containing the target. Then, when a goto is executed, the static chain is followed by the number of links indicated in the nesting_depth of the goto target. The stack top pointer is reset to the top of the activation record at the end of the chain.
9. The static-chain method could be expanded slightly by using two static links in each activation record instance where the second points to the static grandparent activation record instance. How would this approach affect the time required for subprogram linkage and nonlocal references?
Answer : Including two static links would reduce the access time to nonlocals that are defined in scopes two steps away to be equal to that for nonlocals that are one step away. Overall, because most nonlocal references are relatively close, this could significantly increase the execution efficiency of many programs.
10. Design a skeletal program and a calling sequence that results in an acti- vation record instance in which the static and dynamic links point to dif- ferent activation-recorded instances in the run-time stack.
Answer :
>\verb+ + X : Integer;\\
\verb+ +procedure Bigsub is\\
\verb+ +\verb+ + A, B, C : Integer;\\
\verb+ +\verb+ + procedure Sub1 is\\
\verb+ +\verb+ +\verb+ + A, D : Integer;\\
\verb+ +\verb+ +\verb+ + begin — of Sub1\\
\verb+ +\verb+ +\verb+ + A := B + C; $\longleftarrow$ 1\\
\verb+ +\verb+ +\verb+ + …\\
\verb+ + end; — of Sub1\\
\verb+ + procedure Sub2(X : Integer) is\\
\verb+ +\verb+ + B, E : Integer;\\
\verb+ +\verb+ + procedure Sub3 is\\
\verb+ +\verb+ +\verb+ + C, E : Integer;\\
\verb+ +\verb+ +\verb+ + begin — of Sub3\\
\verb+ +\verb+ +\verb+ + …\\
\verb+ +\verb+ +\verb+ + Sub1;\\
\verb+ +\verb+ +\verb+ + …\\
\verb+ +\verb+ +\verb+ + E := B + A; $\longleftarrow$ 2\\
\verb+ +\verb+ + end; — of Sub3\\
\verb+ +\verb+ + begin — of Sub2\\
\verb+ +\verb+ + …\\
\verb+ +\verb+ + Sub3;\\
\verb+ +\verb+ + …\\
\verb+ +\verb+ + A := D + E; $\longleftarrow$ 3\\
\verb+ + end; — of Sub2\\
\verb+ + begin — of Bigsub\\
\verb+ +\verb+ + …\\
\verb+ +\verb+ + Sub2(7);\\
\verb+ +\verb+ + …\\
\verb+ + end; — of Bigsub\\
begin — of Main\_2\\
\verb+ + …\\
\verb+ + Bigsub;\\
\verb+ + …\\
end; — of Main\_2\\
\\
The sequence of procedure calls is:\\
Main\_2 calls Bigsub\\
Bigsub calls Sub2\\
Sub2 calls Sub3\\
Sub3 calls Sub1\\
\\
The activation records with static and dynamic links is as follows:\\
\begin{figure}
\centering
\includegraphics[scale=0.5]{ari}
\end{figure}
11. If a compiler uses the static chain approach to implementing blocks, which of the entries in the activation records for subprograms are needed in the activation records for blocks?
Answer : There are two options for implementing blocks as parameterless subprograms: One way is to use the same activation record as a subprogram that has no parameters. This is the most simple way, because accesses to block variables will be exactly like accesses to local variables. Of course, the space for the static and dynamic links and the return address will be wasted. The alternative is to leave out the static and dynamic links and the return address, which saves space but makes accesses to block variables different from subprogram locals.
12. Examine the subprogram call instructions of three different architec- tures, including at least one CISC machine and one RISC machine, and write a short comparison of their capabilities. (The design of these instructions usually determines at least part of the compiler writer’s design of subprogram linkage.)
Answer :
Rabu, 17 Desember 2014
CHAPTER 9
Nama : Aditya Isnugraha
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER 9
- REVIEW QUESTIONS
11. What are the design issues for subprograms?
Answer :
The design issues for subprograms are:
- Are local variables statically or dynamically allocated?
- Can subprogram definitions appear in other subprogram definitions?
- What parameter-passing method or methods are used?
- Are the types of the actual parameters checked against the types of the
formal parameters?
- If subprograms can be passed as parameters and subprograms can be nested,
what is the referencing environment of a passed subprogram?
- Can subprograms be overloaded?
- Can subprograms be generic?
- If the language allows nested subprograms, are closures supported?
12. What are the advantages and disadvantages of dynamic local variables?
Answer :
There are several advantages of stack-dynamic local variables, the primary one being the flexibility they provide to the subprogram. It is essential that recursive subprograms have stack-dynamic local variables. Another advantage of stack-dynamic locals is that the storage for local variables in an active subprogram can be shared with the local variables in all inactive subprograms.
The main disadvantages of stack-dynamic local variables are the following:
First, there is the cost of the time required to allocate, initialize (when necessary), and deallocate such variables for each call to the subprogram. Second, accesses to stack-dynamic local variables must be indirect, whereas accesses to static variables can be direct. This indirectness is required because the place in the stack where a particular local variable will reside can be determined only during execution. Finally, when all local variables are stack dynamic, subprograms cannot be history sensitive; that is, they cannot retain data values of local variables between calls.
13. What are the advantages and disadvantages of static local variables?
Answer :
The primary advantage of static local variables over stack-dynamic local variables is that they are slightly more efficient—they require no run-time overhead for allocation and deallocation. Also, if accessed directly, these accesses are obviously more efficient. And, of course, they allow subprograms to be history sensitive. The greatest disadvantage of static local variables is their inability to support recursion. Also, their storage cannot be shared with the local variables of other inactive subprograms.
14. What languages allow subprogram definitions to be nested?
Answer :
For a long time, the only languages that allowed nested subprograms were those directly descending from Algol 60, which were Algol 68, Pascal, and Ada. JavaScript, Python, Ruby, and Lua are also, most functional programming languages allow subprograms to be nested.
15. What are the three semantic models of parameter passing?
Answer :
Formal parameters are characterized by one of three distinct semantics models:
(1) They can receive data from the corresponding actual parameter; (2) they can transmit data to the actual parameter; or (3) they can do both. These models are called in mode, out mode, and inout mode, respectively.
- PROBLEM SET
11. C# supports out-mode parameters, but neither Java nor C++ does. Explain the difference.
Answer :
12. Research Jensen’s Device, which was a widely known use of pass-by- name parameters, and write a short description of what it is and how it can be used.
Answer :
Implementing a pass-by-name parameter requires a subprogram to be passed to the called subprogram to evaluate the address or value of the formal parameter. The referencing environment of the passed subprogram must also be passed. This subprogram/referencing environment is a closure. Pass-by-name parameters are both complex to implement and inefficient. They also add significant complexity to the program, thereby lowering its readability and reliability. Because pass-by-name is not part of any widely used language, it is not discussed further here. However, it is used at compile time by the macros in assembly languages and for the generic parameters of the generic subprograms in C++, Java 5.0, and C# 2005.
13. Study the iterator mechanisms of Ruby and CLU and list their similari- ties and differences.
Answer :
14. Speculate on the issue of allowing nested subprograms in programming languages—why are they not allowed in many contemporary languages?
Answer :
Because it is concerned to lead to ambiguity and some error compilation.
15. What are at least two arguments against the use of pass-by-name parameters?
Answer :
Ada compilers are able to determine the defined size of the dimensions of all arrays that are used as parameters at the time subprograms are compiled. In Ada, unconstrained array types can be formal parameters. An unconstrained array type is one in which the index ranges are not given in the array type definition. Definitions of variables of unconstrained array types must include index ranges. The code in a subprogram that is passed an unconstrained array can obtain the index range information of the actual parameter associated with such parameters
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Share this:
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER 9
- REVIEW QUESTIONS
11. What are the design issues for subprograms?
Answer :
The design issues for subprograms are:
- Are local variables statically or dynamically allocated?
- Can subprogram definitions appear in other subprogram definitions?
- What parameter-passing method or methods are used?
- Are the types of the actual parameters checked against the types of the
formal parameters?
- If subprograms can be passed as parameters and subprograms can be nested,
what is the referencing environment of a passed subprogram?
- Can subprograms be overloaded?
- Can subprograms be generic?
- If the language allows nested subprograms, are closures supported?
12. What are the advantages and disadvantages of dynamic local variables?
Answer :
There are several advantages of stack-dynamic local variables, the primary one being the flexibility they provide to the subprogram. It is essential that recursive subprograms have stack-dynamic local variables. Another advantage of stack-dynamic locals is that the storage for local variables in an active subprogram can be shared with the local variables in all inactive subprograms.
The main disadvantages of stack-dynamic local variables are the following:
First, there is the cost of the time required to allocate, initialize (when necessary), and deallocate such variables for each call to the subprogram. Second, accesses to stack-dynamic local variables must be indirect, whereas accesses to static variables can be direct. This indirectness is required because the place in the stack where a particular local variable will reside can be determined only during execution. Finally, when all local variables are stack dynamic, subprograms cannot be history sensitive; that is, they cannot retain data values of local variables between calls.
13. What are the advantages and disadvantages of static local variables?
Answer :
The primary advantage of static local variables over stack-dynamic local variables is that they are slightly more efficient—they require no run-time overhead for allocation and deallocation. Also, if accessed directly, these accesses are obviously more efficient. And, of course, they allow subprograms to be history sensitive. The greatest disadvantage of static local variables is their inability to support recursion. Also, their storage cannot be shared with the local variables of other inactive subprograms.
14. What languages allow subprogram definitions to be nested?
Answer :
For a long time, the only languages that allowed nested subprograms were those directly descending from Algol 60, which were Algol 68, Pascal, and Ada. JavaScript, Python, Ruby, and Lua are also, most functional programming languages allow subprograms to be nested.
15. What are the three semantic models of parameter passing?
Answer :
Formal parameters are characterized by one of three distinct semantics models:
(1) They can receive data from the corresponding actual parameter; (2) they can transmit data to the actual parameter; or (3) they can do both. These models are called in mode, out mode, and inout mode, respectively.
- PROBLEM SET
11. C# supports out-mode parameters, but neither Java nor C++ does. Explain the difference.
Answer :
12. Research Jensen’s Device, which was a widely known use of pass-by- name parameters, and write a short description of what it is and how it can be used.
Answer :
Implementing a pass-by-name parameter requires a subprogram to be passed to the called subprogram to evaluate the address or value of the formal parameter. The referencing environment of the passed subprogram must also be passed. This subprogram/referencing environment is a closure. Pass-by-name parameters are both complex to implement and inefficient. They also add significant complexity to the program, thereby lowering its readability and reliability. Because pass-by-name is not part of any widely used language, it is not discussed further here. However, it is used at compile time by the macros in assembly languages and for the generic parameters of the generic subprograms in C++, Java 5.0, and C# 2005.
13. Study the iterator mechanisms of Ruby and CLU and list their similari- ties and differences.
Answer :
14. Speculate on the issue of allowing nested subprograms in programming languages—why are they not allowed in many contemporary languages?
Answer :
Because it is concerned to lead to ambiguity and some error compilation.
15. What are at least two arguments against the use of pass-by-name parameters?
Answer :
Ada compilers are able to determine the defined size of the dimensions of all arrays that are used as parameters at the time subprograms are compiled. In Ada, unconstrained array types can be formal parameters. An unconstrained array type is one in which the index ranges are not given in the array type definition. Definitions of variables of unconstrained array types must include index ranges. The code in a subprogram that is passed an unconstrained array can obtain the index range information of the actual parameter associated with such parameters
About these ads
Share this:
CHAPTER 8
Nama : Aditya Isnugraha
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER 8
- REVIEW QUESTION
11. What is unusual about C’s multiple-selection statement?
Answer :
the C switch statement has virtually no restrictions on the placement of the case expressions, which are treated as if they were normal statement labels. This laxness can result in highly complex structure eithin the switch body.
12. On what previous language was C’s switch statement based?
Answer :
ALGOL
13. Explain how C#’s switch statement is safer than that of C.
Answer :
C# has a static semantics rule that disallows the implicit execution of more than one segment. Every segment must end with an explicit unconditional branch statement which transfer control out of the switch statement, or a goto, which can transfer control to one of the selectable segments
14. What are the design issues for all iterative control statements?
Answer :
The design issues are how the iteration is controlled and where the control mechanism should appear in the loop statement.
15. What are the design issues for counter-controlled loop statements?
Answer :
The design issues for counter-controlled loop statement are what the type and scope of the loop variable are, whether it should be legal for the loop variable or loop parameters to be changed in the loop and if so, whether the change affects loop control or not, and if the lop parameters should be evaluated only once or once for every iteration.
- PROBLEM SET
10. In Ada, the choice lists of the case statement must be exhaustive, so that there can be no unrepresented values in the control expression. In C++, unrepresented values can be caught at run time with the default selec- tor. If there is no default, an unrepresented value causes the whole statement to be skipped. What are the pros and cons of these two designs (Ada and C++)?
Answer :
Ada was designed for military grade software development. The idea is that whenever you modify code in such a way that a new case emerges (for example adding a new value for an enumeration type), you are forced to manually revisit (and therefore re-validate) all the case statements that analyze it. Having a "default" is risky: you may forget that there is a case somewhere where the new case should not have been handled by the default.
11. Explain the advantages and disadvantages of the Java for statement, compared to Ada’s for.
Answer :
Java’s variable in the argument of a switch statement can be of integeral type (byte, short, int, etc), char, and String (JDK 1.7 and newer versions), but C++ can only be int or char.
12. Describe a programming situation in which the else clause in Python’s for statement would be convenient.
Answer :
13. Describe three specific programming situations that require a posttest loop.
Answer :
14. Speculate as to the reason control can be transferred into a C loop statement.
Answer :
goto statements can be used to have one cleanup sction in routine rather than multiple return statements, or used to exit out a nested loop that is very long
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER 8
- REVIEW QUESTION
11. What is unusual about C’s multiple-selection statement?
Answer :
the C switch statement has virtually no restrictions on the placement of the case expressions, which are treated as if they were normal statement labels. This laxness can result in highly complex structure eithin the switch body.
12. On what previous language was C’s switch statement based?
Answer :
ALGOL
13. Explain how C#’s switch statement is safer than that of C.
Answer :
C# has a static semantics rule that disallows the implicit execution of more than one segment. Every segment must end with an explicit unconditional branch statement which transfer control out of the switch statement, or a goto, which can transfer control to one of the selectable segments
14. What are the design issues for all iterative control statements?
Answer :
The design issues are how the iteration is controlled and where the control mechanism should appear in the loop statement.
15. What are the design issues for counter-controlled loop statements?
Answer :
The design issues for counter-controlled loop statement are what the type and scope of the loop variable are, whether it should be legal for the loop variable or loop parameters to be changed in the loop and if so, whether the change affects loop control or not, and if the lop parameters should be evaluated only once or once for every iteration.
- PROBLEM SET
10. In Ada, the choice lists of the case statement must be exhaustive, so that there can be no unrepresented values in the control expression. In C++, unrepresented values can be caught at run time with the default selec- tor. If there is no default, an unrepresented value causes the whole statement to be skipped. What are the pros and cons of these two designs (Ada and C++)?
Answer :
Ada was designed for military grade software development. The idea is that whenever you modify code in such a way that a new case emerges (for example adding a new value for an enumeration type), you are forced to manually revisit (and therefore re-validate) all the case statements that analyze it. Having a "default" is risky: you may forget that there is a case somewhere where the new case should not have been handled by the default.
11. Explain the advantages and disadvantages of the Java for statement, compared to Ada’s for.
Answer :
Java’s variable in the argument of a switch statement can be of integeral type (byte, short, int, etc), char, and String (JDK 1.7 and newer versions), but C++ can only be int or char.
12. Describe a programming situation in which the else clause in Python’s for statement would be convenient.
Answer :
13. Describe three specific programming situations that require a posttest loop.
Answer :
14. Speculate as to the reason control can be transferred into a C loop statement.
Answer :
goto statements can be used to have one cleanup sction in routine rather than multiple return statements, or used to exit out a nested loop that is very long
Rabu, 26 November 2014
CHAPTER 7
Nama : Aditya Isnugraha
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER 7
- REVIEW QUESTION
11. What is an overloaded operator?
Answer :
Operator that has different implementation depending on its arguments.
12. Define narrowing and widening conversions.
Answer :
Widening conversion is a conversion that converts a value to a type that can include least approximations of all values of the original type. Narrowing conversion is a conversion of data that might cause a loss of precision
13. In JavaScript, what is the difference between == and ===?
Answer :
“==” operator is known as type coercion operator and anytime if both values are same and compared using ==operator, type coercion happens. On the other hand === is known as strictly equality operator. It’s much similar Java’s equality operator (==), which gives compilation error if you compare two variables, whose types are not compatible to each other.
14. What is a mixed-mode expression?
Answer :
Mixed-mode expression: The expressions that allow them to design decisions concerning arithmetic expressions is whether an operator can have operands of different types.
15. What is referential transparency?
Answer :
two expressions in the program that have the same value can be substituted for one
another anywhere in the program, without affecting the action of the program
- PROBLEM SET
11. Write a BNF description of the precedence and associativity rules defined for the expressions in Problem 9. Assume the only operands are the names a,b,c,d, and e.
Answer :
<expr> → <expr> or <e1> | <expr> xor <e1> | <e1>
<e1> → <e1> and <e2> | <e2>
<e2> → <e2> = <e3> | <e2> /= <e3> | <e2> < <e3>
| <e2> <= <e3> | <e2> > <e3> | <e2> >= <e3> | <e3>
<e3> → <e4>
<e4> → <e4> + <e5> | <e4> – <e5> | <e4> & <e5> | <e4> mod <e5> | <e5>
<e5> → <e5> * <e6> | <e5> / <e6> | not <e5> | <e6>
<e6> → a | b | c | d | e | const | ( <expr> )
12. Using the grammar of Problem 11, draw parse trees for the expressions of Problem 9.
Answer :
13. Let the function fun be defined as
int fun(int *k) {
*k += 4;
return 3 * (*k) - 1;
}
Suppose fun is used in a program as follows:
void main() {
int i = 10, j = 10, sum1, sum2;
sum1 = (i / 2) + fun(&i);
sum2 = fun(&j) + (j / 2); }
What are the values of sum1 and sum2
a. if the operands in the expressions are evaluated left to right?
b. if the operands in the expressions are evaluated right to left?
Answer :
(a) (left -> right)
sum 1 = 46
sum 2 = 48
(b) (right -> left)
sum 1 = 48
sum 2 = 46
14. What is your primary argument against (or for) the operator precedence rules of APL?
Answer :
The operator precedence rules of the common imperative languages are nearly all the same, because they are based on those of mathematics.
15. Explain why it is difficult to eliminate functional side effects in C.
Answer :
One reason functional side effects would be difficult to remove from C is that all of C’s subprograms are functions, providing the ability of returning only a single data value.
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER 7
- REVIEW QUESTION
11. What is an overloaded operator?
Answer :
Operator that has different implementation depending on its arguments.
12. Define narrowing and widening conversions.
Answer :
Widening conversion is a conversion that converts a value to a type that can include least approximations of all values of the original type. Narrowing conversion is a conversion of data that might cause a loss of precision
13. In JavaScript, what is the difference between == and ===?
Answer :
“==” operator is known as type coercion operator and anytime if both values are same and compared using ==operator, type coercion happens. On the other hand === is known as strictly equality operator. It’s much similar Java’s equality operator (==), which gives compilation error if you compare two variables, whose types are not compatible to each other.
14. What is a mixed-mode expression?
Answer :
Mixed-mode expression: The expressions that allow them to design decisions concerning arithmetic expressions is whether an operator can have operands of different types.
15. What is referential transparency?
Answer :
two expressions in the program that have the same value can be substituted for one
another anywhere in the program, without affecting the action of the program
- PROBLEM SET
11. Write a BNF description of the precedence and associativity rules defined for the expressions in Problem 9. Assume the only operands are the names a,b,c,d, and e.
Answer :
<expr> → <expr> or <e1> | <expr> xor <e1> | <e1>
<e1> → <e1> and <e2> | <e2>
<e2> → <e2> = <e3> | <e2> /= <e3> | <e2> < <e3>
| <e2> <= <e3> | <e2> > <e3> | <e2> >= <e3> | <e3>
<e3> → <e4>
<e4> → <e4> + <e5> | <e4> – <e5> | <e4> & <e5> | <e4> mod <e5> | <e5>
<e5> → <e5> * <e6> | <e5> / <e6> | not <e5> | <e6>
<e6> → a | b | c | d | e | const | ( <expr> )
12. Using the grammar of Problem 11, draw parse trees for the expressions of Problem 9.
Answer :
13. Let the function fun be defined as
int fun(int *k) {
*k += 4;
return 3 * (*k) - 1;
}
Suppose fun is used in a program as follows:
void main() {
int i = 10, j = 10, sum1, sum2;
sum1 = (i / 2) + fun(&i);
sum2 = fun(&j) + (j / 2); }
What are the values of sum1 and sum2
a. if the operands in the expressions are evaluated left to right?
b. if the operands in the expressions are evaluated right to left?
Answer :
(a) (left -> right)
sum 1 = 46
sum 2 = 48
(b) (right -> left)
sum 1 = 48
sum 2 = 46
14. What is your primary argument against (or for) the operator precedence rules of APL?
Answer :
The operator precedence rules of the common imperative languages are nearly all the same, because they are based on those of mathematics.
15. Explain why it is difficult to eliminate functional side effects in C.
Answer :
One reason functional side effects would be difficult to remove from C is that all of C’s subprograms are functions, providing the ability of returning only a single data value.
Rabu, 05 November 2014
CHAPTER 6
Nama : Aditya Isnugraha
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER 6
- REVIEW QUESTION
10. What happens when a nonexistent element of an array is referenced in Perl?
Answer :
If you try to append non-existent elements from an array to another one, the initial array will grow as needed, even though the elements to append do not exist.
11. How does JavaScript support sparse arrays?
Answer :
JavaScript objects are sparse, and arrays are just specialized objects with an auto-maintained length property (which is actually one larger than the largest index, not the number of defined elements) and some additional methods.
12. What languages support negative subscripts?
Answer : Ruby and Lua support negative subscripts.
13. What languages support array slices with Stepsizes?
Answer : Ruby, Python, Perl.
14. What array initialization feature is available in Ada that is not available in other common imperative languages?
Answer :
Ada provides two mechanisms for initializing arrays in the declarations statements: by listing them in the order in which they are to be stored, or by directly assigning them to an index position using the => operator, which in Ada is called an arrow.
15. What is an aggregate constant?
Answer : A parenthesized lists of values.
- PROBLEM SET
11. In the Burroughs Extended ALGOL language, matrices are stored as a single-dimensioned array of pointers to the rows of the matrix, which are treated as single-dimensioned arrays of values. What are the advantages and disadvantages of such a scheme?
Answer :
The advantage of this scheme is that accesses that are done in order of the rows can be made very fast; once the pointer to a row is gotten, all of the elements of the row can be fetched very quickly. If, however, the elements of a matrix must be accessed in column order, these accesses will be much slower; every access requires the fetch of a row pointer and an address computation from there. Note that this access technique was devised to allow multidimensional array rows to be segments in a virtual storage management technique. Using this method, multidimensional arrays could be stored and manipulated that are much larger than the physical memory of the computer.
12. Analyze and write a comparison of C’s malloc and free functions with C++’s new and delete operators. Use safety as the primary consider- ation in the comparison.
Answer : new and delete are type safe (no need for casts), malloc and free are not. Also malloc returns a void* which then has to be cast to the appropriate pointer type. new returns the correct pointer type itself; type safety. malloc requires you to tell the number of bytes to allocate, new figures it out itself.
13. Analyze and write a comparison of using C++ pointers and Java reference variables to refer to fixed heap-dynamic variables. Use safety and conve- nience as the primary considerations in the comparison.
Answer :
-In c and C++ pointer can be use the same way as addresses. This design offer
no solutions to the dangling pointer or lost heap-dynamic variable problems.
-Reference types, such as those in Java and C#, provide heap management
without the dangers of pointers.
so it's clear that Java has more safety for the heap-dynamic variables.
14. Write a short discussion of what was lost and what was gained in Java’s designers’ decision to not include the pointers of C++.
Answer :
There's a copious amount of documentation out there on this subject so there's no reason to write at length. I should also point out that pointers are a feature of both C and C++.
It's obvious that preventing the use of pointers significantly bolstered the amount of security that programmers could get 'for free' given that issues like stack overruns, memory corruption and inadequate/incorrect freeing of memory (though this is more an attribute of GC than lack of pointers) were largely nullified.
Given that programmers tend to be kind of ornery and resistant to change, preventing a means for direct memory access and easy manipulation of the data within rubbed a lot of programmers the wrong way. Many pointed out that the use of pointers was one of the ways that C/C++ were able to be such a high-level language while still maintaining considerable speed. While assembly language and machine code were still de rigeur amongst people that were looking to squeeze the most amount of speed and memory out of a given program, things like pointers and compilers that effectively optimized C and C++ code did a lot to win people over, not to mention that it's a hell of a lot easier to read.
With Java, however, a lot of those same programmers felt that the security that was gained was offset by the decreases in speed and the fact that a lot of the earlier JIT compilers were pretty pokey didn't help that. For all their faults, Sun's been pretty adamant about listening to the community so they've made a lot of improvements regarding in-place optimization, HotSpot, conditional compilation, etc.
Another problem that came about is the fact that removing pointers made existing C and C++ code hard to port. That's generally pretty true of any new or significantly changed language, however. Eventually people got a handle on how to best replicate pointer usage, tips and tricks got passed around and things improved..
15. What are the arguments for and against Java’s implicit heap stor- age recovery, when compared with the explicit heap storage recovery required in C++? Consider real-time systems.
Answer :
Implicit eliminates the creation of dangling pointers. Disadv: cpu-time to do recovery, sometimes when there’s plenty of heap storage so recovery isn’t necessary.
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER 6
- REVIEW QUESTION
10. What happens when a nonexistent element of an array is referenced in Perl?
Answer :
If you try to append non-existent elements from an array to another one, the initial array will grow as needed, even though the elements to append do not exist.
11. How does JavaScript support sparse arrays?
Answer :
JavaScript objects are sparse, and arrays are just specialized objects with an auto-maintained length property (which is actually one larger than the largest index, not the number of defined elements) and some additional methods.
12. What languages support negative subscripts?
Answer : Ruby and Lua support negative subscripts.
13. What languages support array slices with Stepsizes?
Answer : Ruby, Python, Perl.
14. What array initialization feature is available in Ada that is not available in other common imperative languages?
Answer :
Ada provides two mechanisms for initializing arrays in the declarations statements: by listing them in the order in which they are to be stored, or by directly assigning them to an index position using the => operator, which in Ada is called an arrow.
15. What is an aggregate constant?
Answer : A parenthesized lists of values.
- PROBLEM SET
11. In the Burroughs Extended ALGOL language, matrices are stored as a single-dimensioned array of pointers to the rows of the matrix, which are treated as single-dimensioned arrays of values. What are the advantages and disadvantages of such a scheme?
Answer :
The advantage of this scheme is that accesses that are done in order of the rows can be made very fast; once the pointer to a row is gotten, all of the elements of the row can be fetched very quickly. If, however, the elements of a matrix must be accessed in column order, these accesses will be much slower; every access requires the fetch of a row pointer and an address computation from there. Note that this access technique was devised to allow multidimensional array rows to be segments in a virtual storage management technique. Using this method, multidimensional arrays could be stored and manipulated that are much larger than the physical memory of the computer.
12. Analyze and write a comparison of C’s malloc and free functions with C++’s new and delete operators. Use safety as the primary consider- ation in the comparison.
Answer : new and delete are type safe (no need for casts), malloc and free are not. Also malloc returns a void* which then has to be cast to the appropriate pointer type. new returns the correct pointer type itself; type safety. malloc requires you to tell the number of bytes to allocate, new figures it out itself.
13. Analyze and write a comparison of using C++ pointers and Java reference variables to refer to fixed heap-dynamic variables. Use safety and conve- nience as the primary considerations in the comparison.
Answer :
-In c and C++ pointer can be use the same way as addresses. This design offer
no solutions to the dangling pointer or lost heap-dynamic variable problems.
-Reference types, such as those in Java and C#, provide heap management
without the dangers of pointers.
so it's clear that Java has more safety for the heap-dynamic variables.
14. Write a short discussion of what was lost and what was gained in Java’s designers’ decision to not include the pointers of C++.
Answer :
There's a copious amount of documentation out there on this subject so there's no reason to write at length. I should also point out that pointers are a feature of both C and C++.
It's obvious that preventing the use of pointers significantly bolstered the amount of security that programmers could get 'for free' given that issues like stack overruns, memory corruption and inadequate/incorrect freeing of memory (though this is more an attribute of GC than lack of pointers) were largely nullified.
Given that programmers tend to be kind of ornery and resistant to change, preventing a means for direct memory access and easy manipulation of the data within rubbed a lot of programmers the wrong way. Many pointed out that the use of pointers was one of the ways that C/C++ were able to be such a high-level language while still maintaining considerable speed. While assembly language and machine code were still de rigeur amongst people that were looking to squeeze the most amount of speed and memory out of a given program, things like pointers and compilers that effectively optimized C and C++ code did a lot to win people over, not to mention that it's a hell of a lot easier to read.
With Java, however, a lot of those same programmers felt that the security that was gained was offset by the decreases in speed and the fact that a lot of the earlier JIT compilers were pretty pokey didn't help that. For all their faults, Sun's been pretty adamant about listening to the community so they've made a lot of improvements regarding in-place optimization, HotSpot, conditional compilation, etc.
Another problem that came about is the fact that removing pointers made existing C and C++ code hard to port. That's generally pretty true of any new or significantly changed language, however. Eventually people got a handle on how to best replicate pointer usage, tips and tricks got passed around and things improved..
15. What are the arguments for and against Java’s implicit heap stor- age recovery, when compared with the explicit heap storage recovery required in C++? Consider real-time systems.
Answer :
Implicit eliminates the creation of dangling pointers. Disadv: cpu-time to do recovery, sometimes when there’s plenty of heap storage so recovery isn’t necessary.
Rabu, 29 Oktober 2014
CHAPTER 5
Nama : Aditya Isnugraha
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER5
- REVIEW QUESTION
11. What are the advantages and disadvantages of dynamic type binding?
Answer :
Advantages: flexibility and no definite types declared(PHP, JavaScript).
Disadvantages: adds to the cost of implementation and type error detection by the compiler is difficult.
12. Define static, stack-dynamic, explicit heap-dynamic, and implicit heap- dynamic variables. What are their advantages and disadvantages?
Answer :
- Static: bound to memory cells before execution begins and remains bound to the same memory cell throughout the execution.
- Stack-dynamic: storage bindings are created for variables when their declaration statements are elaborated.
- Explicit heap-dynamic: allocated and deallocated by explicit directives, specified by the programmer, which take effect during execution.
- Implicit heap-dynamic variables: Allocation and deallocation caused by assignment statements.
13. Define lifetime, scope, static scope, and dynamic scope.
Answer :
- Lifetime: A time during which the variable is bound to a specific memory location. The lifetime begins when it is bound to a specific cell and ends when it is unbound from that cell.
- Scope: The range of statements in which the variable is visible. A variable is visible in a statement if it can be referenced in that statement.
- Static scope: is based on program text and to connect a name reference to a variable , you (or the compiler) must find the declaration.
- Dynamic scope: Based on calling sequences of program units, not their textual layout (temporal versus spatial). References to variables are connected to declarations by searching back through the chain of subprogram calls that forced execution to this point.
14. How is a reference to a nonlocal variable in a static-scoped program con- nected to its definition?
Answer :
A reference to a non-locally variable in a static-scoped language with nested subprograms requires a two step access process:
1. Find the correct activation record instance
2. Determine the correct offset within that activation record instance
15. What is the general problem with static scoping?
Answer :
Usually too much access. Scope structure destroyed as program evolves.
- PROBLEM SET
1. Which of the following identifier forms is most readable? Support your decision.
SumOfSales
sum_of_sales
SUMOFSALES
Answer :
Sum Of Sales is the most readable. It is because it’s doesn’t have any problem with case sensitive, because the following identifier doesn’t use any caps lock.
2. Some programming languages are typeless. What are the obvious advan- tages and disadvantages of having no types in a language?
Answer :
- Advantages : allow users to write sloppy programs faster.
- Disadvantages : cannot control the data and variables, compiler cannot detect any mistakes.
3. Write a simple assignment statement with one arithmetic operator in some language you know. For each component of the statement, list the various bindings that are required to determine the semantics when the statement is executed. For each binding, indicate the binding time used for the language.
Answer :
(C++)
int count;count = count + 5;
Possible types for count: set at language design time. Type of count: bound at compile time.
Set of possible values of count: bound at compiler design time. Value of count: bound at execution time with this statement. Set of possible meanings for the operator symbol ““:*bound at language definition time.*Meaning of the operator symbol “” in this statement: bound at compile time.
Internal representation of the literal “5”: bound at compiler design time.
4. Dynamic type binding is closely related to implicit heap-dynamic vari- ables. Explain this relationship.
Answer :
Implicit heap-dynamic variables acquire types only when assigned value, which must be at runtime. Therefore, this variable are always dynamically bound to types.
5. Describe a situation when a history-sensitive variable in a subprogram is useful.
Answer :
To describe a situation when a history-sensitive variable in a subprogram is useful, suppose that a FORTRAN subroutine is used to implement a data structure as an abstraction. In this situation, it is essential that the structure persists between different calls to the managing subroutine.
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER5
- REVIEW QUESTION
11. What are the advantages and disadvantages of dynamic type binding?
Answer :
Advantages: flexibility and no definite types declared(PHP, JavaScript).
Disadvantages: adds to the cost of implementation and type error detection by the compiler is difficult.
12. Define static, stack-dynamic, explicit heap-dynamic, and implicit heap- dynamic variables. What are their advantages and disadvantages?
Answer :
- Static: bound to memory cells before execution begins and remains bound to the same memory cell throughout the execution.
- Stack-dynamic: storage bindings are created for variables when their declaration statements are elaborated.
- Explicit heap-dynamic: allocated and deallocated by explicit directives, specified by the programmer, which take effect during execution.
- Implicit heap-dynamic variables: Allocation and deallocation caused by assignment statements.
13. Define lifetime, scope, static scope, and dynamic scope.
Answer :
- Lifetime: A time during which the variable is bound to a specific memory location. The lifetime begins when it is bound to a specific cell and ends when it is unbound from that cell.
- Scope: The range of statements in which the variable is visible. A variable is visible in a statement if it can be referenced in that statement.
- Static scope: is based on program text and to connect a name reference to a variable , you (or the compiler) must find the declaration.
- Dynamic scope: Based on calling sequences of program units, not their textual layout (temporal versus spatial). References to variables are connected to declarations by searching back through the chain of subprogram calls that forced execution to this point.
14. How is a reference to a nonlocal variable in a static-scoped program con- nected to its definition?
Answer :
A reference to a non-locally variable in a static-scoped language with nested subprograms requires a two step access process:
1. Find the correct activation record instance
2. Determine the correct offset within that activation record instance
15. What is the general problem with static scoping?
Answer :
Usually too much access. Scope structure destroyed as program evolves.
- PROBLEM SET
1. Which of the following identifier forms is most readable? Support your decision.
SumOfSales
sum_of_sales
SUMOFSALES
Answer :
Sum Of Sales is the most readable. It is because it’s doesn’t have any problem with case sensitive, because the following identifier doesn’t use any caps lock.
2. Some programming languages are typeless. What are the obvious advan- tages and disadvantages of having no types in a language?
Answer :
- Advantages : allow users to write sloppy programs faster.
- Disadvantages : cannot control the data and variables, compiler cannot detect any mistakes.
3. Write a simple assignment statement with one arithmetic operator in some language you know. For each component of the statement, list the various bindings that are required to determine the semantics when the statement is executed. For each binding, indicate the binding time used for the language.
Answer :
(C++)
int count;count = count + 5;
Possible types for count: set at language design time. Type of count: bound at compile time.
Set of possible values of count: bound at compiler design time. Value of count: bound at execution time with this statement. Set of possible meanings for the operator symbol ““:*bound at language definition time.*Meaning of the operator symbol “” in this statement: bound at compile time.
Internal representation of the literal “5”: bound at compiler design time.
4. Dynamic type binding is closely related to implicit heap-dynamic vari- ables. Explain this relationship.
Answer :
Implicit heap-dynamic variables acquire types only when assigned value, which must be at runtime. Therefore, this variable are always dynamically bound to types.
5. Describe a situation when a history-sensitive variable in a subprogram is useful.
Answer :
To describe a situation when a history-sensitive variable in a subprogram is useful, suppose that a FORTRAN subroutine is used to implement a data structure as an abstraction. In this situation, it is essential that the structure persists between different calls to the managing subroutine.
Kamis, 16 Oktober 2014
CHAPTER 4
Nama : Aditya Isnugraha
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER 4
- REVIEW QUESTION
11. Describe the parsing problem for a bottom-up parser.
Answer :
12. Explain why compilers use parsing algorithms that work on only a subset of all grammars.
Answer :
13. Why are named constants used, rather than numbers, for token codes?
Answer :
- for the sake of readability of lexical and syntax analyzers.
14. Describe how a recursive-descent parsing subprogram is written for a rule with a single RHS.
Answer :
-A recursive-descent subprogram for a rule with a single RHS is relatively
simple. For each terminal symbol in the RHS, that terminal symbol is compared
with nextToken. If they do not match, it is a syntax error. If they match,
the lexical analyzer is called to get the next input token. For each non terminal,
the parsing subprogram for that nonterminal is called.
15. Explain the two grammar characteristics that prohibit them from being used as the basis for a top-down parser.
Answer : Two grammar characteristics that prohibit top-down parsing:
Direct or indirect Left Recursion.
- PROBLEM SET
1. Perform the pairwise disjointness test for the following grammar rules.
a. A→aB b cBB
b. B→aB bA aBb
c. C→aaA b caB
Answer :
(a) FIRST(aB) = {a}, FIRST(b) = {b}, FIRST(cBB) = {c}, Passes the test
(b) FIRST(aB) = {a}, FIRST(bA) = {b}, FIRST(aBb) = {a}, Fails the test
(c) FIRST(aaA) = {a}, FIRST(b) = {b}, FIRST(caB) = {c}, Passes the test
2. Perform the pairwise disjointness test for the following grammar rules.
a. S→aSb bAA
b. A→b{aB} a
c. B→aB a
Answer :
a. FIRST(aSb)=a
FIRST(bAA)=b
b. FIRST(b{aB}) = b
FIRST (a) = a
c. FIRST(aB)=a
FIRST(a) = a
3. Show a trace of the recursive descent parser given in Section 4.4.1 for the string a + b * c.
Answer :
a + b * c
Call lex /* returns a */
Enter <expr>
Enter <term>
Enter <factor>
Call lex /* returns + */
Exit <factor>
Exit <term>
Call lex /* returns b */
Enter <term>
Enter <factor>
Call lex /* returns * */
Exit <factor>
Call lex /* returns c */
Enter <factor>
Call lex /* returns end-of-input */
Exit <factor>
Exit <term>
Exit <expr>
4. Show a trace of the recursive descent parser given in Section 4.4.1 for the string a * (b + c).
Answer :
call lex // return a
enter <expr>
enter <term>
enter <factor>
call lex // return *
exit <factor>
call lex // return (
enter <factor>
call lex // return b
enter <expr>
enter <term>
enter <factor>
call lex // return +
exit <factor>
exit <term>
call lex // return c
enter <term>
enter <factor>
call lex // return )
exit <factor>
exit <term>
exit <expr>
call lex // return EOF
exit <factor>
exit <term>
exit <expr>
5. Given the following grammar and the right sentential form, draw a parse tree and show the phrases and simple phrases, as well as the handle. S→aAb bBA A→ab aAB B→aB b
a. aaAbb
b. bBab
c. aaAbBb
Answer :
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER 4
- REVIEW QUESTION
11. Describe the parsing problem for a bottom-up parser.
Answer :
12. Explain why compilers use parsing algorithms that work on only a subset of all grammars.
Answer :
13. Why are named constants used, rather than numbers, for token codes?
Answer :
- for the sake of readability of lexical and syntax analyzers.
14. Describe how a recursive-descent parsing subprogram is written for a rule with a single RHS.
Answer :
-A recursive-descent subprogram for a rule with a single RHS is relatively
simple. For each terminal symbol in the RHS, that terminal symbol is compared
with nextToken. If they do not match, it is a syntax error. If they match,
the lexical analyzer is called to get the next input token. For each non terminal,
the parsing subprogram for that nonterminal is called.
15. Explain the two grammar characteristics that prohibit them from being used as the basis for a top-down parser.
Answer : Two grammar characteristics that prohibit top-down parsing:
Direct or indirect Left Recursion.
- PROBLEM SET
1. Perform the pairwise disjointness test for the following grammar rules.
a. A→aB b cBB
b. B→aB bA aBb
c. C→aaA b caB
Answer :
(a) FIRST(aB) = {a}, FIRST(b) = {b}, FIRST(cBB) = {c}, Passes the test
(b) FIRST(aB) = {a}, FIRST(bA) = {b}, FIRST(aBb) = {a}, Fails the test
(c) FIRST(aaA) = {a}, FIRST(b) = {b}, FIRST(caB) = {c}, Passes the test
2. Perform the pairwise disjointness test for the following grammar rules.
a. S→aSb bAA
b. A→b{aB} a
c. B→aB a
Answer :
a. FIRST(aSb)=a
FIRST(bAA)=b
b. FIRST(b{aB}) = b
FIRST (a) = a
c. FIRST(aB)=a
FIRST(a) = a
3. Show a trace of the recursive descent parser given in Section 4.4.1 for the string a + b * c.
Answer :
a + b * c
Call lex /* returns a */
Enter <expr>
Enter <term>
Enter <factor>
Call lex /* returns + */
Exit <factor>
Exit <term>
Call lex /* returns b */
Enter <term>
Enter <factor>
Call lex /* returns * */
Exit <factor>
Call lex /* returns c */
Enter <factor>
Call lex /* returns end-of-input */
Exit <factor>
Exit <term>
Exit <expr>
4. Show a trace of the recursive descent parser given in Section 4.4.1 for the string a * (b + c).
Answer :
call lex // return a
enter <expr>
enter <term>
enter <factor>
call lex // return *
exit <factor>
call lex // return (
enter <factor>
call lex // return b
enter <expr>
enter <term>
enter <factor>
call lex // return +
exit <factor>
exit <term>
call lex // return c
enter <term>
enter <factor>
call lex // return )
exit <factor>
exit <term>
exit <expr>
call lex // return EOF
exit <factor>
exit <term>
exit <expr>
5. Given the following grammar and the right sentential form, draw a parse tree and show the phrases and simple phrases, as well as the handle. S→aAb bBA A→ab aAB B→aB b
a. aaAbb
b. bBab
c. aaAbBb
Answer :
Kamis, 09 Oktober 2014
CHAPTER 3
Nama : Aditya Isnugraha
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER 3
- REVIEW QUESTION
11. How is the order of evaluation of attributes determined for the trees of a
given attribute grammar?
Answer : Expr → Expr + Term
Expr → Term
Term → Term * Factor
Term → Factor
Factor → "(" Expr ")"
Factor → integer
12. What is the primary use of attribute grammars?
Answer : An attribute grammar is an extension to a context-free grammar. The primary purpose of an attribute grammar is it allows certain language rules to be described, such as type of compatibility. An attribute grammar is a formal way to define attributes for the productions of a formal grammar, associating these attributes to values. The evaluation occurs in the nodes of the abstract syntax tree, when the language is processed by some parser or compiler.
13. Explain the primary uses of a methodology and notation for describing
the semantics of programming languages.
Answer : Object modeling techniques and notations presents a methodology using a completely different diagrams with other techniques commonly used for data modeling and process modeling.
14. Why can machine languages not be used to define statements in operational
semantics?
Answer : Machine language can not be used to define statements in operational semantics because of some problems. First, the individual steps in the execution of machine
language and the resulting changes to the state of the machine are too small and too numerous. Second, the storage of a real computer is too large and complex.
15. Describe the two levels of uses of operational semantics.
Answer : There are different levels of uses in operational semantics. At the highest level, the focus is on the final result of the execution of a program, this is sometime called natural operational semantics. At the lowest level, operational semantics can be used to determine the precise meaning of a program through an examination of the complete sequence.
- PROBLEM SET
11. Consider the following grammar:
<S> → <A> a <B> b
<A> → <A> b | b
<B> → a <B> | a
Which of the following sentences are in the language generated by this
grammar?
a. baab
b. bbbab
c. bbaaaaa
d. bbaab
Answer : None of those are the language generated by the grammar, it should be k<A>aj<B>b,(A contains b and B contains a) in which the end will be a single b, the list of answer does not include an answer that ends with a single b
12. Consider the following grammar:
<S> → a <S> c <B> | <A> | b
<A> → c <A> | c
<B> → d | <A>
Which of the following sentences are in the language generated by this
grammar?
a. abcd
b. acccbd
c. acccbcc
d. acd
e. accc
Answer : No one of the following statements are generated.
13. Write a grammar for the language consisting of strings that have n
copies of the letter a followed by the same number of copies of the
letter b, where n > 0. For example, the strings ab, aaaabbbb, and
aaaaaaaabbbbbbbb are in the language but a, abb, ba, and aaabb are not.
Answer : <S> => a<S>bb | abb
14. Draw parse trees for the sentences aabb and aaaabbbb, as derived from
the grammar of Problem 13.
Answer : We first derive a grammar in order to know how to draw the tree.
<stmt> -> <A>
<A> -> a<A>b | ab
tree for aabb
tree for aaaabbbb
15. Convert the BNF of Example 3.1 to EBNF.
Answer : <program> -> begin <stmt_list> end
. <stmt_list> -> stmt[stmt_list]
. <stmt> -> <var> = <expressions>
. <var> -> A| B | C
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER 3
- REVIEW QUESTION
11. How is the order of evaluation of attributes determined for the trees of a
given attribute grammar?
Answer : Expr → Expr + Term
Expr → Term
Term → Term * Factor
Term → Factor
Factor → "(" Expr ")"
Factor → integer
12. What is the primary use of attribute grammars?
Answer : An attribute grammar is an extension to a context-free grammar. The primary purpose of an attribute grammar is it allows certain language rules to be described, such as type of compatibility. An attribute grammar is a formal way to define attributes for the productions of a formal grammar, associating these attributes to values. The evaluation occurs in the nodes of the abstract syntax tree, when the language is processed by some parser or compiler.
13. Explain the primary uses of a methodology and notation for describing
the semantics of programming languages.
Answer : Object modeling techniques and notations presents a methodology using a completely different diagrams with other techniques commonly used for data modeling and process modeling.
14. Why can machine languages not be used to define statements in operational
semantics?
Answer : Machine language can not be used to define statements in operational semantics because of some problems. First, the individual steps in the execution of machine
language and the resulting changes to the state of the machine are too small and too numerous. Second, the storage of a real computer is too large and complex.
15. Describe the two levels of uses of operational semantics.
Answer : There are different levels of uses in operational semantics. At the highest level, the focus is on the final result of the execution of a program, this is sometime called natural operational semantics. At the lowest level, operational semantics can be used to determine the precise meaning of a program through an examination of the complete sequence.
- PROBLEM SET
11. Consider the following grammar:
<S> → <A> a <B> b
<A> → <A> b | b
<B> → a <B> | a
Which of the following sentences are in the language generated by this
grammar?
a. baab
b. bbbab
c. bbaaaaa
d. bbaab
Answer : None of those are the language generated by the grammar, it should be k<A>aj<B>b,(A contains b and B contains a) in which the end will be a single b, the list of answer does not include an answer that ends with a single b
12. Consider the following grammar:
<S> → a <S> c <B> | <A> | b
<A> → c <A> | c
<B> → d | <A>
Which of the following sentences are in the language generated by this
grammar?
a. abcd
b. acccbd
c. acccbcc
d. acd
e. accc
Answer : No one of the following statements are generated.
13. Write a grammar for the language consisting of strings that have n
copies of the letter a followed by the same number of copies of the
letter b, where n > 0. For example, the strings ab, aaaabbbb, and
aaaaaaaabbbbbbbb are in the language but a, abb, ba, and aaabb are not.
Answer : <S> => a<S>bb | abb
14. Draw parse trees for the sentences aabb and aaaabbbb, as derived from
the grammar of Problem 13.
Answer : We first derive a grammar in order to know how to draw the tree.
<stmt> -> <A>
<A> -> a<A>b | ab
tree for aabb
tree for aaaabbbb
15. Convert the BNF of Example 3.1 to EBNF.
Answer : <program> -> begin <stmt_list> end
. <stmt_list> -> stmt[stmt_list]
. <stmt> -> <var> = <expressions>
. <var> -> A| B | C
Rabu, 08 Oktober 2014
CHAPTER 2
Nama : Aditya Isnugraha
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER 2
- REVIEW QUESTION
11. What control flow statements were added to Fortran IV to get Fortran
77?
Answer : Character string handling, logical loop control statement and an if without an optional else clause were added to Fortran IV to get Fortran 77.
12. Which version of Fortran was the first to have any sort of dynamic
variables?
Answer : Fortran 90 is the first version of Fortran which support dynamic variables.
13. Which version of Fortran was the first to have character string handling?
Answer : The first version of Fortran supporting character string handling is Fortran 77.
14. Why are linguists interested in artificial intelligence in the late 1950s?
Answer : Because they were concerned with natural language processing, when in 1950’s, there are only computers with computation based on numeric data in arrays.
15. Where was LISP developed? By whom?
Answer : It was developed at MIT by John McCarthy.
- PROBLEM SET
11. Was IBM’s assumption, on which it based its decision to develop PL/I, correct, given the history of computers and language developments since 1964?
Answer : The assumption is correct because in that 1970’s, PL/I is widely used for both business and scientific applications although it suffers a lot in previous years and afterwards.
12. Describe, in your own words, the concept of orthogonality in programming language design.
Answer : It appears that orthogonality means the simplicity of programming constructs, or a minimal number of control and data structures in a language. Each additional construct increases the complexity, removing orthogonality.
13. What is the primary reason why PL/I became more widely used than ALGOL 68?
Answer :
- PL/I included the best of ALGOL 60 (recursion and block structure), FORTRAN IV (separate compilation with communication through global data), and COBOL (data structures, input/output, and report generating facilities), along with a few new constructs
- PL/I was the first language to have programs allowed to create concurrently executing tasks, the possibility to detect and handle 23 different types of exceptions, procedures allowed to be use recursively, pointers included as a data type, and reference to the cross sections of arrays
14. What are the arguments both for and against the idea of a typeless language?
Answer : Arguments for are obvious flexibility and ease of use. Without having to define a data type the programmer is free to develop code that is generated quickly and without much thought. Learning the language is much simpler because one doesn’t have to determine size or how the compiler will interpret the type later on, only what information must be included.
Arguments against include data insecurity, such as the assignment of a character type ‘A’ that could in fact be “defined” as a HEX value by the programmer. The compiler would also have trouble interpreting floating point values compared to integers. The resulting arithmetic would also cause serious problems; like adding 5 + “happy” and how they are interpreted different than perhaps the programmer intended.
15. Are there any logic programming languages other than Prolog?
Answer : Yes there are some non procedural languages other than Prolog, for example, Visual Basic, SQL.
-FORTRAN
-LISP
-ALGOL 60
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER 2
- REVIEW QUESTION
11. What control flow statements were added to Fortran IV to get Fortran
77?
Answer : Character string handling, logical loop control statement and an if without an optional else clause were added to Fortran IV to get Fortran 77.
12. Which version of Fortran was the first to have any sort of dynamic
variables?
Answer : Fortran 90 is the first version of Fortran which support dynamic variables.
13. Which version of Fortran was the first to have character string handling?
Answer : The first version of Fortran supporting character string handling is Fortran 77.
14. Why are linguists interested in artificial intelligence in the late 1950s?
Answer : Because they were concerned with natural language processing, when in 1950’s, there are only computers with computation based on numeric data in arrays.
15. Where was LISP developed? By whom?
Answer : It was developed at MIT by John McCarthy.
- PROBLEM SET
11. Was IBM’s assumption, on which it based its decision to develop PL/I, correct, given the history of computers and language developments since 1964?
Answer : The assumption is correct because in that 1970’s, PL/I is widely used for both business and scientific applications although it suffers a lot in previous years and afterwards.
12. Describe, in your own words, the concept of orthogonality in programming language design.
Answer : It appears that orthogonality means the simplicity of programming constructs, or a minimal number of control and data structures in a language. Each additional construct increases the complexity, removing orthogonality.
13. What is the primary reason why PL/I became more widely used than ALGOL 68?
Answer :
- PL/I included the best of ALGOL 60 (recursion and block structure), FORTRAN IV (separate compilation with communication through global data), and COBOL (data structures, input/output, and report generating facilities), along with a few new constructs
- PL/I was the first language to have programs allowed to create concurrently executing tasks, the possibility to detect and handle 23 different types of exceptions, procedures allowed to be use recursively, pointers included as a data type, and reference to the cross sections of arrays
14. What are the arguments both for and against the idea of a typeless language?
Answer : Arguments for are obvious flexibility and ease of use. Without having to define a data type the programmer is free to develop code that is generated quickly and without much thought. Learning the language is much simpler because one doesn’t have to determine size or how the compiler will interpret the type later on, only what information must be included.
Arguments against include data insecurity, such as the assignment of a character type ‘A’ that could in fact be “defined” as a HEX value by the programmer. The compiler would also have trouble interpreting floating point values compared to integers. The resulting arithmetic would also cause serious problems; like adding 5 + “happy” and how they are interpreted different than perhaps the programmer intended.
15. Are there any logic programming languages other than Prolog?
Answer : Yes there are some non procedural languages other than Prolog, for example, Visual Basic, SQL.
-FORTRAN
-LISP
-ALGOL 60
Sabtu, 27 September 2014
CHAPTER 1
Nama : Aditya Isnugraha
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER 1
- REVIEW QUESTIONS
11. What primitive control statement is used to build more complicated
control statements in languages that lack them?
Answer :
The selection statement plus GOTO is used to build more complicated control statements such as FOR loop.
12. What construct of a programming language provides process
abstraction?
Answer :
Subprograms
13. What does it mean for a program to be reliable?
Answer :
A program is said to be reliable if it performs to its specifications under all conditions.
14. Why is type checking the parameters of a subprogram important?
Answer : It can lead to lots of hard to debug errors.
15. What is aliasing?
Answer : 2 or more distinct names that can be used to access the same memory cell.
Assignment 2
- PROBLEM SET
11. Describe some design trade-offs between efficiency and safety in some
language you know.
Answer : C is what assembly language was 10 or 15 years ago. It's the language you use when efficiency or low level access to hardware really matters to you. You can, of course, still use assembly language today. But optimizing C/C++ compilers are so good today, that they can often trump hand coded assembly language.
12. In your opinion, what major features would a perfect programming language
include?
Answer : High-level of abstraction with low-level efficiency. Such languages already exist, but none are "perfect" in every situation. That's one of the reasons we have so many languages at our disposal. It would be impractical to combine them into a single language, you simply choose the best tool for the job at hand.
13. Was the first high-level programming language you learned implemented
with a pure interpreter, a hybrid implementation system, or a
compiler? (You may have to research this.)
Answer : My first high-level programming language I learned is C++ which is implemented with visual C++ and it is a compiler.
14. Describe the advantages and disadvantages of some programming environment
you have used.
Answer :The advantages of VB are the ease of learning - the syntax is simpler than other languages (although it can be argued that C has more flexibility). The visual environment is excellent (although that's common to all the visual languages). It's widely used and, therefore, well understood.
Disadvantages compared with C: C has better declaration of arrays - its possible to initialise an array of structures in C at declaration time; this is impossible in VB.
15. How do type declaration statements for simple variables affect the readability
of a language, considering that some languages do not require
them?
Answer : The use of type declaration statements for simple scalar variables may have very little effect on the readability of programs. If a language has no type declarations at all, it may be an aid to readability, because regardless of where a variable is seen in the program text, its type can be determined without looking elsewhere. Unfortunately, most languages that allow implicitly declared variables also include explicit declarations. In a program in such a language, the declaration of a variable must be found before the reader can determine the type of that variable when it is used in the program.
Class : LM01
NIM : 1801419606
Assignment from Tri Djoko Wahjono
CHAPTER 1
- REVIEW QUESTIONS
11. What primitive control statement is used to build more complicated
control statements in languages that lack them?
Answer :
The selection statement plus GOTO is used to build more complicated control statements such as FOR loop.
12. What construct of a programming language provides process
abstraction?
Answer :
Subprograms
13. What does it mean for a program to be reliable?
Answer :
A program is said to be reliable if it performs to its specifications under all conditions.
14. Why is type checking the parameters of a subprogram important?
Answer : It can lead to lots of hard to debug errors.
15. What is aliasing?
Answer : 2 or more distinct names that can be used to access the same memory cell.
Assignment 2
- PROBLEM SET
11. Describe some design trade-offs between efficiency and safety in some
language you know.
Answer : C is what assembly language was 10 or 15 years ago. It's the language you use when efficiency or low level access to hardware really matters to you. You can, of course, still use assembly language today. But optimizing C/C++ compilers are so good today, that they can often trump hand coded assembly language.
12. In your opinion, what major features would a perfect programming language
include?
Answer : High-level of abstraction with low-level efficiency. Such languages already exist, but none are "perfect" in every situation. That's one of the reasons we have so many languages at our disposal. It would be impractical to combine them into a single language, you simply choose the best tool for the job at hand.
13. Was the first high-level programming language you learned implemented
with a pure interpreter, a hybrid implementation system, or a
compiler? (You may have to research this.)
Answer : My first high-level programming language I learned is C++ which is implemented with visual C++ and it is a compiler.
14. Describe the advantages and disadvantages of some programming environment
you have used.
Answer :The advantages of VB are the ease of learning - the syntax is simpler than other languages (although it can be argued that C has more flexibility). The visual environment is excellent (although that's common to all the visual languages). It's widely used and, therefore, well understood.
Disadvantages compared with C: C has better declaration of arrays - its possible to initialise an array of structures in C at declaration time; this is impossible in VB.
15. How do type declaration statements for simple variables affect the readability
of a language, considering that some languages do not require
them?
Answer : The use of type declaration statements for simple scalar variables may have very little effect on the readability of programs. If a language has no type declarations at all, it may be an aid to readability, because regardless of where a variable is seen in the program text, its type can be determined without looking elsewhere. Unfortunately, most languages that allow implicitly declared variables also include explicit declarations. In a program in such a language, the declaration of a variable must be found before the reader can determine the type of that variable when it is used in the program.
Senin, 26 November 2012
Sejak blog ini dibuat, blum kepikiran
posting hal2 yg berhubungan dengan liverpool, pdhl kan kita nge fans
amat, sampe2 saat liverpool juara champion 04/05 air mata secara gk
sangaja kluar,,,,,hehehe,,,alay. Pendukung The Red, Liverpool , pasti
tau hymne/ lagu kebangsaan-nya pendukung liverpool (the Kop) yang
judulnya you’ll never walk alone. Lagu ini pasti dinyanyikan oleh
pendukung setia the kop, di anfield atau di mana saja mereka berada
mendukung liverpool bertanding. lagu ini Diciptakan dua orang musisi
Amerika Serikat untuk pementasan opera di Broadway, Carousel, pada 1945.
Mereka adalah pemusik Richard Rodgers dan pencipta lirik Oscar
Hammerstein II. Pada 5 Oktober 1965, grup band asal Liverpool, Gerry
& The Pacemakers, melansir albumnya yang salah satu lagunya berisi
You’ll Never Walk Alone.
Sebagian sejarah menyebut, gara-gara Gerry & The Pacemakers, suporter Liverpool lantas kerap menyanyikannya. Tapi sebagian kesaksian menyebut The Kop sudah demen melagukannya beberapa pekan sebelum launching album band itu.
Sampai sekarang suporter Liverpool masih bertengkar dengan pendukung Glasgow Celtic, Skotlandia, soal siapa yang lebih berhak “memiliki” anthem tersebut. Pasalnya, para pendukung Celtic juga mengklaim sebagai suporter pertama yang menjadikannya sebagai lagu kebangsaan.
Satu yang tak bisa dimungkiri, pada akhir 1950-an dan awal 1960-an, Liverpool adalah garda depan urusan musik. Dipimpin The Beatles, yang semua anggotanya tak ada yang suka sepak bola, Kota Liverpool dikenal banyak menghasilkan band dan musisi dunia. Inilah gelombang Merseybeat–lagu dari Merseyside. Aksen orang-orang Liverpool, scouser, berbeda dengan aksen Inggris kebanyakan.
Penjelasan itu mungkin bisa menerangkan bila You’ll Never Walk Alone ala Gerry & The Pacemakers turut mendunia bersamaan dengan gelombang Merseybeat. Ipswich Town juga mengklaim lagu itu sebagai miliknya.
Banyak klub luar Inggris juga menyanyikannya: CSKA Sofia (Bulgaria); Rapid Vienna (Austria); Dinamo Zagreb (Kroasia); Ajax dan Twente (Belanda); Dortmund, Schalke, Bremen, St Pauli, Aachen, Mainz 05, dan Kaiserslautern (Jerman); AEK Athens (Yunani); FC Tokyo (Jepang); serta Brugge, Antwerp, dan Mechelen (Belgia).
Entah siapa yang sebenarnya lebih berhak
ini dia liriknya …
Keep your head up high
And don’t be afraid of the dark.
At he end of the storm
Is a golden sky
And the sweet silver song of a lark.
Walk on through the wind,
Walk on through the rain,
Tho’ your dreams be tossed and blown.Walk on, walk on
With hope in your heart
And you’ll never walk alone,
You’ll never walk alone.
Salam buat liverpudlian..
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