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Kali ini saya akan menjawab assigntment #6 dari chapter 6 Programming Language Concepts R Sebesta E-book :
Review Questions #6-10 :
6. What are the advantages of user-defined enumeration types?
Enumeration types provide a way of defining and grouping collections of named constants,
which are called enumeration constants.
7. In what ways are the user-defined enumeration types of C# more reliable than those of C++?
enum days {Mon, Tue, Wed, Thu, Fri, Sat, Sun};
8. What are the design issues for arrays?
• What types are legal for subscripts?
• Are subscripting expressions in element references range checked?
• When are subscript ranges bound?
• When does array allocation take place?
• Are ragged or rectangular multidimensioned arrays allowed, or both?
• Can arrays be initialized when they have their storage allocated?
• What kinds of slices are allowed, if any?
9. Define static, fixed stack-dynamic, stack-dynamic, fixed heap-dynamic, and heap-dynamic arrays. What are the advantages of each?
A static array is one in which the subscript ranges are statically bound and storage allocation is static (done before run time). The advantage of static arrays is efficiency: No dynamic allocation or deallocation is required. The disadvantage is that the storage for the array is fixed for the entire execution time of the program.
A fixed stack-dynamic array is one in which the subscript ranges are staticallybound, but the allocation is done at declaration elaboration time during execution. The advantage of fixed stack-dynamic arrays over static arrays is space efficiency.
A stack-dynamic array is one in which both the subscript ranges and the storage allocation are dynamically bound at elaboration time. Once the subscript ranges are bound and the storage is allocated, however, they remain fixed during the lifetime of the variable. The advantage of stack-dynamic arrays over static and fixed stack-dynamic arrays is flexibility. The size of an array need not
be known until the array is about to be used.
A fixed heap-dynamic array is similar to a fixed stack-dynamic array, in that the subscript ranges and the storage binding are both fixed after storage is allocated. The differences are that both the subscript ranges and storage bindings are done when the user program requests them during execution, and the storage is allocated from the heap, rather than the stack. The advantage of fixed heap-dynamic arrays is flexibility—the array’s size always fits the problem.
A heap-dynamic array is one in which the binding of subscript ranges and storage allocation is dynamic and can change any number of times during the array’s lifetime. The advantage of heap-dynamic arrays over the others is flexibility: Arrays can grow and shrink during program execution as the need for space changes.
10. What happens when a nonexistent element of an array is referenced in Perl?
A reference to a nonexistent element in Perl yields undef, but no error is reported.
Problem Set#6-10 :
6. Explain all of the differences between Ada’s subtypes and derived types.
A subtype is compatible with its base type, so you can mix operands of the base type with operands of the base type. For example:
subtype Week_Days is Integer range 1..7;
Since this is a subtype, you can (for example) add 1 to a weekday to get the next weekday.
A derived type is a completely separate type that has the same characteristics as its base type. You cannot mix operands of a derived type with operands of the base type. If, for example, you used:
type Week_Day is new Integer range 1..7;
Then you would not be able to add an integer to a weekday to get another weekday. To do manipulations on a derived type, you'd normally define those manipulations yourself (e.g., create a package). At the same time, a derived type does "inherit" all the operations of its base type (even some that may not make sense) so you do still get addition.
7. What significant justification is there for the -> operator in C and C++?
The only justification for the -> operator in C and C++ is writability. It is slightly easier to write p -> q than (*p).q.
8. What are all of the differences between the enumeration types of C++ and those of Java?
In C/C++, enums are more or less integers (that's the way they are handled under the hood). You can't give your enum methods or anything like that, basically you only capable of making variables of the type and giving them the values you list.
In Java, your enums can have methods attached.
9. The unions in C and C++ are separate from the records of those languages, rather than combined as they are in Ada. What are the advantages and disadvantages to these two choices?
Ada advantage:
Unconstrained variant records in Ada allow the values of their variants to change types during execution.
disadvantage:
7. What significant justification is there for the -> operator in C and C++?
The only justification for the -> operator in C and C++ is writability. It is slightly easier to write p -> q than (*p).q.
8. What are all of the differences between the enumeration types of C++ and those of Java?
In C/C++, enums are more or less integers (that's the way they are handled under the hood). You can't give your enum methods or anything like that, basically you only capable of making variables of the type and giving them the values you list.
In Java, your enums can have methods attached.
9. The unions in C and C++ are separate from the records of those languages, rather than combined as they are in Ada. What are the advantages and disadvantages to these two choices?
Ada advantage:
Unconstrained variant records in Ada allow the values of their variants to change types during execution.
disadvantage:
The type of the variant can be changed only by assigning the entire record, including the discriminant. This disallows inconsistent records because if the newly assigned record is a constant data aggregate, the value of the tag and the type of the variant can be statically checked for consistency.
10. Multidimensional arrays can be stored in row major order, as in C++, or in column major order, as in Fortran. Develop the access functions for both of these arrangements for three-dimensional arrays.
Let the subscript ranges of the three dimensions be named min(1), min(2), min(3), max(1), max(2), and max(3). Let the sizes of the subscript ranges be size(1), size(2), and size(3). Assume the element size is 1.
Row Major Order:
location(a[i,j,k]) = (address of a[min(1),min(2),min(3)]) + ((i-min(1))*size(3) + (j-min(2)))*size(2) + (k-min(3))
Column Major Order:
location(a[i,j,k]) = (address of a[min(1),min(2),min(3)]) + ((k-min(3))*size(1) + (j-min(2)))*size(2) + (i-min(1))
10. Multidimensional arrays can be stored in row major order, as in C++, or in column major order, as in Fortran. Develop the access functions for both of these arrangements for three-dimensional arrays.
Let the subscript ranges of the three dimensions be named min(1), min(2), min(3), max(1), max(2), and max(3). Let the sizes of the subscript ranges be size(1), size(2), and size(3). Assume the element size is 1.
Row Major Order:
location(a[i,j,k]) = (address of a[min(1),min(2),min(3)]) + ((i-min(1))*size(3) + (j-min(2)))*size(2) + (k-min(3))
Column Major Order:
location(a[i,j,k]) = (address of a[min(1),min(2),min(3)]) + ((k-min(3))*size(1) + (j-min(2)))*size(2) + (i-min(1))
