Chapter 6 Programming Language Concepts R Sebesta
Nama: Billy
NIM: 1801374785
NIM: 1801374785
Review Questions
6. What are the advantages of user-defined enumeration types?
The advantages are readability and reliability.
7. In what ways are the user-defined enumeration types of C# more reliable than those of C++?
C# enumeration types are like those of C++, except that they are never coerced to integer. So, operations on enumeration types are restricted to those that make sense. Also, the range of values is restricted to that of the particular enumeration type.
8. What are the design issues for arrays?
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 allocation take place?
What is the maximum number of subscripts?
Can array objects be initialized?
Are any kind of slices supported?
9. Define static, fixed stack-dynamic, stack-dynamic, fixed heap-dynamic, and heap-dynamic arrays. What are the advantages of each?
Static: subscript ranges are statically bound and storage allocation is static (before run-time)
Advantage: efficiency (no dynamic allocation).
Fixed stack-dynamic: subscript ranges are statically bound, but the allocation is done at declaration time.
Advantage: space efficiency .
Stack-dynamic: subscript ranges are dynamically bound and the storage allocation is dynamic (done at run-time)
Advantage: flexibility (the size of an array need not be known until the array is to be used).
Fixed heap-dynamic: similar to fixed stack-dynamic: storage binding is dynamic but fixed after allocation (i.e., binding is done when requested and storage is allocated from heap, not stack).
Heap-dynamic: binding of subscript ranges and storage allocation is dynamic and can change any number of times.
Advantage: flexibility (arrays can grow or shrink during program execution).
10. What happens when a nonexistent element of an array is referenced in Perl?
If an r-value is required, undef is returned. If an l-value is required, the array is extended, then the newly created but undefined element is returned. No error is reported.
Problem Set
6. Explain all of the differences between Ada’s subtypes and derived types.
Ada’s subtype is compatible with its base type, so you can mix operands of the base type with operands of the base type. While Ada’s derived type is a completely separate type that has the same characteristics as its base type. We can’t mix operands of a derived type with operands of the base type.
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++, an enumeration is just a set of named, integral constants. Also, C++ will implicitly convert enum values to their integral equivalent. In Java, an enumeration is more like a named instance of a class. You have the ability to customize the members available on the enumeration. Java will explicitly convert enum values to their integral equivalent.
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:
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))