Chapter 6 : Problem Set

Q6 : Explain all of the differences between Ada’s subtypes and derived types.

A   : 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.

Q7 : What significant justification is there for the -> operator in C and C++?

A   : The only justification for the -> operator in C and C++ is writability. It is slightly easier to write p -> q than (*p).q.

Q8 : What are all of the differences between the enumeration types of C++ and those of Java?

A   : 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.

Q9 : 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?

A   :

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.

Q10 : 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.

A      :

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))