ICS 313-03
Fundamentals of Programming Languages
Exam II
(20%)
(75 Minutes)
Student ID:
Name: KEY
Question
Max
1
14
2
26
3
20
4
20
5
20
Total
100
Mark
21 December 2002
(021)
Question 1
[14]
Assume the following program was compiled and executed using static scoping rules :
program main;
var x : integer;
procedure sub1;
begin { sub1 }
writeln ( ‘x = ‘ , x}
end; {sub1 }
procedure sub2;
var x : integer;
begin { sub2 }
x := 5;
sub1
end; { sub2 }
begin
{ main }
x := 10;
sub2
end. { main }
1.1 What value of x is printed in procedure sub1?
x = 10
1.2 Under dynamic scoping rules, what value of x is printed in procedure sub1?
x=5
Question 2
[16, 10]
2.1 Define static, fixed stack-dynamic, stack-dynamic, and dynamic arrays. What are the
advantages of each?
Static: range of subscripts and storage bindings are static
Advantage: execution efficiency (no allocation or deallocation)
Fixed stack dynamic: range of subscripts is statically bound, but storage is bound at
elaboration time
Advantage: space efficiency
Stack-dynamic: range and storage are dynamic, but fixed from then on for the variable’s
lifetime
Advantage: flexibility (size need not be known until the array is about to be used)
Dynamic: subscript range and storage bindings are dynamic and not fixed
Advantage: flexibility (arrays can grow and shrink during program execution as needed)
2.2 Suppose that a language includes user-defined enumeration types and that the enumeration
values could be overloaded; that is, the same literal value could appear in two different
enumeration types, as in:
type
colors = (red, blue, green, white);
mood
= (happy, angry, blue);
Use of the constant blue cannot be type checked.
Propose a method of allowing such type checking without completely disallowing such
overloading. Give an example.
Require all references to enumeration constants to be qualified with the type name,
as in colors.blue
Question 3
[10, 10]
3.1 Describe a situation in which the add operator in a programming language would not be
commutative.
An expression such as a + fun(b), and fun(b) changes a.
3.2 Consider the following C program:
int
fun(int *i) {
*i += 5;
return 4;
}
void main () {
int x = 3;
x = x + fun(&x);
}
What is the value of x after the assignment statement in main, assuming
a. Operands are evaluated left to right.
7
b. Operands are evaluated right to left.
12
Question 4
[20]
The following code segment was used as evidence that the readability of some code with gotos
is better that the equivalent code without gotos. This code finds the first row of an n × n
integer matrix named x that has nothing but zero values.
for i := 1 to n do
begin
for j := 1 to n do
if x[i, j] <> 0
then goto reject;
writln (‘First all-zero row is:’, i);
break;
reject:
end;
Rewrite this code without gotos in C or Java
for (i = 1; i <= n; i++) {
flag = 1;
for (j = 1; j <= n; j++)
if (x[i][j] != 0) {
flag = 0;
break;
}
if (flag == 1) {
printf("First all-zero row is: %d\n", i);
break;
}
}
Question 5
[20]
Consider the following program written in C syntax:
void main() {
int value = 2, list[5] = {1, 3, 5, 7, 9};
swap(value, list[0]);
swap(list[0], list[1]);
swap(value, list[value]);
}
void swap(int a, int b) {
int temp;
temp = a;
a = b;
b = temp;
}
For each of the following parameter-passing methods, what are all of the values of the
variables value and list after each of the three calls to swap?
a. Passed by value
b. Passed by reference
c. Passed by name
d. Passed by value-result
a. Passed by value
value = 2, list = {1, 3, 5, 7, 9}
value = 2, list = {1, 3, 5, 7, 9}
value = 2, list = {1, 3, 5, 7, 9}
[ after first swap ]
[ after second swap ]
[ after third swap ]
b. Passed by reference
value = 1, list = {2, 3, 5, 7, 9}
value = 1, list = {3, 2, 5, 7, 9}
value = 2, list = {3, 1, 5, 7, 9}
[ after first swap ]
[ after second swap ]
[ after third swap ]
c. Passed by name
value = 1, list = {2, 3, 5, 7, 9}
value = 1, list = {3, 2, 5, 7, 9}
value = 2, list = {3, 2, 1, 7, 9}
[ after first swap ]
[ after second swap ]
[ after third swap ]
a. Passed by value-result
value = 1, list = {2, 3, 5, 7, 9}
value = 1, list = {3, 2, 5, 7, 9}
value = 2, list = {3, 2, 1, 7, 9}
[ after first swap ]
[ after second swap ]
[ after third swap ], assume that we change first parameter
first