Programming Languages

Assignment 2

1. Consider the following code written in a language with nested subroutines:

procedure main()

a : integer

procedure B(a : integer)

x : integer

procedure A(n : integer)

a := n

procedure R(m : integer)

write_integer(x)

x = x – 2

if x > 1

R(m + 1)

else

A(m)

–– body of B

x := a × a

R(1)

–– body of main

B(2)

write_integer(a)

1. What does this program print assuming static scope, respectively dynamic scope? Explain your answer.
2. Show the frames on the stack when procedure A has just been called. For each frame, show the values of the local variables and the static links.

2. Consider the following pseudocode with nested subroutines:

1. procedure main()
2. a : integer := 1
3. b : integer := 2
4. c : integer := 3

5. procedure first()
6. a : integer := c
7. c : integer := b

8. procedure second()
9. c : integer := 4

10. –-body of second
11. print a, b, c

12. b : integer := 5

13. –body of first
14. second()
15. print a, b, c

16. –body of main
17. first()
18. print a, b, c

What does the program print (or will the compiler identify static semantic errors) if we assume that:

1. names must be declared before use, and the scope of a name extends from its declaration through the end of the block?
2. names must be declared before use, but the scope of a name is the entire block in which it is declared?

3. Consider the following program written in generic pseudocode:

int x = 1;
int a[3] = { 2, 5, 3 };

void foo(int u, int v) {
v = a[x];
v = u * 9;
x = 2;
a[2] = u;
u = u – 2;
print u;
print v;
print x;
}
int main() {
foo(a[x], a[2]);
print x;
print a[0];
print a[1];
print a[2];
}

What is the output printed when this program is executed using call by value-result, respectively, call by macro expansion? Explain what happens in each case.