GATE IT 2006 | Question: 50

Question

Which one of the choices given below would be printed when the following program is executed?

#include <stdio.h>
void swap (int *x, int *y)
{
    static int *temp;
    temp = x;
    x = y;
    y = temp;
}
void printab ()
{
    static int i, a = -3, b = -6;
    i = 0;
    while (i <= 4)
    {
        if ((i++)%2 == 1) continue;
        a = a + i;
        b = b + i;
    }
    swap (&a, &b);
    printf("a =  %d, b = %d\n", a, b);
}
main()
{
    printab();
    printab();
}

• A. $a = 0, b = 3$
  $a = 0, b = 3$
• B. $a = 3, b = 0$
  $a = 12, b = 9$
• C. $a = 3, b = 6$
  $a = 3, b = 6$
• D. $a = 6, b = 3$
  $a = 15, b = 12$

Comments

the first printab(); execution is enough to know the answer we don't need to solve the whole question but this particualr question tests so many concepts that we already know in c programming

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Solution

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Never saw more clearer and better solution than this
I dont know why it doesnt have more upvotes

Answer 1

Answer:- 

Answer 2

It prints:

a = 6, b = 3
a = 15, b = 12

Why:

• a and b are static, so they retain their values between calls to printab().
• i is reset to 0 at the start of each call.
• The loop adds 1, 3, and 5 to both a and b, for a total of 9.
• swap(&a, &b) does not swap the values. It only swaps local pointer variables because x and y are passed by value.

First call:

a = -3 + 9 = 6
b = -6 + 9 = 3

Second call:

a = 6 + 9 = 15
b = 3 + 9 = 12

Answer 3

Answer: (D)

a = 6, b = 3
a = 15, b = 12

Why:

1. swap does nothing visible. It receives copies of the two pointers and swaps those copies. The variables a and b are never touched, so the call has no effect on what gets printed.
2. The loop in printab. i++ is a post-increment, so the comparison uses the old value of i, and i is incremented afterward. Tracing it:
   • i=0: the check uses 0, so no continue. i becomes 1, then a += 1, b += 1.
   • i=1: the check uses 1, so continue. i becomes 2.
   • i=2: no continue. i becomes 3, then a += 3, b += 3.
   • i=3: continue. i becomes 4.
   • i=4: no continue. i becomes 5, then a += 5, b += 5.
   • i=5: the loop condition 5 <= 4 fails, so the loop ends.
   Each call therefore adds 1 + 3 + 5 = 9 to both a and b.
3. First call. a = -3 + 9 = 6 and b = -6 + 9 = 3, so it prints a = 6, b = 3.
4. Second call. a, b and i are static, so they keep their values between calls. The initializers a = -3 and b = -6 run only once. i is reset to 0 by the explicit i = 0; statement, so the loop runs the same way again. Then a = 6 + 9 = 15 and b = 3 + 9 = 12, so it prints a = 15, b = 12.

The other options fail because they assume the swap worked (A, B) or that the static values reset on the second call (C).

 

Answer 4

Answer: (D)

a = 6, b = 3
a = 15, b = 12

Why:

1. swap does nothing visible. It receives copies of the two pointers and swaps those copies. The variables a and b are never touched, so the call has no effect on what gets printed.
2. The loop in printab. i++ is a post-increment, so the comparison uses the old value of i, and i is incremented afterward. Tracing it:
   • i=0: the check uses 0, so no continue. i becomes 1, then a += 1, b += 1.
   • i=1: the check uses 1, so continue. i becomes 2.
   • i=2: no continue. i becomes 3, then a += 3, b += 3.
   • i=3: continue. i becomes 4.
   • i=4: no continue. i becomes 5, then a += 5, b += 5.
   • i=5: the loop condition 5 <= 4 fails, so the loop ends.
   Each call therefore adds 1 + 3 + 5 = 9 to both a and b.
3. First call. a = -3 + 9 = 6 and b = -6 + 9 = 3, so it prints a = 6, b = 3.
4. Second call. a, b and i are static, so they keep their values between calls. The initializers a = -3 and b = -6 run only once. i is reset to 0 by the explicit i = 0; statement, so the loop runs the same way again. Then a = 6 + 9 = 15 and b = 3 + 9 = 12, so it prints a = 15, b = 12.