GATE | GATE-CS-2003 | Question 90

Consider the C program shown below.

C

#include <stdio.h>
#define print(x) printf(\"%d \", x)
int x;
void Q(int z)
{
    z += x;
    print(z);
}
void P(int *y)
{
    int x = *y + 2;
    Q(x);
    *y = x - 1;
    print(x);
}
main(void)
{
    x = 5;
    P(&x);
    print(x);
}

The output of this program is

(A)

12 7 6

(B)

22 12 11

(C)

14 6 6

(D)

7 6 6

Answer: (A)

Explanation:

x is global so first x becomes 5 by the first line in main(). Then main() calls P() with address of x.

// in main(void)

x = 5 // Change global x to 5
P(&x)

P() has a local variable named \’x\’ that hides global variable. P() then calls Q() by passing value of local \’x\’.

// In P(int *y)

int x = *y + 2; // Local x = 7
Q(x);

In Q(int z), z uses x which is global

// In Q(int z)

z += x; // z becomes 5 + 7 
printz(); // prints 12

After end of Q(), control comes back to P(). In P(), *y (y is address of global x) is changed to x – 1 (x is local to P()).

// Back in P()

 *y = x - 1; // *y = 7-1
 print(x); // Prints 7

After end of Q(), control comes back to main(). In main(), global x is printed.

// Back in main()

print(x); // prints 6 (updated in P()
          //           by *y = x - 1 )

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