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What does the following program print?
#include < stdio.h >
void f (int *p, int *q)
p = q;
*p = 2;
}
int i = 0, j = 1;
int main () {
f(&i, &j);
printf(“%d%d/n”, i, j);
}
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- 2 2
- 2 1
- 0 1
- 0 2
- 2 2
Correct Option: D
* p points to i and q points to j */
void f(int *p, int *q)
{
p = q; /* p also points to j now */
*p = 2; /* Value of j is changed to 2 now */
}