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Explain this code.
#include <stdio.h>
void f1(int *k)
{
*k = *k + 10;
}
main ( ){
int i;
i = 0;
printf (" The value of i before call %d \n", i);
f1 (&i);
printf (" The value of i after call %d \n", i);
}
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Answer Posted / ankita sharma
answer will be 10. as k has the address of i so when we write *k=*k+10; *k meand that value to which k is pointing so it is pointing to i and i has the value 0 as intial value so 10 would be added to the value of i. so output will be 10.
| Is This Answer Correct ? | 0 Yes | 0 No |
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