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GO Classes DPP | GATE CS | C Programming | Pointers

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8 8 votes

What is the output of the following code?

#include <stdio.h>

int main() {
    int a = 8, b = 12;
    int *p = &a;
    int *q = &b;

    *p = *p + *q;
    q = p;
    *q = *q - 5;

    printf("%d %d", a, b);

    return 0;
}
  1. $\texttt{15\ 12}$
     
  2. $\texttt{20\ 12}$
     
  3. $\texttt{15\ 15}$
     
  4. $\texttt{8\ 12}$

1 Answer

2 2 votes

Initially, $\texttt{a = 8}$ and $\texttt{b = 12}$.

Pointer $\texttt{p}$ points to $\texttt{a}$ and pointer $\texttt{q}$ points to $\texttt{b}$.

The statement $\texttt{*p = *p + *q}$ changes $\texttt{a}$.

So, $\texttt{a = 8 + 12 = 20}$.

Now, $\texttt{q = p}$ means $\texttt{q}$ also starts pointing to $\texttt{a}$.

So now both $\texttt{p}$ and $\texttt{q}$ point to $\texttt{a}$.

The statement $\texttt{*q = *q - 5}$ again changes $\texttt{a}$.

So, $\texttt{a = 20 - 5 = 15}$.

The value of $\texttt{b}$ remains $\texttt{12}$.

Therefore, the output is $\texttt{15\ 12}$.

Answer: A.

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