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GO Classes DPP | GATE CS | C Programming | Functions and Pointers

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7 7 votes

What is the output of the following code?

#include <stdio.h>

void change(int *p, int *q) {
    *p = *p + 4;
    *q = *p + *q;
    p = q;
    *p = *p - 3;
}

int main() {
    int a = 6, b = 10;

    change(&a, &b);

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

    return 0;
}
  1. $\texttt{10\ 17}$
     
  2. $\texttt{10\ 20}$
     
  3. $\texttt{6\ 17}$
     
  4. $\texttt{13\ 17}$

1 Answer

0 0 votes

Initially, $\texttt{a = 6}$ and $\texttt{b = 10}$.

The function call is $\texttt{change(\&a, \&b)}$.

Inside the function, pointer $\texttt{p}$ points to $\texttt{a}$ and pointer $\texttt{q}$ points to $\texttt{b}$.

The statement $\texttt{*p = *p + 4}$ changes $\texttt{a}$.

So, $\texttt{a = 6 + 4 = 10}$.

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

Here, $\texttt{*p = 10}$ and $\texttt{*q = 10}$.

So, $\texttt{b = 10 + 10 = 20}$.

Now, $\texttt{p = q}$ means local pointer $\texttt{p}$ starts pointing to $\texttt{b}$.

The statement $\texttt{*p = *p - 3}$ changes $\texttt{b}$.

So, $\texttt{b = 20 - 3 = 17}$.

Final values are $\texttt{a = 10}$ and $\texttt{b = 17}$.

Therefore, the output is $\texttt{10\ 17}$.

Answer: A.

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