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GO Classes DPP | GATE CS | C Programming | Memory Leak

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5 5 votes

What is the main issue in the following code? Assume both $\texttt{malloc()}$ calls succeed.

#include <stdio.h>
#include <stdlib.h>

int main() {
    int *p = (int *)malloc(sizeof(int));
    *p = 5;

    p = (int *)malloc(sizeof(int));
    *p = 10;

    printf("%d", *p);

    free(p);

    return 0;
}
  1. It gives compilation error
     
  2. It prints $\texttt{10}$, but the first allocated memory block is leaked
     
  3. It prints $\texttt{5}$, and no memory leak occurs
     
  4. It gives undefined behavior because $\texttt{p}$ is reassigned

2 Answers

0 0 votes

First:

$\texttt{int *p = (int *)malloc(sizeof(int));}$

allocates one integer block on the heap.

Then:

$\texttt{*p = 5;}$

stores $\texttt{5}$ in that first heap block.

Now:

$\texttt{p = (int *)malloc(sizeof(int));}$

allocates a second heap block and stores its address in $\texttt{p}$.

But the address of the first heap block is lost.

Since no pointer now stores the address of the first block, we cannot free it later.

This is called a memory leak.

Then:

$\texttt{*p = 10;}$

stores $\texttt{10}$ in the second heap block.

So:

$\texttt{printf("\%d", *p);}$

prints $\texttt{10}$

Finally:

$\texttt{free(p);}$

frees only the second block.

The first block remains leaked.


Answer: B

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