How to deallocate dynamically allocate memory without using “free()” function.

NOTE: this is actually puzzle type question nobody  use it realloc() function to free the memory. But I heard, in many interviews this question had been asked.

Standard library function realloc() can be used to deallocate previously allocated memory. Below is function declaration of “realloc()” from “stdlib.h”

void *realloc(void *ptr, size_t size);

If “size” is zero, then call to realloc is equivalent to “free(ptr)”. And if “ptr” is NULL and size is non-zero then call to realloc is equivalent to “malloc(size)”.

Let us check with simple example.

/* code with memory leak */ #include <stdio.h> #include <stdlib.h>   int main(void) {     int *ptr = (int*)malloc(10);       return 0; }

Check the leak summary with valgrind tool. It shows memory leak of 10 bytes, which is highlighed in red colour.

 valgrind –leak-check=full ./free
  ==1238== LEAK SUMMARY:
  ==1238==    definitely lost: 10 bytes in 1 blocks.
  ==1238==      possibly lost: 0 bytes in 0 blocks.
  ==1238==    still reachable: 0 bytes in 0 blocks.
  ==1238==         suppressed: 0 bytes in 0 blocks.

Let us modify the above code.

#include <stdio.h> #include <stdlib.h>   int main(void) {     int *ptr = (int*) malloc(10);       /* we are calling realloc with size = 0 */     realloc(ptr, 0);            return 0; }

Check the valgrind’s output. It shows no memory leaks are possible, highlighted in red color.

  >valgrind –leak-check=full ./a.out
  ==1435== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 11 from 1)
  ==1435== malloc/free: in use at exit: 0 bytes in 0 blocks.
  ==1435== malloc/free: 1 allocs, 1 frees, 10 bytes allocated.
  ==1435== For counts of detected errors, rerun with: -v
  ==1435== All heap blocks were freed — no leaks are possible.