Someone said that using the original malloc()/calloc() pointer as the variable that holds the return value of realloc() is wrong. Here is an example:
#include <stdlib.h>
int main()
{
int *malloc_ptr = malloc(5 * sizeof(int));
malloc_ptr = realloc(malloc_ptr, 10 * sizeof(int));
return 0;
}
Their reasoning was that if realloc() failed it would return NULL thus making malloc_ptr == NULL. That would make you lose access to the memory you requested with malloc(). Their solution was to do this:
#include <stdlib.h>
int main()
{
int *malloc_ptr = malloc(5 * sizeof(int));
int *realloc_ptr = realloc(malloc_ptr, 10 * sizeof(int));
if (realloc_ptr == NULL) {
// Whatever you want to do
} else
malloc_ptr = realloc_ptr;
return 0;
}
However if they are correct doesn’t that mean that some software could be incorrect because I’ve seen a lot of people use the malloc()/calloc() pointer as the variable that holds the return value of realloc(). Is this true?
>Solution :
As per man realloc, this would be correct way of realloc handling:
#include <stdlib.h>
int main()
{
int *malloc_ptr = malloc(5 * sizeof(int));
if (malloc_ptr == NULL) {
fprintf(stderr, "Could not allocate memory\n");
/* here goes error handling */
}
/* ... */
int *realloc_ptr = realloc(malloc_ptr, 10 * sizeof(int));
if (realloc_ptr == NULL) {
fprintf(stderr, "Could not reallocate memory, keeping the old allocated block\n");
/* here goes error handling */
} else {
malloc_ptr = realloc_ptr;
}
/* ... */
free(malloc_ptr);
return 0;
}