Question

Reading some literature, I was able to grasp that realloc takes in a void pointer and a size variable and re-allocates the memory of the block the void pointer points to.

• What will happen if realloc is called on an integer pointer (int *)
  with a size of character? And vice versa.

• What can be a possible application of this? (An example would definitely help.)

Answer 1

The realloc() function is the all-in-one memory management system.

• If called with a null pointer and a non-zero size, it allocates memory.
• If called with a valid pointer and a zero size, it frees memory.
• If called with a valid pointer and a non-zero size, it changes the size of the allocated memory.

If you call realloc() with an invalid pointer — one which was not obtained from malloc(), calloc() or realloc() — then you get undefined behaviour.

You could pass realloc() an integer pointer to an allocated space of sizeof(char) bytes (1 byte), but you’d be in danger of invoking undefined behaviour. The problem is not with realloc(); it is with the code that was given an unusable integer pointer. Since only 1 byte was allocated but sizeof(int) is greater than 1 (on essentially all systems; there could be exceptions, but not for someone asking this question), there is no safe way to use that pointer except by passing it to free() or realloc().

Given:

int *pointer = malloc(sizeof(char));

you cannot do *pointer = 0; because there isn’t enough space allocated (formally) for it to write to. You cannot do int x = *pointer; because there isn’t enough space allocated (formally) for it to read from. The word ‘formally’ is there because in practice, the memory allocators allocate a minimum size chunk, which is often 8 or 16 bytes, so there actually is space after the one byte. However, your are stepping outside the bounds of what the standard guarantees, and it is possible to conceive of memory allocators that would hand you exactly one byte. So, don’t risk it. An integer pointer to a single byte of allocated memory is unusable except as an argument to the memory allocation functions.

The first argument to realloc() is a void *. Since you’re going to have a prototype in scope (#include <stdlib.h>), the compiler will convert the int * to a void * (if there’s anything to do for such a cast), and as long as the space pointed at was allocated, everything will be fine; realloc() will change the allocation size, possibly returning the same pointer or possibly returning a different pointer, or it will release the space if the new size is zero bytes.