What will happen in this case of malloc and realloc

Question

malloc take some contiguous part of heap for some variable,

 -------------
|     p       |
 -------------

again malloc happens for some other pointer

 ------------- -----------------
|      P      |      S          |
 ------------- -----------------

Now realloc happens for p

 ------------------- -----------
|       p           |    S      |
 ------------------- -----------

So S space is reduced??? Does it happen?

I ran a program like this,

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

int main(void)
{
int *p;
char *s;
p = malloc(sizeof(*p));
printf("1:p points to %p\n", (void *)p);
//printf("1:p points to %p\n", (void *)(p+1));
s = malloc(sizeof(*s));
printf("1:s points to %p\n", (void *)s);
p = realloc(p, sizeof(*p)*2);
printf("2:p+2 points to %p\n", (void *)(p+2));
p = realloc(p, sizeof(*p)*7);
printf("3:p points to %p\n", (void *)(p));
printf("3:p+7 points to %p\n", (void *)(p+7));//should run over s's memory
printf("2:s points to %p\n", (void *)s);
free(p);
p =  NULL;
free(s);
s =  NULL;   
return 0;
}
output:
1:p points to 0x1291010
1:s points to 0x1291030
2:p+2 points to 0x1291018
3:p points to 0x1291050 --> location of p pointed to by p is changed.
3:p+7 points to 0x129106c
2:s points to 0x1291030

So first malloc gave 0x20 = 32 bytes on its own?? 0x1291030 - 0x1291010 = 0x20 = 32 bytes.

Is it possible that if I keep increasing size of memory pointed by 'p' will lead me to s's pointed memory and then i can access s using p?

P.S

edited the question to correctly say what I wanted to say. realloc will allocate some other place in memory once the memory of s is threatened to be stomped. Its clearly visible in output.

`realloc` is not guaranteed to keep the original address. It will allocate a larger region elsewhere and bit-wise copy the contents of P. — StoryTeller - Unslander Monica, Mar 27 '17 at 08:57
`realloc` will return a pointer to a new piece of memory (where it has copied to old contents to). The heap may reserve space after an allocation to cope with reallocs. No "bits" are allocated, only bytes and probably suitably alligned on ` 4 or 8 byte boundary. — Paul Ogilvie, Mar 27 '17 at 08:57
You talk about `realloc` but in your code you use `calloc`, and never `free` your first allocation. I struggle to make sense from your reasoning. — Kami Kaze, Mar 27 '17 at 09:03

score 3 · Answer 1 · answered Mar 27 '17 at 08:58

All of this is implementation-defined, and thus impossible to reason about in general.

The only thing that can be said is that no, of course realloc() on one pointer will never affect the memory returned by another call to malloc() or any other allocation function. This is why realloc() returns the resulting pointer, it can be different from the one you passed in.

The inner workings of malloc() and friends depends on the allocator, and there are many allocators. You can work around the standard library's if you don't agree with its choices, by using some other implementation.

Also, this:

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

makes little sense, you're allocating the size of the pointer which is not the same as the size of the data the pointer points at (i.e. an integer). To do that, you should use:

p = malloc(sizeof *p);

Finally, while printing pointers with %p is correct, that's only correct for void * and there's no automatic conversion since the varargs system doesn't know; you should cast to void * or you get undefined behavior.

changed to sizeof(*p). – acidlategamer Mar 27 '17 at 09:01 — acidlategamer, Mar 27 '17 at 09:01

score 2 · Accepted Answer · answered Mar 27 '17 at 09:01

There is no defined "least allocation unit" for malloc. It is implementation defined. It also depends on how much memory malloc tries to use for book-keeping purposes.

Regarding realloc, realloc will return the same address provided it has contiguous space to allocate for newly requested size, else, it might allocate the memory from other address and will return it and will free the memory pointed to by original pointer.

One thing which is guaranteed is that a space allocated once will not be resized without the user requesting for it i.e. in your question, the space for "S" will never automatically be reduced or resized.

I did some mistake in asking the question, edited it. Apparently your answer behaviour is shown and this question should have never been asked if I had checked myself properly. — acidlategamer, Mar 27 '17 at 09:24

score 1 · Answer 3 · answered Mar 27 '17 at 09:05

1

the memory location for p will be changed if there isn't enough space for reallocating. so the thing you show never happens.

answered Mar 27 '17 at 09:05

Fatemeh Karimi

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What will happen in this case of malloc and realloc

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