what does malloc(0) return? [duplicate]

随声附和 提交于 2019-11-26 04:21:48


This question already has an answer here:

  • what's the point in malloc(0)? 17 answers

What does malloc(0) returns? Would the answer be same for realloc(malloc(0),0) ?

int main()
        printf(\"%p\\n\", malloc(0));
        printf(\"%p\\n\", realloc(malloc(0), 0));
        return 0;

Output from linux gcc:

manav@manav-workstation:~$ gcc -Wall mal.c
manav@manav-workstation:~$ ./a.out

The output keep changing everytime for malloc(0). Is this a standard answer? And why would anyone be interested in getting such a pointer, other than academic research?


If malloc(0) returns dummy pointer, then how does following works:

int main()
    void *ptr = malloc(0);
    printf(\"%p\\n\", realloc(ptr, 1024));
    return 0;


The following code outputs \"possible\" for every iteration. Why should it not fail ?

int main()

        int i;
        void *ptr;
        printf(\"Testing using BRUTE FORCE\\n\");
        for (i=0; i<65000; i++)
                ptr = malloc(0);
                if (ptr == realloc(ptr, 1024))
                        printf(\"Iteration %d: possible\\n\", i);
                        printf(\"Failed for iteration %d\\n\", i);
        return 0;


Others have answered how malloc(0) works. I will answer one of the questions that you asked that hasn't been answered yet (I think). The question is about realloc(malloc(0), 0):

What does malloc(0) return? Would the answer be same for realloc(malloc(0),0)?

The standard says this about realloc(ptr, size):

  • if ptr is NULL, it behaves like malloc(size),
  • otherwise (ptr is not NULL), it deallocates the old object pointer to by ptr and returns a pointer to a new allocated buffer. But if size is 0, C89 says that the effect is equivalent to free(ptr). Interestingly, I can't find that statement in C99 draft (n1256 or n1336). In C89, the only sensible value to return in that case would be NULL.

So, there are two cases:

  • malloc(0) returns NULL on an implementation. Then your realloc() call is equivalent to realloc(NULL, 0). That is equivalent to malloc(0) from above (and that is NULL in this case).
  • malloc(0) returns non-NULL. Then, the call is equivalent to free(malloc(0)). In this case, malloc(0) and realloc(malloc(0), 0) are not equivalent.

Note that there is an interesting case here: in the second case, when malloc(0) returns non-NULL on success, it may still return NULL to indicate failure. This will result in a call like: realloc(NULL, 0), which would be equivalent to malloc(0), which may or may not return NULL.

I am not sure if the omission in C99 is an oversight or if it means that in C99, realloc(ptr, 0) for non-NULL ptr is not equivalent to free(ptr). I just tried this with gcc -std=c99, and the above is equivalent to free(ptr).

Edit: I think I understand what your confusion is:

Let's look at a snippet from your example code:

ptr = malloc(0);
if (ptr == realloc(ptr, 1024))

The above is not the same as malloc(0) == realloc(malloc(0), 1024). In the second, the malloc() call is made twice, whereas in the first, you're passing a previously allocated pointer to realloc().

Let's analyze the first code first. Assuming malloc(0) doesn't return NULL on success, ptr has a valid value. When you do realloc(ptr, 1024), realloc() basically gives you a new buffer that has the size 1024, and the ptr becomes invalid. A conforming implementation may return the same address as the one already in ptr. So, your if condition may return true. (Note, however, looking at the value of ptr after realloc(ptr, 1024) may be undefined behavior.)

Now the question you ask: malloc(0) == realloc(malloc(0), 1024). In this case, let's assume that both the malloc(0) on the LHS and RHS returns non-NULL. Then, they are guaranteed to be different. Also, the return value from malloc() on the LHS hasn't been free()d yet, so any other malloc(), calloc(), or realloc() may not return that value. This means that if you wrote your condition as:

if (malloc(0) == realloc(malloc(0), 1024)

you won't see possible on the output (unless both malloc() and realloc() fail and return NULL).

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

int main(void)
    void *p1;
    void *p2;

    p1 = malloc(0);
    p2 = realloc(p1, 1024);
    if (p1 == p2)
        puts("possible, OK");

    /* Ignore the memory leaks */
    if (malloc(0) == realloc(malloc(0), 1024))
        puts("shouldn't happen, something is wrong");
    return 0;

On OS X, my code didn't output anything when I ran it. On Linux, it prints possible, OK.


malloc(0) is Implementation Defined as far as C99 is concerned.

From C99 [Section 7.20.3]

The order and contiguity of storage allocated by successive calls to the calloc, malloc, and realloc functions is unspecified. The pointer returned if the allocation succeeds is suitably aligned so that it may be assigned to a pointer to any type of object and then used to access such an object or an array of such objects in the space allocated (until the space is explicitly deallocated). The lifetime of an allocated object extends from the allocation until the deallocation. Each such allocation shall yield a pointer to an object disjoint from any other object. The pointer returned points to the start (lowest byte address) of the allocated space. If the space cannot be allocated, a null pointer is returned. If the size of the space requested is zero, the behavior is implementation- defined: either a null pointer is returned, or the behavior is as if the size were some nonzero value, except that the returned pointer shall not be used to access an object.


In C89, malloc(0) is implementation dependent - I don't know if C99 has fixed this or not. In C++, using:

char * p = new char[0];

is well defined - you get a valid, non-null pointer. Of course, you can't use the pointer to access what it points to without invoking undefined behaviour.

As to why this exists, it is convenient for some algorithms, and means you don't need to litter your code with tests for zero values.


C99 standard

If the space cannot be allocated, a nullpointer is returned. If the size of the space requested is zero, the behavior is implementation-defined: either a null pointer is returned, or the behavior is as if the size were some nonzero value, except that the returned pointer shall not be used to access an object.


The comp.lang.c FAQ has the following to say:

The ANSI/ISO Standard says that it may do either; the behavior is implementation-defined (see question 11.33). Portable code must either take care not to call malloc(0), or be prepared for the possibility of a null return.

So, it's probably best to avoid using malloc(0).


See C99, section 7.20.3:

If the size of the space requested is zero, the behavior is implementationdefined: either a null pointer is returned, or the behavior is as if the size were some nonzero value, except that the returned pointer shall not be used to access an object.

This is valid for all three allocation functions (ie calloc(), malloc() and realloc()).


One point nobody cared to talk about yet, in your first program is that realloc with length 0 is the same thing as free.

from the Solaris man page:

The realloc() function changes the size of the block pointed to by ptr to size bytes and returns a pointer to the (possibly moved) block. The contents will be unchanged up to the lesser of the new and old sizes. If ptr is NULL, realloc() behaves like malloc() for the specified size. If size is 0 and ptr is not a null pointer, the space pointed to is made available for further allocation by the application, though not returned to the system. Memory is returned to the system only upon termination of the application.

If one doesn't know that it can be a source of bad surprise (happened to me).


I think it depends. I checked the Visual Studio 2005 sources and saw this in the _heap_alloc function:

if (size == 0)
    size = 1;

I think that in many cases you may want a valid pointer, even when asking for zero bytes. This is because this consistent behavior makes it easier to check your pointers because: if you have a non-NULL pointer it's OK; if you have a NULL pointer you probably have a problem. That's why I think that most implementations will return a valid pointer, even when asking for zero bytes.


If malloc(0) returns dummy pointer, then how does following works:

void *ptr = malloc(0);

printf("%p\n", realloc(ptr, 1024));

I don't know what you mean by "dummy pointer". If malloc(0) returns non-NULL, then ptr is a valid pointer to a memory block of size zero. The malloc implementation saves this information in an implementation-specific way. realloc knows the (implementation-specific) way to figure out that ptr points to a memory block of size zero.

(How malloc/realloc/free do this is implementation-specific. One possibility is to allocate 4 bytes more than requested and store the size just before the memory block. In that case, ((int *)ptr)[-1] would give the memory block size, which is 0. You should never do this from your code, it's only for use by realloc and free).