MEM36-C. Do not modify the alignment of objects by calling realloc()

Do not invoke realloc() to modify the size of allocated objects that have stricter alignment requirements than those guaranteed by malloc(). Storage allocated by a call to the standard aligned_alloc() function, for example, can have stricter than normal alignment requirements. The C standard requires only that a pointer returned by realloc() be suitably aligned so that it may be assigned to a pointer to any type of object with a fundamental alignment requirement.

Noncompliant Code Example

This noncompliant code example returns a pointer to allocated memory that has been aligned to a 4096-byte boundary.  If the resize argument to the realloc() function is larger than the object referenced by ptr, then realloc() will allocate new memory that is suitably aligned so that it may be assigned to a pointer to any type of object with a fundamental alignment requirement but may not preserve the stricter alignment of the original object.

#include <stdlib.h>
 
void func(void) {
  size_t resize = 1024;
  size_t alignment = 1 << 12;
  int *ptr;
  int *ptr1;
  
  if (NULL == (ptr = (int *)aligned_alloc(alignment, sizeof(int)))) {
    /* Handle error */
  }

  if (NULL == (ptr1 = (int *)realloc(ptr, resize))) {
    /* Handle error */
  }
}

Implementation Details

When compiled with GCC 4.1.2 and run on the x86_64 Red Hat Linux platform, the following code produces the following output:

CODE

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

int main(void) {
  size_t  size = 16;
  size_t resize = 1024;
  size_t align = 1 << 12;
  int *ptr;
  int *ptr1;

  if (posix_memalign((void **)&ptr, align , size) != 0) {
    exit(EXIT_FAILURE);
  }

  printf("memory aligned to %zu bytes\n", align);
  printf("ptr = %p\n\n", ptr);

  if ((ptr1 = (int*) realloc((int *)ptr, resize)) == NULL) {
    exit(EXIT_FAILURE);
  }

  puts("After realloc(): \n");
  printf("ptr1 = %p\n", ptr1);

  free(ptr1);
  return 0;
}

OUTPUT

memory aligned to 4096 bytes
ptr = 0x1621b000

After realloc():
ptr1 = 0x1621a010

ptr1 is no longer aligned to 4096 bytes.

Compliant Solution

This compliant solution  allocates resize bytes of new memory with the same alignment as the old memory, copies the original memory content, and then frees the old memory. This solution has implementation-defined behavior because it depends on whether extended alignments in excess of _Alignof (max_align_t) are supported and the contexts in which they are supported. If not supported, the behavior of this compliant solution is undefined.

#include <stdlib.h>
#include <string.h>
 
void func(void) {
  size_t resize = 1024;
  size_t alignment = 1 << 12;
  int *ptr;
  int *ptr1;

  if (NULL == (ptr = (int *)aligned_alloc(alignment,
                                          sizeof(int)))) {
    /* Handle error */
  }

  if (NULL == (ptr1 = (int *)aligned_alloc(alignment,
                                           resize))) {
    /* Handle error */
  }
  
  if (NULL == memcpy(ptr1, ptr, sizeof(int))) {
    /* Handle error */
  }
  
  free(ptr);
}

Compliant Solution (Windows)

Windows defines the _aligned_malloc() function to allocate memory on a specified alignment boundary.  The _aligned_realloc() [