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If ptr was allocated with an alignment returned from 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() and if realloc() reallocates memory with a different alignment then, the behavior is undefined.

This rule only applies to compilers that conform to the (emerging) C1X standard \[[Jones 09|AA. C References#Jones 09]\].

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 aligns ptr returns a pointer to allocated memory that has been aligned to a 4096-byte boundary whereas .  If the resize argument to the realloc() function aligns the memory to a different alignmentis 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))) == NULL) {
    /* handleHandle error */
  }

/* ... */

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

The resulting program has undefined behavior as the alignment that realloc() enforces is different from aligned_alloc() function's alignment.

}

Implementation Details

When compiled with gcc version GCC 4.1.2 and run on the x86_64 -redhat-linux platform 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 %d%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

Unfortunately, ptr1 is no longer aligned to 4096 bytes.

Compliant Solution

This compliant solution implements an aligned realloc() function. It allocates new memory of resize bytes with an alignment equal to that of old memory and copies old memory into it. It then frees the old memory.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))) == NULL) {
    /* handleHandle error */
  }

/* ... */

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

Compliant Solution (Windows)

Windows defines the _aligned_malloc() function to allocate memory on a specified alignment boundary.  The _aligned_realloc() [MSDN] can be used to change the size of this memory. This compliant solution demonstrates one such usage:

#include <malloc.h>

void func(void) {
  size_t alignment = 1 << 12;
  int *ptr;
  int *ptr1;

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

  /* Reallocation */
  if (NULL == (ptr1 = (int *)_aligned_realloc(ptr, 1024,
                                              alignment))) {
    _aligned_free(ptr);
    /* Handle(memcpy(ptr1, ptr, sizeof(int)) == NULL) {
  /* handle error */
  }

  _aligned_free(ptrptr1);
}

The size and alignment arguments for _aligned_malloc() are provided in reverse order of the C Standard aligned_alloc() function.

Risk Assessment

Improper alignment can lead to accessing arbitrary memory locations and write into itbeing accessed and written to.

References

Automated Detection

Related Vulnerabilities

Search for vulnerabilities resulting from the violation of this rule on the CERT website.

Bibliography

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Image Added Image Added Image Added Wiki Markup\[[Jones 09|AA. C References#Jones 09]\] Section 7.21.3 \[[aligned_alloc()|http://www.open-std.org/Jtc1/sc22/wg14/www/docs/n1397.htm]\] 7.20.3.5