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

Do not use realloc() to increase the size of a block that has strict alignment requirements.

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 space referenced by ptr, 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 requirements of the original space.

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

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

  if ((ptr1 = realloc(ptr, resize)) == NULL) {
    /* 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 %d bytes\n", align);
  printf("ptr = %p\n\n", ptr);

  if ((ptr1 = 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:

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

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

  if ((ptr1 = aligned_alloc(alignment, resize)) == NULL) {
    /* Handle error */
  }

  if ((memcpy(ptr1, ptr, sizeof(int)) == NULL) {
    /* Handle error */
  }

  free(ptr);
}

Compliant Solution (Windows)

Windows defines the _aligned_malloc() 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 ((ptr = _aligned_malloc(sizeof(int), alignment))
      == NULL) {
    /* Handle error */
  }
 
  /* Reallocation */
  if ((ptr1 = _aligned_realloc(ptr, 1024, alignment))
      == NULL) {
    _aligned_free(ptr);
    /* Handle error */
  }

  _aligned_free(ptr1);
}

Note that 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 arbitrary memory locations being accessed and written to.

Recommendation	Severity	Likelihood	Remediation Cost	Priority	Level
MEM36-C	Low	Probable	High	P2	L3

Related Vulnerabilities

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

Bibliography

[ISO/IEC 9899:2011] Subclause 7.22.3.1, "The aligned_alloc Function"

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