Page History

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 alignment (assuming that the sizeof(double) = 8 and sizeof(float) = 4.)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 sizeresize = 161024;
  size_t alignment = 1 << 12;
float  int *ptr;
double  int *ptr1;
  
ptr  =if aligned_alloc(alignNULL , size);
ptr1== (ptr = realloc(ptr,int size);

...

*)aligned_alloc(

...

Compliant Solution

This compliant solution checks that aligned_alloc() has the same alignment as the alignment realloc() function enforces on the memory pointed to by ptr (again assuming that the sizeof(double) = 8 and sizeof(float) = 4).

size_t size = 16;
size_t alignment = 1 << 12;
float *ptr;
double *ptr1;

ptr = aligned_alloc(align , size);

if(align == alignof(ptr1)) {
  ptr1 = realloc(ptr, size);
}
alignment, sizeof(int)))) {
    /* Handle error */
  }

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

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 = 101024;
  size_t align = 1 << 12;
  floatint *ptr;
  doubleint *ptr1;

  if (posix_memalign((void **)&ptr, align , 8size) != 0) {
    exit(0EXIT_FAILURE);
  }

ptr[0] = 12.5;
ptr[1] = 25.5;

printf("memory aligned to %d%zu bytes\n\n", align);
printf("ptr[0]   : %p = %f\n",ptr, ptr[0]);
printf("ptr[1]   : = %p = %f\n\n",&ptr[1], ptr[1]);

  if ((ptr1 = (int*) realloc((int *)ptr, 1024resize)) == NULL) {
    exit(0EXIT_FAILURE);
  }

printf  puts("After realloc(): \n");
printf("ptr1[0]   : %p = %lf\n",ptr1, ptr[0]);
printf("ptr1[1]  := %p = %lf\n\n",&ptr1[1], ptr1[1]);

  free(ptr1);
  return 0;
}

...

OUTPUT

memory aligned to 4096 bytes

ptr[0] = 0x1621b000

After realloc(): 0xb43c000 = 12.500000
ptr[1]   : 0xb43c004 = 25.500000

After realloc():
ptr1[0]   : 0xb43c000 = 12.500000
ptr1[1]  : 0xb43c008 = 0.000000

The value at ptr[1] remains the same after realloc()
ptr[1]  : 0xb43c004 = 25.500000


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() [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 error */
  }

  _aligned_free(ptr1);
}

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

...

Image Added Image Added Image Added Wiki Markup\[[Jones 09|AA. C References#Jones 09]\] Section 7.21.3