Page History

Do not convert convert a pointer value to a pointer type that is more strictly aligned than the referenced type the value actually points to. Different . Different alignments are possible for different types of objects. If the type-checking system is overridden by an explicit cast or the pointer is converted to a void pointer (void *) and then to a different type, the alignment of an object may be changed.

Subclause The C Standard, 6.3.2.3, paragraph 7 , of the C Standard [ISO/IEC 9899:20112024], states:

A pointer to an object or incomplete type may be converted to a pointer to a different object or incomplete type. If the resulting pointer is not correctly aligned for the pointed-to referenced type, the behavior is undefined.

(See also undefined behavior 25  in Annex J of the C Standard.)

If the misaligned pointer is dereferenced, the program may terminate abnormally. The On some architectures, the cast alone may cause a loss of information even if the value is not dereferenced . For if the types involved have differing alignment requirements.

Noncompliant Code Example

...

In this noncompliant example, the

...

char pointer &c is converted to the more strictly aligned int pointer ip. On some implementations, cp will not match &c. As a result, if a pointer to one object type is converted to a pointer to a different object type, the second object type must not require stricter alignment than the first.

#include <assert.h>
 
void func(void) {
  char c = 'x';
  int *ip = (int *)&c; /* This can lose information. */
  char *cp = (char *)ip;

  assert(cp == &c);    /* Will fail on some conforming implementations */

  /* ... */
}

On some implementations, cp will not match &c. As a result, if a pointer to one object type is converted to a pointer to a different object type, the second object type must not require stricter alignment than the first.

Noncompliant Code Example

In this noncompliant example, the char pointer &c is converted to the more strictly aligned int pointer i_ptr:

void f(void) {
  int *i_ptr;
  char c;
 
  i_ptr = (int *)&c;  /* Violation */
  /* ... */
assert(cp == &c);
}

Compliant Solution (Intermediate Object)

In this compliant solution, the value referenced by the char pointer c_ptr has the alignment of type intthe char value is stored into an object of type int so that the pointer's value will be properly aligned:

#include <assert.h>
 
void ffunc(void) {
  char *c_ptrc = 'x';
  int *i_ptr;
 = int ic;
 
 int c_ptr*ip = (char *)&i;

  i_ptrassert(ip = (int *)c_ptr;
  /* ... */
= &i);
}

Noncompliant Code Example

The C Standard allows a any object pointer to be cast into to and out of from void *. As a result, it is possible to silently convert from one pointer type to another without the compiler diagnosing the problem by storing or casting a pointer to void * and then storing or casting it to the final type. In this noncompliant code example, loop_function() is passed the char pointer loopchar_ptr but returns an object of type int pointer:

char *loop_ptr;
int *int_ptr;

int *loop_function(void *v_pointer) {
  /* ... */
  return v_pointer;
}
 
void func(voidchar *char_ptr) {
  int *int_ptr = loop_function(loopchar_ptr);

  /* ... */
}

This example compiles without warning using GCC 4.8 on Ubuntu Linux 14.04. However, vint_pointer can be aligned on a 1-byte boundarymore strictly aligned than an object of type char *.

Compliant Solution

Because the input parameter directly influences the return value, and loop_function() returns an object of type int *, the formal parameter v_pointer is redeclared to accept only an object of type int *:

int *loop_ptr;
int *int_ptr;

int *loop_function(int *v_pointer) {
  /* ... */
  return v_pointer;
}
 
void func(voidint *loop_ptr) {
  int *int_ptr = loop_function(loop_ptr);

  /* ... */
}

...

Noncompliant Code Example

Many Some architectures require that pointers are correctly aligned when accessing objects larger than a byte. There are, however, many places However, it is common in system code where you receive that unaligned data (for example, the network stacks) that needs to must be copied to a properly aligned memory location, such as in this noncompliant code example:

#include <string.h>
 
void func(void) {
  char *data;struct foo_header {
  int len;
  /* ... */
};
 
void func(char *data, size_t offset) {
  struct foo_header *tmp;
  struct foo_header *header;

  tmp = (struct foo_header *)(data + offset);
  memcpy(&header, tmp, sizeof(header));

  if (header.len < FOO) {
    /* ... */

  }
}

Unfortunately, the behavior is undefined when you assign Assigning an unaligned value to a pointer that points to references a type that needs to be aligned is undefined behavior. An implementation may notice, for example, that tmp and header must be aligned , so it could and use an inlined inline memcpy() that uses instructions that assume aligned data.

Compliant Solution

This compliant solution does not avoids the use of the foo_header pointer:

#include <string.h>
 
char *data;
struct foo_header header;
 {
  int len;
  /* ... */
};
  
void func(voidchar *data, size_t offset) {
  struct foo_header header; 
  memcpy(&header, data + offset, sizeof(header));

  if (header.len < FOO) {
    /* ... */

  }
}

Noncompliant Code Example

Exceptions

EXP36-C-EX1: Some hardware architectures have relaxed requirements with regard to pointer alignment. Using a pointer that is not properly aligned is correctly handled by the architecture, although there might be a performance penalty. On such an architecture, improper pointer alignment is permitted but remains an efficiency problem.

The x86 32- and 64-bit architectures usually impose only a performance penalty for violations of this rule, but under some circumstances, noncompliant code can still exhibit undefined behavior. Consider the following programFor objects declared on the stack, the C Standard provides alignas to declare an object to have a stricter alignment. It can be used to resolve the following noncompliant code example:

#include <stdio.h>
#include <assert.h>
 
void func(void) {
  char c = 'x';
  int *ip = (int *)&c; /* This can lose information. */
  char *cp = (char *)ip;
  assert(cp == &c);    /* Will fail on some conforming implementations */

  /* ... */
}

Compliant Solution

<stdint.h>

#define READ_UINT16(ptr)       (*(uint16_t *)(ptr))
#define WRITE_UINT16(ptr, val) (*(uint16_t *)(ptr) = (val))

void compute(unsigned char *b1, unsigned char *b2,
             int value, int range) {
  int i;
  for (i = 0; i < range; i++) {
    int newval = (int)READ_UINT16(b1) + value;
    WRITE_UINT16(b2, newval);
    b1 += 2;
    b2 += 2;
  }
}

int main() {
  unsigned char buffer1[1024];
  unsigned char buffer2[1024];
  printf("Compute something\n");
  compute(buffer1 + 3, buffer2 + 1, 42, 500);
  return 0;
}

This code tries to read short ints (which are 16 bits long) from odd pairs in a character array, which violates this rule. On 32- and 64-bit x86 platforms, this program should run to completion without incident. However, the program aborts with a SIGSEGV due to the unaligned reads on a 64-bit platform running Debian Linux, when compiled with GCC 4.9.4 using the flags  -O3   or  -O2 -ftree-loop-vectorize -fvect-cost-model.

If a developer wishes to violate this rule and use undefined behavior, they must not only ensure that the hardware guarantees the behavior of the object code, but they must also ensure that their compiler, along with its optimizer, also respect these guarantees.

EXP36-C-EX2: If a pointer is known to be correctly aligned to the target type, then a cast to that type is permitted. There are several cases where a pointer is known to be correctly aligned to the target type. The pointer could point to an object declared with a suitable alignment specifier. It could point to an object returned by aligned_alloc(), calloc(), malloc(), or realloc(), as per the C standard, section 7.22.3, paragraph 1  [ISO/IEC 9899:2011].

This compliant solution uses the alignment specifier, which is new to C11, to declare the char object c with the same alignment as that of an object of type intThis compliant solution uses alignas to align the the character c to the alignment of an integer. As a result, the two pointers point to reference equally aligned pointer types:

#include <stdalign.h>  /* For alignas() */
#include <assert.h>
 
void func(void) {
  alignas(int) char c = 'x'; /* Align c to the alignment of an int */
  alignas(int) char c = 'x';
  int *ip = (int *)&c; 
  char *cp = (char *)ip;
  assert(cp == &c);    /* Both cp and &c point to equally aligned objects. */
  /* ... */assert(cp == &c);
}

Risk Assessment

Accessing a pointer or an object that is no longer on the correct access boundary can not properly aligned can cause a program to crash or give wrong erroneous information, or it can cause slow pointer accesses (if the architecture allows misaligned accesses).

Automated Detection

Related Vulnerabilities

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

Related Guidelines

Key here (explains table format and definitions)

Bibliography

...

...

Image Modified Image Modified Image Modified