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Typically, there are several different possible alignments used for the fundamental types of C. If the C type Do not convert a pointer value to a pointer type that is more strictly aligned than the referenced type. Different alignments are possible for different types of objects. If the type-checking system is overridden by an explicit cast , it is possible the alignment of the underlying object or type may not match up with the object to which it was cast. Therefore, the alignment must always be the same if a pointer is to be cast into another.

Non-compliant Code Example

or the pointer is converted to a void pointer (void *) and then to a different type, the alignment of an object may be changed.

The C Standard, 6.3.2.3, paragraph 7 [ISO/IEC 9899:2024], states

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

See undefined behavior 25.

If the misaligned pointer is dereferenced, the program may terminate abnormally. On some architectures, the cast alone may cause a loss of information even if the value is not dereferenced 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;

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

Compliant Solution (Intermediate Object)

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

#include <assert.h>
 
void func(void) {
  char c = 'x';
  int i = c;
  int *ip = &i;

  assert(ip == &i);
}

Noncompliant Code Example

The C Standard allows any object pointer to be cast to and 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 By definition of C99, a pointer may be cast into and out of void* validly. Thus it is possible to silently switch from one type of pointer to another without flagging a compiler warning by first storing or casting the initial pointer to void * and then storing or casting it to the final type. In the following non-compliant code, the type checking system is circumvented due to the caveats of void pointers.this noncompliant code example, loop_function() is passed the char pointer char_ptr but returns an object of type int pointer:

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

  /* ... */
}

This example compiles without warning using GCC 4.8 on Ubuntu Linux 14.04. However, int_pointer can be more 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_function(voidint *v_pointer) {
  /* ... */
  return v_pointer;
}
 
void func(int *loop_ptr) {
  int *int_ptr = loop_function(loop_ptr);

This example should compile without warning. However, v_pointer might be aligned on a 1 byte boundary. Once it is cast to an int, some architectures will require it to be on 4 byte boundaries. If int_ptr is then later dereferenced, abnormal termination of the program may result.

Compliant Solution

In this compliant solution, the parameter is changed to only accept other int* pointers since the input parameter directly influences the output parameter.

int *loop_ptr;
int *int_ptr;

int *loopFunction(int *v_pointer) {
  return v_pointer;
}
int_ptr = loopFunction(loop_ptr);

Implementation Details

List of common alignments for Microsoft, Borland, and GNU compilers to x86

  /* ... */
}

Noncompliant Code Example

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

#include <string.h>
 
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));

  /* ... */
}

Assigning an unaligned value to a pointer that references a type that needs to be aligned is undefined behavior. An implementation may notice, for example, that tmp and header must be aligned and use an inline memcpy() that uses instructions that assume aligned data.

Compliant Solution

This compliant solution avoids the use of the foo_header pointer:

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

  /* ... */
}

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 program:

#include <stdio.h>
#include <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 int. As a result, the two pointers reference equally aligned pointer types:

#include <stdalign.h>
#include <assert.h>
 
void func(void) {
  /* Align c to the alignment of an int */
  alignas(int) char c = 'x';
  int *ip = (int *)&c; 
  char *cp = (char *)ip;
  /* 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 may it can cause slow pointer accesses (if the architecture does not care about alignmentallows misaligned accesses).

Automated Detection

Related Vulnerabilities

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

References

Related Guidelines

Key here (explains table format and definitions)

Bibliography

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Image Added Image Added Image Added Wiki Markup\[[Bryant 03|AA. C References#Bryant 03]\] \[[ISO/IEC 9899-1999:TC2|AA. C References#ISO/IEC 9899-1999TC2]\] Section 6.2.5, "Types"