EXP36-C. Do not convert pointers into more strictly aligned pointer types

Do not convert a pointer value to a pointer type that is more strictly aligned than the type the value actually points to. 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.

Section 6.3.2.3, para. 7, of the C Standard [ISO/IEC 9899:2011] 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 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 cast alone may cause a loss of information, even if the value is not dereferenced. For example, the assertion in the following code example will fail on some conforming implementations, even though no pointers are dereferenced:

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 */
  /* ... */
}

Compliant Solution

In this compliant solution, the value referenced by the char pointer c_ptr has the alignment of type int.

void f(void) {
  char *c_ptr;
  int *i_ptr;
  int i;
 
  c_ptr = (char *)&i;
  i_ptr = (int *)c_ptr;
  /* ... */
}

Noncompliant Code Example

The C Standard allows a pointer to be cast into and out of 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  loop_ptr but returns an int pointer.

char *loop_ptr;
int *int_ptr;

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

This example compiles without warning. However, v_pointer can be aligned on a 1-byte boundary.

Compliant Solution

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

int *loop_ptr;
int *int_ptr;

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

Another solution is to ensure that loop_ptr points to an object returned by malloc() because this object is guaranteed to be aligned properly for any need. However, this is a subtlety that is easily missed when the program is modified in the future. It is easier and safer to let the type system document the alignment needs.

Noncompliant Code Example

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

char *data;
struct foo_header *tmp;
struct foo_header *header;

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

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

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

Compliant Solution

This compliant solution does not use the foo_header pointer.

char *data;
struct foo_header header;

memcpy(&header, data + offset, sizeof(header));

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

Noncompliant Code Example

For 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.

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

The compliant solution uses alignas to align the the character c to the alignment of an integer. As a result, the two pointers point to equally aligned pointer types.

alignas(int) char c = 'x'; /* align c to the alignment of an int */
int *ip = (int *)&c; 
char *cp = (char *)ip;
assert(cp == &c);    /* both cp and &c point to equally aligned objects */

Risk Assessment

Accessing a pointer or an object that is no longer on the correct access boundary can cause a program to crash or give wrong 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

Bibliography