EXP09-C. Use sizeof to determine the size of a type or variable

Do not hard code the size of a type into an application. Because of alignment, padding, and differences in basic types (e.g., 32-bit versus 64-bit pointers), the size of most types can vary between compilers and even versions of the same compiler. Using the sizeof operator to determine sizes improves the clarity of what is meant and ensures that changes between compilers or versions will not affect the code.

Type alignment requirements can also affect the size of structures. For example, the size of the following structure is implementation-defined :

struct s {
  int i;
  double d;
};

Assuming 32-bit integers and 64-bit doubles, for example, the size can range from 12 to 16 bytes, depending on alignment rules.

Noncompliant Code Example

This noncompliant code example attempts to declare a two-dimensional array of integers with variable length rows. On a platform with 64-bit integers, the loop will access memory outside the allocated memory section.

int f(void) { /* Assuming 32-bit pointer, 32-bit integer */
  size_t i;
  int **matrix = (int **)calloc(100, 4);
  if (matrix == NULL) {
    return -1; /* Indicate calloc() failure */
  }

  for (i = 0; i < 100; i++) {
    matrix[i] = (int *)calloc(i, 4);
    if (matrix[i] == NULL) {
      return -1; /* Indicate calloc() failure */
    }
  }
 return 0;
}

Compliant Solution

This compliant solution replaces the hard-coded value 4 with sizeof(int *) :

int f(void) {
  size_t i;
  int **matrix = (int **)calloc(100, sizeof(*matrix));
  if (matrix == NULL) {
    return -1; /* Indicate calloc() failure */
  }

  for (i = 0; i < 100; i++) {
    matrix[i] = (int *)calloc(i, sizeof(**matrix));
    if (matrix[i] == NULL) {
      return -1; /* Indicate calloc() failure */
    }
  }

  return 0;
}

Also see MEM02-C. Immediately cast the result of a memory allocation function call into a pointer to the allocated type for a discussion on the use of the sizeof operator with memory allocation functions.

Exceptions

EXP09-C-EX1: The C Standard explicitly declares sizeof(char) == 1 , so any sizes based on characters or character arrays may be evaluated without using sizeof . This does not apply to char* or any other data types.

Risk Assessment

Porting code with hard-coded sizes can result in a buffer overflow or related vulnerability .

Automated Detection

Related Vulnerabilities

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

Related Guidelines