ARR01-C. Do not apply the sizeof operator to a pointer when taking the size of an array

The sizeof operator yields the size (in bytes) of its operand, which can be an expression or the parenthesized name of a type. However, using the sizeof operator to determine the size of arrays is error prone.

The sizeof operator is often used in determining how much memory to allocate via malloc(). However using an incorrect size is a violation of MEM35-C. Allocate sufficient memory for an object.

Noncompliant Code Example

In this noncompliant code example, the function clear() zeros the elements in an array. The function has one parameter declared as int array[] and is passed a static array consisting of 12 int as the argument. The function clear() uses the idiom sizeof(array) / sizeof(array[0]) to determine the number of elements in the array. However, array has a pointer type because it is a parameter. As a result, sizeof(array) is equal to the sizeof(int *). For example, on an architecture (such as IA-32) where the sizeof(int) == 4 and the sizeof(int *) == 4, the expression sizeof(array) / sizeof(array[0]) evaluates to 1, regardless of the length of the array passed, leaving the rest of the array unaffected.

void clear(int array[]) {
  for (size_t i = 0; i < sizeof(array) / sizeof(array[0]); ++i) {
     array[i] = 0;
   }
}

void dowork(void) {
  int dis[12];

  clear(dis);
  /* ... */
}

Footnote 103 in subclause 6.5.3.4 of the C Standard [ISO/IEC 9899:2011] applies to all array parameters:

When applied to a parameter declared to have array or function type, the sizeof operator yields the size of the adjusted (pointer) type.

Compliant Solution

In this compliant solution, the size of the array is determined inside the block in which it is declared and passed as an argument to the function:

void clear(int array[], size_t len) {
  for (size_t i = 0; i < len; i++) {
    array[i] = 0;
  }
}

void dowork(void) {
  int dis[12];

  clear(dis, sizeof(dis) / sizeof(dis[0]));
  /* ... */
}

This sizeof(array) / sizeof(array[0]) idiom will succeed provided the original definition of array is visible.

Noncompliant Code Example

In this noncompliant code example, sizeof(a) does not equal 100 * sizeof(int), because the sizeof operator, when applied to a parameter declared to have array type, yields the size of the adjusted (pointer) type even if the parameter declaration specifies a length:

enum {ARR_LEN = 100};

void clear(int a[ARR_LEN]) {
  memset(a, 0, sizeof(a)); /* Error */
}

int main(void) {
  int b[ARR_LEN];
  clear(b);
  assert(b[ARR_LEN / 2]==0); /* May fail */
  return 0;
}

Compliant Solution

In this compliant solution, the size is specified using the expression len * sizeof(int):

enum {ARR_LEN = 100};

void clear(int a[], size_t len) {
  memset(a, 0, len * sizeof(int));
}

int main(void) {
  int b[ARR_LEN];
  clear(b, ARR_LEN);
  assert(b[ARR_LEN / 2]==0); /* Cannot fail */
  return 0;
}

Risk Assessment

Incorrectly using the sizeof operator to determine the size of an array can result in a buffer overflow, allowing the execution of arbitrary code.

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