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];

  /* ... */

Footnote 103 in subclause 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];
  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.




Remediation Cost









Automated Detection





Fully checked
Axivion Bauhaus Suite


CertC-ARR01Fully implemented


sizeof Array Parameter

Can detect violations of the recommendation but cannot distinguish between incomplete array declarations and pointer declarations

LDRA tool suite

401 S

Fully implemented

Parasoft C/C++test


Do not call 'sizeof' on a pointer type

PC-lint Plus


682, 882

Fully supported

Polyspace Bug Finder


CERT C: Rec. ARR01-C

Checks for:

  • Wrong type used in sizeof
  • Possible misuse of sizeof

Rec, fully covered.




V511, V512, V514, V568, V579, V604, V697, V1086
sizeof-array-parameterFully checked

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)


Taxonomy item


CERT CCTR01-CPP. Do not apply the sizeof operator to a pointer when taking the size of an arrayPrior to 2018-01-12: CERT: Unspecified Relationship
CWE 2.11CWE-467, Use of sizeof() on a pointer typePrior to 2018-01-12: CERT:
ISO/IEC TS 17961Taking the size of a pointer to determine the size of the pointed-to type [sizeofptr]Prior to 2018-01-12: CERT: Unspecified Relationship
MITRE CWECWE-569Prior to 2018-01-12:
MITRE CWECWE-783Prior to 2018-01-12:


[Drepper 2006]Section 2.1.1, "Respecting Memory Bounds"
[ISO/IEC 9899:2011]Subclause, "The sizeof and _Alignof Operators"


  1. We may want to show another NCCE, where sizeof(array) / sizeof(array[0]) is used in other dangerous ways. If not, we may want to just make these example to be part of the test suite.

  2. Per the discussion between me, David, and Robert:

    We first thought that this title should be "Do not apply the sizeof operator to array parameters" to make it more concrete and to help with static tool analysis.  However, we thought of an example that would fit the current title:

    char arg[SIZE] = /* ... */
    char *ptr = arg;
    printf("%d\n", sizeof(ptr));

    Thoughts on changing the title, including this NCCE, splitting this into two recommendations, or (my favorite) making a recommendation that says "Do not use sizeof on pointers"?

    1. I thought of a more likely scenario:

      char *str = "Hello, world";
      /* I can never remember if it's strlen(str) + 1 or -1. */
      char *str2 = (char *)malloc(sizeof(str));  /* Woops, got 4 instead of strlen(str) + 1. */
  3. Should this rule also detect this kind of error?

            unsigned char   string[SIZE];
            memset(&string, 0, sizeof(string));
    1. What makes you think the code is in error? The address of an array is the address of its first element so the snippet simply zeroes out the array.

  4. Why is this not a rule but a recommendation? Clearly applying the "sizeof operator to a pointer" does not work when one is "taking the size of an array", i.e. the title is a bit of a contradiction in terms in my non-native ears. In any case this easily leads to miscalculated sizes of objects and thus potential OOB accesses, so why not make it a rule?

    1. Because violations of this recommendation also violate MEM35-C. Allocate sufficient memory for an object.

  5. The coderwall explains in Some detail the theory of how best to get the size of an array in C++

    1. Interesting. I would guess that, like the sizeof trick, the template only applies to arrays whose sizes are known at compile time. Trying to pass a pointer to that template would yield a compiler error. Which is admittedly better than having the computer return the "wrong" value for the szieof ratio.

      1. An even better approach is to use std::size() since it's defined for you (in C++17 and later).

  6. Note that there's also the much less common pointer-to-array type you could use here, as it preserve the array size.

    #include <stddef.h>
    #include <stdio.h>

    void clear(int (*parray)[12]) {

        printf("sizeof parray: %d\n", sizeof(parray));
        printf("sizeof *parray: %d\n", sizeof(*parray));
        printf("sizeof parray[0]: %d\n", sizeof(parray[0]));
        printf("sizeof *parray[0]: %d\n", sizeof(*parray[0]));
        printf("element count: %d\n", sizeof(*parray) / sizeof(*parray[0]));

        for (size_t i = 0; i < sizeof(*parray) / sizeof(*parray[0]); ++i) {
            *parray[i] = 0;

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

    I wouldn't recommend it though, as the element counter is easy to mess up, especially when compared to the normal idiom.

    1. Agreed. I suspect the compiler is inferring some of those sizes at compile timel If you change the parray type  to **int, you get completely different sizes.

      1. Remember, though, that int ** and int (*)[12] are incompatible types, and changing one to the other would break things even if they don't use sizeof.