V511. The sizeof() operator returns size of the pointer, and not of the array, in given expression

23.05.2011

There is one specific feature of the language you might easily forget about and make a mistake. Look at the following code fragment:

char A[100];
void Foo(char B[100])
{
}

In this code, the A object is an array and the sizeof(A) expression will return value 100.

The B object is simply a pointer. Value 100 in the square brackets indicates to the programmer that he is working with an array of 100 items. But it is not an array of a hundred items which is passed into the function - it is only the pointer. So, the sizeof(B) expression will return value 4 or 8 (the size of the pointer in a 32-bit/64-bit system).

The V511 warning is generated when the size of a pointer is calculated which is passed as an argument in the format "TypeName ArrayName[N]". Such code is most likely to have an error. Look at the sample:

void Foo(float array[3])
{
  size_t n = sizeof(array) / sizeof(array[0]);
  for (size_t i = 0; i != n; i++)
    array[i] = 1.0f;
}

The function will not fill the whole array with value 1.0f but only 1 or 2 items depending on the system's capacity.

Win32: sizeof(array) / sizeof(array[0]) = 4/4 = 1.

Win64: sizeof(array) / sizeof(array[0]) = 8/4 = 2.

To avoid such errors, we must explicitly pass the array's size. Here is correct code:

void Foo(float *array, size_t arraySize)
{
  for (size_t i = 0; i != arraySize; i++)
    array[i] = 1.0f;
}

Another way is to use a reference to the array:

void Foo(float (&array)[3])
{
  size_t n = sizeof(array) / sizeof(array[0]);
  for (size_t i = 0; i != n; i++)
    array[i] = 1.0f;
}
You can look at examples of errors from real projects which were detected by this diagnostic message.