# V511\. The sizeof\(\) operator returns pointer size instead of array size\.

The 'sizeof' operator returns size of a pointer, not of an array, when the array was passed by value to a function\.

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

```cpp
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:

```cpp
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:

```cpp
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:

```cpp
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;
}
```