C++ array size determination

My current project has me working with C-style C++ arrays. In order to make things easier and safer I created some helper templates.

Lets start with finding the the number of elements in an array. The conventional C way would be to use sizeof like:

  sizeof(array)/sizeof(array[0])

I'm not a big fan of this approach. I have seen production code by seasoned C++ developers that looks like this:

  const char* tmp = 0;
  // ...
  int size = sizeof(tmp)/sizeof(tmp[0]);

Of course this is sometimes hidden behind a macro:

  #define array_size(array) (sizeof(array)/sizeof(array[0]))
  // ...
  int size = array_size(tmp);

Either way, this gives the wrong answer. The developer, of course, is not interested in what the size of a pointer to an element divided by the size of an element is.

So, can we do better? Actually, yes, we can.

  template <typename T, size_t N>
  inline
  size_t array_size(const T (&lhs)[N])
  {
    return N;
  }

Here we pass an array by reference to array_size() that extracts the size of the array and returns it. I talked briefly about array passing a few years ago.

We can now use it like so:

  int ia[10];
  size_t s = array_size(ia); // s == 10

And if we try to use this with a pointer like so:

  const char* tmp = 0;
  // ...
  size_t s = array_size(tmp); // error

We get a compiler error similar to:

  error: no matching function for call to 'array_size(const char*&)'

Very nice.

However, sizeof can also be used with a type instead of a variable. In my next post, I'll show how we can create a template that will work with a type instead of a variable.