Thursday, April 10, 2008

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:
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>
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.

1 comment:

Anonymous said...

Oh, that is wicked cool.... Gotta love templating