C++ array size determination - Part 2

In my last post I showed how to create a template function to determine the size of a C++ array and promised to show how to create a template that will work with a type instead of a variable.

Lets start by looking at a struct template that through specialization we can use to capture the number of elements in an array.

  template <typename T>
  struct array_info
  {
  };

This just names a template that we will specialize for arrays, like so:

  template <typename T, size_t N>
  struct array_info<T[N]>
  {
    typedef T type;
    enum { size = N };
  };

Now we have a template that for arrays will enable us to get to the size and type of elements in the array. This can be used like so:

  typedef int IntArray[10];
  size_t s = array_info<IntArray>::size; // s == 10

Or like so:

  float floatArray[array_info<IntArray>::size];

If we try something other than an array we get a compiler error.

  size_t s = array_info<int*>::size; // error

Generates a compiler error similar to:

  error: 'size' is not a member of '<unnamed>::array_info<int*>'

This is nice, but the real power is in using this with other templates. Even though I'm working with arrays one of the apis I'm using returns a pointer instead of an array reference. I know the type (which includes its length) so I wanted to take advantage of that. Here I use the array_info in an equal method to determine the length:

  template <typename A, typename T>
  bool equal(const T* lhs, const T* rhs)
  {
    // Use 'typename array_info<A>::type' instead of 'T' so that compiler
    // verifies A and T types match up.
    const typename array_info<A>::type* const end = &lhs[array_info<A>::size];
    return std::equal(&lhs[0], end, rhs, &equal_to<T>);
  }

This template can then be used like so:

  equal<IntArray>(lhs.getVal(), rhs.getVal());

That is not syntax you get to use everyday.