Example
#include <stddef.h> // size_t, ptrdiff_t
//----------------------------------- Machinery:
using Size = ptrdiff_t;
template< class Item, size_t n >
constexpr auto n_items( Item (&)[n] ) noexcept
-> Size
{ return n; }
//----------------------------------- Usage:
#include <iostream>
using namespace std;
auto main()
-> int
{
int const a[] = {3, 1, 4, 1, 5, 9, 2, 6, 5, 4};
Size const n = n_items( a );
int b[n] = {}; // An array of the same size as a.
(void) b;
cout << "Size = " << n << "\n";
}
The C idiom for array size, sizeof(a)/sizeof(a[0]), will accept a pointer as argument and will then generally yield an incorrect result.
For C++11
using C++11 you can do:
std::extent<decltype(MyArray)>::value;
Example:
char MyArray[] = { 'X','o','c','e' };
const auto n = std::extent<decltype(MyArray)>::value;
std::cout << n << "\n"; // Prints 4
Up till C++17 (forthcoming as of this writing) C++ had no built-in core language or standard library utility to obtain the size of an array, but this can be implemented by passing the array by reference to a function template, as shown above. Fine but important point: the template size parameter is a size_t, somewhat inconsistent with the signed Size function result type, in order to accommodate the g++ compiler which sometimes insists on size_t for template matching.
With C++17 and later one may instead use std::size, which is specialized for arrays.
