Using size of one array in another array

Question

// sizeofarray.cpp
#include <iostream>
template <typename T,int N>
int size(T (&Array)[N])
{
  return N;
}

int main()
{
   char p[]="Je suis trop bon, et vous?";
   char q[size(p)]; // (A)
   return 0;
}

I heard that an array size in C++ must be a constant expression. So char q[size(p)] is invalid, am I right? But I got no errors when I tried

 g++ -Wall sizeofarray.cpp

Why?

Answer 1

Like Prasoon says, it's not a constant expression. For now, you can get a constant-expression value of the size of an array like this:

template <std::size_t N>
struct type_of_size
{
    typedef char type[N];
};

template <typename T, std::size_t Size>
typename type_of_size<Size>::type& sizeof_array_helper(T(&)[Size]);

#define sizeof_array(pArray) sizeof(sizeof_array_helper(pArray))

Explanation here. You basically encode the size of the array into the size of a type, then get the sizeof of that type, giving you:

char q[sizeof_array(p)];