C++ template parameter deduction for std::array with non size_t integer

Question

I'm trying to adapt the solution presented in Avoiding struct in variadic template function to my need. However, I can't understand the the behavior of G++. Consider the following function:

 template <typename T, unsigned Size>
 int nextline(const typename std::array<T, Size> ar) {
    return 0;
 }

Then the call

 nextline(std::array<int, 2> { 1,0 });

doesn't match with GCC complaining with

eslong.cpp: In function ‘int main()’:
eslong.cpp:10:38: error: no matching function for call to ‘nextline(std::array<int, 2ul>)’
   nextline(std::array<int, 2> { 1,0 });
                                      ^
eslong.cpp:10:38: note: candidate is:
eslong.cpp:4:5: note: template<class T, unsigned int Size> int nextline(std::array<T, Size>)
 int nextline(const typename std::array<T, Size> ar) {
     ^
eslong.cpp:4:5: note:   template argument deduction/substitution failed:
eslong.cpp:10:38: note:   mismatched types ‘unsigned int’ and ‘#‘integer_cst’ not supported by dump_type#<type error>’
   nextline(std::array<int, 2> { 1,0 });
                                      ^
eslong.cpp:10:38: note:   ‘std::array<int, 2ul>’ is not derived from ‘std::array<T, Size>’

However it matches if I changes unsigned Size to unsigned long Size or size_t. I'm not sure to understand what's happening here. Isn't the Size parameter in the call to std::array<T, Size> converted to a size_t ?

Answer 1

std::array is templated as:

template<class T, std::size_t N > struct array;

while the size N is required to be the type size_t. But in your function, you are passing an unsigned (int) which cannot be interpreted as size_t. According to SFINAE If a template cannot be deduced, it does not exist, thus your templated function does not exist.

It is NOT the problem with the call line, but your declaration of your function template. To correct this, use the correct type:

template <typename T, size_t Size>
int nextline(const typename std::array<T, Size> ar) {
  return 0;
 }

In this case, even you use:

nextline(std::array<int, 2ul> { 1,0 });

It still works because it can be deduced and casted.

---

Additional explanation by dyp:

[temp.deduct.type]/17 for non-type template parameters that requires the type of the deduced thing (template-argument) to be of the same type as the template-parameter it is deduced for.

Answer 2

Your literal 2 is interpreted as an unsigned long, but you are declaring the Size template to be an unsigned int. Just use this instead:

template <typename T, size_t Size>