How to make compiler infer the right template integer for me

Question

Consider this piece:

template <int N> void fill_with_magic(array<int, N>& whatever){
    for(int i = 0; i < N; i++){
        whatever[i] = magic(i, N);
    }
}

I call it by specific instance so for array of 3 I would have to do:

array<int, 3> some_array_of_3;
fill_with_magic<3>(some_array_of_3);

But do I really have to write <3>? Compiler already knows the array size so theoretically it could of infer the right instance based on that size. Can I really make it do that?

A call to fill_with_magic(some_array_of_3) or fill_with_magic<>(some_array_of_3) will work. — bku_drytt, Sep 19 '15 at 15:33
You must declare the non-type parameter with `size_t` for the deduction to work. If a conversion (int -> size_t for N) is involved, no deduction happens. — Johannes Schaub - litb, Sep 19 '15 at 15:33

score 6 · Accepted Answer · answered Sep 19 '15 at 15:35

The problem is the deduction of the argument: the second template argument of a std::array is not an int so the deduction doesn't take place because it requires a conversion.

You should define your method as

template <array<int, 0>::size_type N> void fill_with_magic(array<int, N>& whatever){
  for(int i = 0; i < N; i++){
    whatever[i] = magic(i, N);
  }
}

So that you are correctly expecting a array<int,0>::size_type as the argument, which will then be correctly deduced. Actually I think most of the implementation use size_t but this should be more portable (unless std::array<T, 0> gets a special specialization).

`size_type` is required to be `size_t`, so using it is portable as well as being simpler. — Alan Stokes, Sep 19 '15 at 15:38

How to make compiler infer the right template integer for me

1 Answers1