Implicit conversion between template type

Question

I created two types : bool_t and number_t and I wanted to convert one into another (in the two ways). However, I got some issues to convert the bool_t into a number_t.

Basically what I want to do is that (but it does not compile) :

template<bool v>
struct bool_t {
    template<template<int> typename T>
    operator T<v ? 1 : 0>() {
        return {};
    }
};

template<int N>
struct number_t {
template<int n1, int n2>
friend number_t<n1 + n2> operator+(number_t<n1>, number_t<n2>) {
  return {};
}
};


int main() {
    number_t<0>{} + bool_t<0>{};   
}

and the error is :

prog.cc:19:19: error: invalid operands to binary expression ('number_t<0>' and 'bool_t<0>')
    number_t<0>{} + bool_t<0>{};   
    ~~~~~~~~~~~~~ ^ ~~~~~~~~~~~
prog.cc:12:26: note: candidate template ignored: could not match 'number_t' against 'bool_t'
friend number_t<n1 + n2> operator+(number_t<n1>, number_t<n2>) {
                         ^
1 error generated.

How to solve this problem?

Answer 1

User-defined conversions are never considered when attempting to match up function argument types with function parameter types for template argument deduction. So this issue is a somewhat more complicated version of this sort of error:

template <int> struct X {};
struct Y {
    operator X<2> () const { return {}; }
};
template <int N>
void f(X<N>) {}

int main()
{
    Y y;
    f(y); // Error: Cannot deduce template argument.
}

Since it seems you're making something along the lines of a template meta-programming library, perhaps you could define a custom mechanism for converting types in your library "to a template"?

#include <type_traits>

template<bool v>
struct bool_t {
    // (Add some appropriate SFINAE.)
    template<template<int> typename T>
    constexpr T<v ? 1 : 0> convert_template() const {
        return {};
    }
};

template<typename T, template<int> class TT>
struct type_specializes : std::false_type {};

template<template<int> class TT, int N>
struct type_specializes<TT<N>, TT> : std::true_type {};

template<int N>
struct number_t {
    // Allow "conversion" to my own template:
    template<template<int> typename T>
    constexpr std::enable_if_t<type_specializes<number_t, T>::value, number_t>
    convert_template() const { return {}; }

private:
    // Used only in decltype; no definition needed.
    template<int n1, int n2>
    static number_t<n1 + n2> sum_impl(number_t<n1>, number_t<n2>);

    template<typename T1, typename T2>
    friend auto operator+(T1&& x, T2&& y)
        -> decltype(number_t::sum_impl(
             x.template convert_template<number_t>(),
             y.template convert_template<number_t>()))
    { return {}; }
};

int main() {
    number_t<0>{} + bool_t<0>{};   
}

If you want to allow both operands to be bool_t specializations, the operator+ will need to be a proper visible namespace member; you can add more SFINAE checks to it if appropriate.

Answer 2

Your issue boils down to that the language allows at most one implicit conversion.

One solution would then be to make a function accepting bool_t as well:

number_t<n1 + n2> operator+(number_t<n1>, bool_t<n2>) {
  //return something
}