Reputation: 4078
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?
Upvotes: 5
Views: 222
Reputation: 72271
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.
Upvotes: 3
Reputation: 10962
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
}
Upvotes: 2