CODEWITHSUNDEEP
c++c++11templatestemplate-argument-deduction
rustyx
rustyx

Reputation: 85351

Template function argument deduction with an implicit conversion

I understand that template function argument deduction does not take implicit conversions into account.

So this code doesn't compile:

#include <iostream>

template<class T>
struct A {};
struct B : public A<int> {};
struct C {
  operator B() { return {}; }
};

template<class X>
void test(A<X> arg1, A<X> arg2) {
  std::cout << "ok1";
}

int main() {
  B b;
  C c;
  test(b, c);  // error: no matching function for call to 'test'
}

What I don't understand is how adding an extra level of indirection with an identity typedef somehow makes it work:

#include <iostream>

template<class T>
struct A {};
struct B : public A<int> {};
struct C {
  operator B() { return {}; }
};

template<typename U> struct magic { typedef U type; };

template<class T> using magic_t = typename magic<T>::type;

template<class X>
void test(A<X> arg1, A<X> arg2) {
  std::cout << "ok1";
}

template<class X>
void test(A<X> arg3, magic_t<A<X>> arg4) {
  std::cout << "ok2";
}

int main() {
  B b;
  C c;
  test(b, c);  // prints "ok2"
}

Live demo on Godbolt

How does magic_t<A<X>> end up matching C?

Upvotes: 3

Views: 1050

Answers (1)

Igor Tandetnik
Igor Tandetnik

Reputation: 52471

The second parameter becomes non-deduced context and doesn't participate in template argument deduction. X is then successfully deduced from the first argument.

Upvotes: 6

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