C ++ 17模板推论指南不用于空参数集吗?

Consider the following reduced example which can also be viewed at https://godbolt.org/g/Et56cm:

#include <utility>

template <class T> struct success
  T value;
  constexpr success(T &&v)
      : value(std::move(v))
  constexpr success(const T &v)
      : value(v)
template <> struct success<void>
template <class T> success(T /*unused*/)->success<T>;

int main(void)
    auto a = success{5};        // works
    auto b = success{};         // works
    auto c = success{"hello"};  // works
    auto d = success(5);        // works
    auto e = success();         // FAILS!
    auto f = success("hello");  // works
    static_assert(std::is_same<decltype(a), success<int>>::value, "");
    static_assert(std::is_same<decltype(b), success<void>>::value, "");
    static_assert(std::is_same<decltype(c), success<const char *>>::value, "");
    static_assert(std::is_same<decltype(d), success<int>>::value, "");
    static_assert(std::is_same<decltype(e), success<void>>::value, "");
    static_assert(std::is_same<decltype(f), success<const char *>>::value, "");
    return 0;

What is surprising to me is that success() does not compile, yet success{} does. I have provided the template deduction guide success() -> success<void>, so I would have thought that success() would work as well.

这是C ++ 17标准中的预期行为,还是我缺少什么?

  • 自娱自乐
    自娱自乐 回复

    This is a gcc bug (just filed 81486). When deducing success(), we synthesize an overload set which consists of:

    // from the constructors
    template <class T> success<T> foo(T&& ); // looks like a forwarding reference
                                             // but is really just an rvalue reference
    template <class T> success<T> foo(T const& );
    // from the deduction guides
    template <class T> success<T> foo(T ); // this one is a bit redundant
    success<void> foo();

    And determine the return type as if it were invoked as foo(), which certainly should give you a type of success<void>. That it doesn't is a bug.