为什么参数包不扩展为正确的类型?

Why doesn't parameter pack expand to correct type?

我有一段有点做作的代码:

#include <functional>

template <typename... ARGs>
auto construct1(std::function<void(ARGs...)> p, const ARGs &...args) {}

template < typename... ARGs>
auto construct2(std::function<void(int)>     p, const ARGs &...args) {}

int main() {
    auto p = [](int) {};
    construct1<int>(p, 0);
    construct2<int>(p, 0);
    return 0;
}

为什么编译器在第一种情况下努力解决 ARGs... = { int }?如果我通过在签名中使用 std::function<void(int)> 来协助编译器(第二种情况),代码可以正常编译。在这两种情况下,编译器都可以毫不费力地推断出 const ARGs&... 应该是 const int&。使用 C++17.

海湾合作委员会:

main.cpp: In function ‘int main()’:
main.cpp:11:25: error: no matching function for call to ‘construct1<int>(main()::<lambda(int)>&, int)’
   11 |     construct1<int>(p, 0);
      |                         ^
main.cpp:4:6: note: candidate: ‘template<class ... ARGs> auto construct1(std::function<void(ARGs ...)>, const ARGs& ...)’
    4 | auto construct1(std::function<void(ARGs...)> p, const ARGs &...args) {}
      |      ^~~~~~~~~~
main.cpp:4:6: note:   template argument deduction/substitution failed:
main.cpp:11:25: note:   ‘main()::<lambda(int)>’ is not derived from ‘std::function<void(ARGs ...)>’
   11 |     construct1<int>(p, 0);
      |  

叮当声:

main.cpp:11:5: error: no matching function for call to 'construct1'
    construct1<int>(p, 0);
    ^~~~~~~~~~~~~~~
main.cpp:4:6: note: candidate template ignored: could not match 'function<void (int, type-parameter-0-0...)>' against '(lambda at main.cpp:10:14)'
auto construct1(std::function<void(ARGs...)> p, const ARGs &...args) {}
     ^
1 error generated.

问题是,construct1 正在使用 std::function,但您正在传递 lambda。当您将参数设置为 std::function<void(ARGs...)>template argument deduction is performed to deduce ARGs on the function parameter p (even template parameter pack is explicitly specified with template arguments 类型时,如果有其他参数,它可能会通过模板参数推导进行扩展),这会失败,因为在推导中不会考虑隐式转换。

Type deduction does not consider implicit conversions (other than type adjustments listed above): that's the job for overload resolution, which happens later.

您可以使用std::type_identity (since C++20) to exclude p from deduction (see non-deduced contexts),例如

template <typename... ARGs>
auto construct1(std::function<void(std::type_identity_t<ARGs>...)> p, const ARGs &...args) {}
//                                 ^^^^^^^^^^^^^^^^^^^^^    ^

LIVE


PS:在 C++20 之前,您可以轻松地赢得 type_identity 为:

template< class T >
struct type_identity {
    using type = T;
};
template< class T >
using type_identity_t = typename type_identity<T>::type;