是什么导致这两个函数模板之间的重载失败? [英] What makes the overload fail between these two function templates?
问题描述
下面是一个非常简短的例子.
Below is the pretty short example.
#include <utility>
template<typename T, typename = void>
struct A {};
template<typename T, typename U>
void f(A<std::pair<T,U>>) {}
template<typename U>
void f(A<std::pair<int,U>, std::enable_if_t<std::is_same_v<int,U>>>) {}
int main() {
A<std::pair<int, int>> x;
f(x);
}
错误很明显
uffa.cpp: In function ‘int main()’:
uffa.cpp:22:4: error: call of overloaded ‘f(A<std::pair<int, int> >&)’ is ambiguous
22 | f(x);
| ~^~~
uffa.cpp:10:6: note: candidate: ‘void f(A<std::pair<_T1, _T2> >) [with T = int; U = int]’
10 | void f(A<std::pair<T,U>>) {}
| ^
uffa.cpp:18:6: note: candidate: ‘void f(A<std::pair<int, U>, typename std::enable_if<is_same_v<int, U>, void>::type>) [with U = int; typename std::enable_if<is_same_v<int, U>, void>::type = void]’
18 | void f(A<std::pair<int,U>, std::enable_if_t<std::is_same_v<int,U>>>) {}
| ^
但我不明白为什么在第二个重载中将 int
作为固定模板参数并没有使它更专业化.毕竟,如果我从中删除 , std::enable_if_t<std::is_same_v<int,U>>
,那么它是首选.
But I don't understand why having int
as a fixed template argument in the second overload doesn't make it more specialized. After all, if I remove , std::enable_if_t<std::is_same_v<int,U>>
from it, then it is preferred.
推荐答案
即使这是 语言律师,我将提供一个外行的解释.
Even though this is language-lawyer, I'm going to provide a layman explanation.
是的,第二个重载将 pair
的第一个参数固定为 int
,而第一个没有.
Yes, the second overload fixes the first parameter of pair
as int
, while the first one doesn't.
但是,另一方面,第一个重载将A
的第二个参数固定为void
,而第二个没有.
But, on the other hand, the first overload fixes the second parameter of A
as void
, while the second one doesn't.
你的功能等同于那些:
template <typename T, typename U>
void f(A<std::pair<T, U>, void>) {}
template <typename U>
void f(A<std::pair<int,U>, blah-blah<U>>) {}
所以他们中没有一个比另一个更专业.
So none of them is more specialized than the other.
如果您使用更传统的 SFINAE,代码将起作用:
The code will work if you use more a conventional SFINAE:
template<typename U, std::enable_if_t<std::is_same_v<U, int>, std::nullptr_t> = nullptr>
void f(A<std::pair<int,U>>) {}
或 C++20 概念:
Or C++20 concepts:
template <std::same_as<int> U>
void f(A<std::pair<int,U>>) {}
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