[英]Why am I getting no matching function call error?
我试图调用一个指向成员函数的指针,但不知道为什么它不起作用。 错误如下:
prog.cpp:在函数
'int main()'
:
prog.cpp:16:19:错误:没有匹配的函数可调用'foo(A, int (A::*)())'
prog.cpp:16:19:注意:候选人是:
prog.cpp:4:6:注意:template<class A, class B> decltype ((declval<A>)().*(declval<B>)()) foo(A&&, B&&)
prog.cpp:4:6:注意:模板参数推导/替换失败:
prog.cpp:代替'template<class A, class B> decltype ((declval<A>)().*(declval<B>)()) foo(A&&, B&&) [with A = A; B = int (A::*)()]'
'template<class A, class B> decltype ((declval<A>)().*(declval<B>)()) foo(A&&, B&&) [with A = A; B = int (A::*)()]'
:
prog.cpp:16:19:从这里开始
prog.cpp:4:6:错误:无效使用非静态成员函数
#include <utility>
template <typename A, typename B>
auto foo(A&& a, B&& b) -> decltype(std::declval<A>().*(std::declval<B>()))
{
return (std::forward<A>(a)).*(std::forward<B>(b));
}
struct A
{
int f() { return 0; }
};
int main()
{
foo(A(), &A::f);
}
更新:
我更改了代码以响应评论/答案,这就是我所拥有的:
template <typename A, typename B>
auto foo(A&& a, B&& b) -> decltype((std::declval<A>().*(std::declval<B>()))())
{
return (std::forward<A>(a)).*(std::forward<B>(b))();
}
但是,当我称之为它时,它正在返回:
prog.cpp:在
'decltype ((declval<A>)().*(declval<B>)()()) foo(A&&, B&&) [with A = A; B = int (A::*)(); decltype ((declval<A>)().*(declval<B>)()()) = int]'
实例中'decltype ((declval<A>)().*(declval<B>)()()) foo(A&&, B&&) [with A = A; B = int (A::*)(); decltype ((declval<A>)().*(declval<B>)()()) = int]'
'decltype ((declval<A>)().*(declval<B>)()()) foo(A&&, B&&) [with A = A; B = int (A::*)(); decltype ((declval<A>)().*(declval<B>)()()) = int]'
'decltype ((declval<A>)().*(declval<B>)()()) foo(A&&, B&&) [with A = A; B = int (A::*)(); decltype ((declval<A>)().*(declval<B>)()()) = int]'
:
prog.cpp:16:19:从这里开始
prog.cpp:6:55:错误:必须使用'.*'
或'->*'
来调用指针成员函数
'std::forward<int (A::*)()>((* & b)) (...)', eg '(... ->* std::forward<int (A::*)()>((* & b))) (...)'
prog.cpp:在函数'decltype ((declval<A>)().*(declval<B>)()()) foo(A&&, B&&) [with A = A; B = int (A::*)(); decltype ((declval<A>)().*(declval<B>)()()) = int]'
'decltype ((declval<A>)().*(declval<B>)()()) foo(A&&, B&&) [with A = A; B = int (A::*)(); decltype ((declval<A>)().*(declval<B>)()()) = int]'
'decltype ((declval<A>)().*(declval<B>)()()) foo(A&&, B&&) [with A = A; B = int (A::*)(); decltype ((declval<A>)().*(declval<B>)()()) = int]'
:
prog.cpp:7:1:警告:控制权到达非无效函数的结尾[-Wreturn-type]
我在做什么错,我该如何解决?
比您编写的代码要简单一些。 而且您忘记了调用函数:
template <typename A, typename B>
auto foo(A&& a, B&& b) -> decltype((a.*b)())
{
return ((std::forward<A>(a)).*(std::forward<B>(b)))();
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.