[英]std::bind and variadic template function
這有可能嗎?
#include <iostream>
#include <functional>
enum class Enum {a, b, c };
class Dispatch {
public:
void check(uint16_t) { std::cout << "check 16\n"; }
void check(uint32_t) { std::cout << "check 32\n"; }
void check(uint64_t) { std::cout << "check 64\n"; }
template<Enum E, typename... A>
void event(A&&... args) {
tag_event(Tag<E>(), std::forward<A>(args)...);
}
private:
template<Enum E> struct Tag {};
void tag_event(Tag<Enum::a>, uint16_t) { std::cout << "a\n"; }
void tag_event(Tag<Enum::b>, uint16_t) { std::cout << "b\n"; }
void tag_event(Tag<Enum::c>, uint16_t) { std::cout << "c\n"; }
};
void exec(std::function<void()>&& func) { func(); }
int main() {
Dispatch d;
// all good
exec(std::bind(static_cast<void(Dispatch::*)(uint16_t)>(&Dispatch::check), &d, uint16_t()));
exec(std::bind(static_cast<void(Dispatch::*)(uint32_t)>(&Dispatch::check), &d, uint32_t()));
exec(std::bind(static_cast<void(Dispatch::*)(uint64_t)>(&Dispatch::check), &d, uint64_t()));
// all good
d.event<Enum::a>(uint16_t());
d.event<Enum::b>(uint16_t());
d.event<Enum::c>(uint16_t());
// but how do we bind an event<> call?
exec(std::bind(static_cast<void(Dispatch::*)(uint16_t)>(&Dispatch::event<Enum::a>), &d, uint16_t()));
}
因此,我試圖將調用綁定到可變參數模板方法,但收到以下編譯器錯誤...
In function 'int main()':
42:86: error: no matches converting function 'event' to type 'void (class Dispatch::*)(uint16_t) {aka void (class Dispatch::*)(short unsigned int)}'
13:10: note: candidate is: template<Enum E, class ... A> void Dispatch::event(A&& ...)
除了公開所有標簽重載以外,還有什么建議嗎?
我建議按照注釋中的建議通過lambda函數。
無論如何,如果您想傳遞給std::bind()
,對我來說,可能的解決方案是
// ..................................................VVVVVVVV <-- ad this
exec(std::bind(static_cast<void(Dispatch::*)(uint16_t const &)>
(&Dispatch::event<Enum::a, uint16_t const &>), &d, uint16_t()));
// ...........................^^^^^^^^^^^^^^^^^^ <-- and this
我的意思是:您必須選擇event()
方法,該方法還應說明接收到的類型; 我建議與您的event()
方法的通用引用簽名兼容的uint16_t const &
(而不是uint16_t
)(我想其他組合也是可行的,但是對於uint16_t
激活move語義...我認為這是多余的)。
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