绑定函数的第一个参数而不知道它的参数 [英] bind first argument of function without knowing its arity
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问题描述
我想要一个函数 BindFirst
,它绑定函数的第一个参数,而不必使用std明确地知道/声明函数的arity: :占位符。我希望客户端代码看起来像这样。
#include< functional>
#include< iostream>
void print2(int a,int b)
{
std :: cout<< a<<的std :: ENDL;
std :: cout<< b<的std :: ENDL;
}
void print3(int a,int b,int c)
{
std :: cout<< a<<的std :: ENDL;
std :: cout<< b<的std :: ENDL;
std :: cout<< c<的std :: ENDL;
}
int main()
{
auto f = BindFirst(print2,1); // std :: bind(print2,1,std :: placeholders :: _ 1);
auto g = BindFirst(print3,1); // std :: bind(print3,1,std :: placeholders :: _ 1,std :: placeholders :: _ 2);
f(2);
g(2,3);
}
任何想法 BindFirst
可以实现吗?
解决方案
在$ C $ 11中:
#include< type_traits>
#include< utility>
模板< typename F,typename T>
结构绑定
{
F f; T t;
模板< typename ... Args>
auto operator()(Args&& ... args)const
- > decltype(f(t,std :: forward< Args>(args)...))
{
return f(t,std :: forward< Args>(args)...);
}
};
模板< typename F,typename T>
binder< typename std :: decay< F> :: type
,typename std :: decay< T> :: type> BindFirst(F& f,T& t)
{
return {std :: forward F(f),std :: forward< T>(t)};
}
在C ++ 14中:
#include< utility>
模板< typename F,typename T>
auto BindFirst(F&& f,T& t)t
{
return [f = std :: forward F(f),t = std :: forward< T>]
(auto& ... args)
{return f(t,std :: forward< decltype(args)>(args)...); };
}
I'd like to have a function BindFirst
that binds the first argument of a function without me having to explicitly know/state the arity of the function by using std::placeholders. I'd like the client code to look something like that.
#include <functional>
#include <iostream>
void print2(int a, int b)
{
std::cout << a << std::endl;
std::cout << b << std::endl;
}
void print3(int a, int b, int c)
{
std::cout << a << std::endl;
std::cout << b << std::endl;
std::cout << c << std::endl;
}
int main()
{
auto f = BindFirst(print2, 1); // std::bind(print2, 1, std::placeholders::_1);
auto g = BindFirst(print3, 1); // std::bind(print3, 1, std::placeholders::_1, std::placeholders::_2);
f(2);
g(2,3);
}
Any ideas how BindFirst
could be implemented?
解决方案
In C++11:
#include <type_traits>
#include <utility>
template <typename F, typename T>
struct binder
{
F f; T t;
template <typename... Args>
auto operator()(Args&&... args) const
-> decltype(f(t, std::forward<Args>(args)...))
{
return f(t, std::forward<Args>(args)...);
}
};
template <typename F, typename T>
binder<typename std::decay<F>::type
, typename std::decay<T>::type> BindFirst(F&& f, T&& t)
{
return { std::forward<F>(f), std::forward<T>(t) };
}
In C++14:
#include <utility>
template <typename F, typename T>
auto BindFirst(F&& f, T&& t)
{
return [f = std::forward<F>(f), t = std::forward<T>(t)]
(auto&&... args)
{ return f(t, std::forward<decltype(args)>(args)...); };
}
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