c ++ 11:模板化包装函数 [英] c++11: Templated wrapper function
问题描述
我尝试创建一个通用包装函数,该函数将任何函数作为参数以及它们的参数。就像 std :: thread
构造函数一样。
I try to create a general wrapper function which takes any function as argument and also their parameters. Just something like the std::thread
constructor.
我当前的代码是:
#include <iostream>
using namespace std;
template<typename FUNCTION, typename... ARGS>
void wrapper(FUNCTION&& func, ARGS&&... args)
{
cout << "WRAPPER: BEFORE" << endl;
auto res = func(args...);
cout << "WRAPPER: AFTER" << endl;
//return res;
}
int dummy(int a, int b)
{
cout << a << '+' << b << '=' << (a + b) << endl;
return a + b;
}
int main(void)
{
dummy(3, 4);
wrapper(dummy, 3, 4);
}
包装函数本身可以工作。它使用给定的参数调用给定的函数对象( std :: function
,函子或仅是正常函数)。但是我也想返回它的返回值。
The wrapper function itself works. It calls the given function object (std::function
, functor or just a "normal" function) with the given arguments. But i also like to return its return value.
这应该与删除的 return
语句一起工作,但不幸的是我不知道如何声明包装函数的返回类型。
This should work with the removed return
-statement, but unfortunately i dont know how to declare the wrapper functions return type.
我尝试了很多事情(例如,使用 decltype
),但没有任何效果。现在的问题是,我如何运行以下代码?
I tried many things (e.g. with decltype
), but nothing worked. My question is now, how i get the following code running?
#include <iostream>
template<typename FUNCTION, typename... ARGS>
??? wrapper(FUNCTION&& func, ARGS&&... args)
{
cout << "WRAPPER: BEFORE" << endl;
auto res = func(args...);
cout << "WRAPPER: AFTER" << endl;
return res;
}
int dummy(int a, int b)
{
cout << a << '+' << b << '=' << (a + b) << endl;
return a + b;
}
int main(void)
{
dummy(3, 4);
cout << "WRAPPERS RES IS: " << wrapper(dummy, 3, 4) << endl;
}
我认为代码应该可以工作,但除外???
。
I think the code should work, except for the ???
.
感谢您的任何想法
感谢
Kevin
Regards Kevin
推荐答案
使用 std :: result_of
:
template <typename F, typename ...Args>
typename std::result_of<F &&(Args &&...)>::type wrapper(F && f, Args &&... args)
{
return std::forward<F>(f)(std::forward<Args>(args)...);
}
在C ++ 14中,您可以使用 result_of_t
别名:
In C++14 you can use the result_of_t
alias:
template <typename F, typename ...Args>
std::result_of_t<F &&(Args &&...)> wrapper(F && f, Args &&... args)
{
return std::forward<F>(f)(std::forward<Args>(args)...);
}
或者您可以使用返回类型推导:
Or you can use return type deduction:
template <typename F, typename ...Args>
decltype(auto) wrapper(F && f, Args &&... args)
{
std::cout << "before\n";
auto && res = std::forward<F>(f)(std::forward<Args>(args)...);
std::cout << "after\n";
return res;
}
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