c++ - 转发参数的可变参数列表

Question

下面的代码都按预期编译并执行，它们有什么不同吗？

template<typename T, typename ...U>
auto time_function(T&& func, U&& ...args)
{
   std::cout << "timing" << std::endl;
   auto val = std::forward<T>(func)(std::forward<U...>(args...));
   std::cout << "timing over" << std::endl;
   return val;
}

template<typename T, typename ...U>
auto time_function(T&& func, U&& ...args)
{
   std::cout << "timing" << std::endl;
   auto val = std::forward<T>(func)(std::forward<U>(args)...);
   std::cout << "timing over" << std::endl;
   return val;
}

查看SO How would one call std::forward on all arguments in a variadic function? ，第二个似乎是推荐的，但是第一个不是做同样的事情吗？

Answer 1

它们不一样。在数量为args的情况下它们是相同的。为1或0。否则将无法编译，请考虑..

#include <iostream>
using namespace std;
template<typename T, typename ...U>
auto time_function_1(T&& func, U&& ...args)
{

    std::cout<<"timing"<<std::endl;
    auto val = std::forward<T>(func)(std::forward<U...>(args...));
    std::cout<<"timing over"<<std::endl;
    return val;
}

    template<typename T, typename ...U>
auto time_function_2(T&& func, U&& ...args)
{

    std::cout<<"timing"<<std::endl;
    auto val = std::forward<T>(func)(std::forward<U>(args)...);
    std::cout<<"timing over"<<std::endl;
    return val;
}



int f (int){return 0;}

int y (int,int){return 0;}

int main() {
    time_function_1(f,1);
    time_function_2(f,1);

    time_function_1(y,1,2); // fail
    time_function_2(y,1,2);
    return 0;
}

Demo

失败案例std::forward<U...>(args...)扩展到forward<int, int>(int&, int&)并且将无法编译。

std::forward<U>(args)...扩展到std::forward<int>(int&),std::forward<int>(int&)