c++ - 使用异步调用模板函数

Question

当函数采用模板参数时，需要异步运行一个函数。下面的代码无法编译，有什么帮助吗？

 template<typename T>
 void say(int n, T t) { 
    cout << " say: " << n << " " << t << endl;
 }

 template<typename F, typename... Ts>
 inline auto reallyAsync(F&& f, Ts&&... params){
   return std::async(
       std::launch::async,
       std::forward<F>(f),
       std::forward<Ts>(params)...);
 }

 int main() {
   int n = 10; float x = 100;
   say(n, x); // works
   reallyAsync(&say, n, x) ; // does not work
 }

Answer 1

say 是一个函数模板，你不能获取一个函数模板的地址，因为它还不是一个函数(见评论):

int main() {
   int n = 10; float x = 100;
   say(n, x);                // works because of template argument deduction
   reallyAsync(&say, n, x);  //fails because say isn't a resolved  function.
 }

但是，您可以传递 say 的实例化:

 int main() {
   int n = 10; float x = 100;
   say(n, x);                     // works
   reallyAsync(&say<decltype(x)>, n, x);
 }

输出:

say: 10 100
say: 10 100

Live Example