c++ - 带有 packaged_task 的模板无法编译

Question

我正在玩 Antony Williams - C++ Concurrency in Action 一书中的示例 4.14，其中使用 std::packaged_task 和 std::thread 模拟 std::async。

为什么当我取消注释行时此代码无法编译以及如何重写模板以使其工作？

#include <iostream>
#include <future>
#include <thread>
#include <string>

void func_string(const std::string &x) {}

void func_int(int x) {}

template <typename F, typename A>
std::future<typename std::result_of<F(A&&)>::type> spawn_task(F &&f, A &&a) {
  typedef typename std::result_of<F(A&&)>::type result_type;
  std::packaged_task<result_type(A&&)> task(std::move(f));
  std::future<result_type> res(task.get_future());
  std::thread t(std::move(task), std::move(a));
  t.detach();
  return res;
}


int main () {
  std::string str = "abc";

  // auto res1 = spawn_task(func_string, str);
  // res1.get();
  auto res2 = spawn_task(func_int, 10);
  res2.get();

  return 0;
}

编译错误:

nnovzver@archer /tmp $ clang++ -std=c++11 -lpthread temp.cpp && ./a.out
In file included from temp.cpp:2:
In file included from /usr/bin/../lib64/gcc/x86_64-unknown-linux-gnu/4.8.2/../../../../include/c++/4.8.2/future:38:
/usr/bin/../lib64/gcc/x86_64-unknown-linux-gnu/4.8.2/../../../../include/c++/4.8.2/functional:1697:56: error: no type
      named 'type' in 'std::result_of<std::packaged_task<void (std::basic_string<char> &)> (std::basic_string<char>)>'
      typedef typename result_of<_Callable(_Args...)>::type result_type;
              ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~~
/usr/bin/../lib64/gcc/x86_64-unknown-linux-gnu/4.8.2/../../../../include/c++/4.8.2/thread:135:41: note: in instantiation
      of template class 'std::_Bind_simple<std::packaged_task<void (std::basic_string<char> &)>
      (std::basic_string<char>)>' requested here
        _M_start_thread(_M_make_routine(std::__bind_simple(
                                        ^
temp.cpp:15:15: note: in instantiation of function template specialization 'std::thread::thread<std::packaged_task<void
      (std::basic_string<char> &)>, std::basic_string<char> >' requested here
  std::thread t(std::move(task), std::move(a));
              ^
temp.cpp:24:15: note: in instantiation of function template specialization 'spawn_task<void (&)(const
      std::basic_string<char> &), std::basic_string<char> &>' requested here
  auto res1 = spawn_task(func_string, str);
              ^
In file included from temp.cpp:2:
In file included from /usr/bin/../lib64/gcc/x86_64-unknown-linux-gnu/4.8.2/../../../../include/c++/4.8.2/future:38:
/usr/bin/../lib64/gcc/x86_64-unknown-linux-gnu/4.8.2/../../../../include/c++/4.8.2/functional:1726:50: error: no type
      named 'type' in 'std::result_of<std::packaged_task<void (std::basic_string<char> &)> (std::basic_string<char>)>'
        typename result_of<_Callable(_Args...)>::type
        ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~~
2 errors generated.

Answer 1

如果您查看 documentation异步，你想要模拟的特性，它会衰减所有 future 的模板参数。它是有效的，因为它像这样工作得更好，如图所示 here :

template <typename T_>
using decay_t = typename std::decay<T_>::type;
template< class T >
using result_of_t = typename std::result_of<T>::type;

template <typename F, typename A>
std::future<result_of_t<decay_t<F>(decay_t<A>)>> spawn_task(F &&f, A &&a) {
    using result_t = result_of_t<decay_t<F>(decay_t<A>)>;
    std::packaged_task< result_t(decay_t<A>)> task(std::forward<F>(f));
    auto res = task.get_future();
    std::thread t(std::move(task), std::forward<A>(a));
    t.detach();
    return res;
}