没有编译带有packaged_task的模板

Template with packaged_task not compile

本文关键字：task packaged 编译      更新时间：2023-10-16

我正在玩Antony Williams - c++ Concurrency in Action一书中的一个示例4.14，其中使用std::packaged_task和std::thread模拟std::async。

为什么这段代码不编译时，我取消注释行和如何重写模板，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.

如果您查看async的文档，您想要模拟的特性，它会衰减所有未来的模板参数。它是有效的，因为它像这样工作得更好，如下所示:

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;
}

首先，std::packaged_task包装任何Callable目标函数。Callable的一个要求是它有一个合适的返回值。void不是一个合适的返回值。我更改了您的函数以返回int用于测试目的:

int func_string(const std::string &x) 
{
    return 1;
}
int func_int(int x) 
{
    return 2;
}

其次，在我看来，您的实际函数签名与您在std::result_of<>模板中指定的内容之间存在不匹配。具体来说你的函数签名是:

int func_string(const std::string &x);
int func_int(int x);

但你正在通过:

std::result_of<F(A&&)>::type result_type

将其改为:

std::result_of<F(A)>::type result_type

一切都编译好了。

这是完整的源代码(减去头):

int func_string(const std::string &x) 
{
    return 1;
}
int func_int(int x) 
{
    return 2;
}
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();
}
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