我正在使用Antony Williams一书中的示例4.14 - C + + Concurrency in Action,其中使用std :: packaged_task和std :: thread模拟std :: async。
当我取消注释行以及如何重写模板以使其工作时,为什么此代码无法编译?
#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.
答案 0 :(得分:1)
首先std::packaged_task
包装任何 Callable 目标函数。 Callable的一个要求是它具有合适的返回值。 void
不是合适的返回值。为了测试目的,我更改了您的函数以返回int
:
int func_string(const std::string &x)
{
return 1;
}
int func_int(int x)
{
return 2;
}
其次,在我看来,您的实际功能签名与您在std::result_of<>
模板中指定的内容之间存在不匹配。具体来说,您的功能签名是:
int func_string(const std::string &x);
int func_int(int x);
但你传递的是:
std::result_of<F(A&&)>::type result_type
将其更改为:
std::result_of<F(A)>::type result_type
一切都在编译。
以下是完整的源代码(减去标题):
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();
}
答案 1 :(得分:1)
如果查看async的documentation,即要模拟的功能,它会衰减所有未来的模板参数。并且它有效,因为它像这样工作得更好,如展示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;
}