使用 std：：async with template functions

Question

如何将模板函数传递给async？

代码如下:

//main.cpp
#include <future>
#include <vector>
#include <iostream>
#include <numeric>

int
main
    ()
{      
    std::vector<double> v(16,1);

    auto r0 =  std::async(std::launch::async,std::accumulate,v.begin(),v.end(),double(0.0));

    std::cout << r0.get() << std::endl;
    return 0;
}

以下是错误消息：

                                                                               ^
a.cpp:13:88: note: candidates are:
In file included from a.cpp:1:0:
/usr/include/c++/4.8/future:1523:5: note: template std::future::type> std::async(std::launch, _Fn&&, _Args&& ...)
     async(launch __policy, _Fn&& __fn, _Args&&... __args)
     ^
/usr/include/c++/4.8/future:1523:5: note:   template argument deduction/substitution failed:
a.cpp:13:88: note:   couldn't deduce template parameter ‘_Fn’
   auto r0 = std::async(std::launch::async,std::accumulate,v.begin(),v.end(),double(0.0));
                                                                                        ^
In file included from a.cpp:1:0:
/usr/include/c++/4.8/future:1543:5: note: template std::future::type> std::async(_Fn&&, _Args&& ...)
     async(_Fn&& __fn, _Args&&... __args)
     ^
/usr/include/c++/4.8/future:1543:5: note:   template argument deduction/substitution failed:
/usr/include/c++/4.8/future: In substitution of ‘template std::future::type> std::async(_Fn&&, _Args&& ...) [with _Fn = std::launch; _Args = {}]’:
a.cpp:13:88:   required from here
/usr/include/c++/4.8/future:1543:5: error: no type named ‘type’ in ‘class std::result_of’

Answer 1

您必须通过显式传递模板参数或使用static_cast来消除可能的实例化之间的歧义，因此：

auto r0 = std::async(std::launch::async
                     , &std::accumulate<decltype(v.begin()), double>
                     , v.begin()
                     , v.end()
                     , 0.0);

或：

auto r0 = std::async(std::launch::async
       , static_cast<double(*)(decltype(v.begin()), decltype(v.end()), double)>(&std::accumulate)
       , v.begin()
       , v.end()
       , 0.0);

Answer 2

问题是，要将第二个参数传递给 std：：async，编译器必须将表达式

正如PiotrS.的答案所说，你可以通过显式模板参数列表或使用强制转换来告诉编译器你想要哪个std：：accumulate，或者你可以只使用lambda表达式：

std::async(std::launch::async,[&] { return std::accumulate(v.begin(), v.end(), 0.0); });

在 lambda 的主体内部，编译器对对 std：：accumulate 的调用执行重载解析，因此它计算出要使用的 std：：accumulate。