带有可变模板的C++异步找不到正确的函数模板专用化

C++ async with variadic template can not find correct function template specialization

本文关键字：找不到 函数模板 专用 异步 C++      更新时间：2023-10-16

我有一个类，它有一个成员函数f，我用可变模板和forward包装它，以制作另一个成员功能rf（只需在f的末尾添加一个特定的参数，就可以做一些不同的事情）。然后，我通过用async包装rf来制作另一个成员函数async_rf，但它不起作用。我试图通过用额外的特定参数包装f来制作async_rf，并且它是有效的。

代码：

#include <future>         // std::async, std::future
#include <iostream>
class test {
public:
    void f(int tmp, bool reverse = 0)
    {
        std::cout << tmp << " | " << reverse << std::endl;
    }
    template<typename... Args>
    void rf(Args... args)
    {
        f(std::forward<Args>(args)..., 1);
    }
    template<typename... Args>
    std::future<void> async_rf(Args... args)
    {
        // doesn't work
        return std::async (&test::rf, this, std::forward<Args>(args)...);
        // work
        return std::async (&test::f, this, std::forward<Args>(args)..., 1);
    }
};

int main()
{
    test s;
    auto tmp = s.async_rf(10);
    tmp.get();
    return 0;
}

这是编译时的错误消息：

$ clang++ --version
clang version 3.6.1 (tags/RELEASE_361/final)
Target: x86_64-unknown-linux-gnu
Thread model: posix
$ clang++ -std=c++14 -Wall -lpthread src/test.cpp -o bin/test
src/test.cpp:23:16: error: no matching function for call to 'async'
        return std::async (&test::rf, this, std::forward<Args>(args)...);
               ^~~~~~~~~~
src/test.cpp:35:18: note: in instantiation of function template specialization 'test::async_rf<int>' requested here
    auto tmp = s.async_rf(10);
                 ^
/usr/bin/../lib64/gcc/x86_64-unknown-linux-gnu/5.1.0/../../../../include/c++/5.1.0/future:1723:5: note: candidate template
      ignored: couldn't infer template argument '_Fn'
    async(_Fn&& __fn, _Args&&... __args)
    ^
/usr/bin/../lib64/gcc/x86_64-unknown-linux-gnu/5.1.0/../../../../include/c++/5.1.0/future:1703:5: note: candidate template
      ignored: substitution failure [with _Fn = test *, _Args = <int>]: no type named 'type' in
      'std::result_of<test *(int)>'
    async(launch __policy, _Fn&& __fn, _Args&&... __args)
    ^
1 error generated.

（gcc）

$ g++ --version
g++ (GCC) 5.1.0
Copyright (C) 2015 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
$ g++ -std=c++14 -Wall -lpthread src/test.cpp -o bin/test
src/test.cpp: In instantiation of ‘std::future<void> test::async_rf(Args ...) [with Args = {int}]’:
src/test.cpp:35:29:   required from here
src/test.cpp:23:27: error: no matching function for call to ‘async(<unresolved overloaded function type>, test*, int)’
         return std::async (&test::rf, this, std::forward<Args>(args)...);
                           ^
In file included from src/test.cpp:1:0:
/usr/include/c++/5.1.0/future:1703:5: note: candidate: template<class _Fn, class ... _Args> std::future<typename std::result_of<_Functor(_ArgTypes ...)>::type> std::async(std::launch, _Fn&&, _Args&& ...)
     async(launch __policy, _Fn&& __fn, _Args&&... __args)
     ^
/usr/include/c++/5.1.0/future:1703:5: note:   template argument deduction/substitution failed:
src/test.cpp:23:27: note:   cannot convert ‘&((test*)this)->*test::rf’ (type ‘<unresolved overloaded function type>’) to type ‘std::launch’
         return std::async (&test::rf, this, std::forward<Args>(args)...);
                           ^
In file included from src/test.cpp:1:0:
/usr/include/c++/5.1.0/future:1723:5: note: candidate: template<class _Fn, class ... _Args> std::future<typename std::result_of<_Functor(_ArgTypes ...)>::type> std::async(_Fn&&, _Args&& ...)
     async(_Fn&& __fn, _Args&&... __args)
     ^
/usr/include/c++/5.1.0/future:1723:5: note:   template argument deduction/substitution failed:
src/test.cpp:23:27: note:   couldn't deduce template parameter ‘_Fn’
         return std::async (&test::rf, this, std::forward<Args>(args)...);
                           ^

有人能提供更多关于为什么它不起作用的细节吗？为什么编译器找不到正确的模板专用化？

要求编译器通过提供给async的参数来推断rf的专门化将需要它研究async的实现，这有点太多了。

只需自己指定模板参数：

return std::async (&test::rf<Args...>, this, std::forward<Args>(args)...);

顺便说一句，您可能希望在任何地方都将Args...更改为Args&&...，否则参数将按值传递。

由于rf是一个函数模板，您不能简单地将其传递给std::async。首先，您必须指定它的模板参数，即简单的Args...。所以你的函数看起来像：

template<typename... Args>
std::future<void> async_rf(Args... args)
{
    // doesn't work
    return std::async (&test::rf<Args...>,this,std::forward<Args>(args)...);
    // work
    return std::async (&test::f, this, std::forward<Args>(args)..., 1);
}

尽管在该函数中有两个返回语句可能不是您想要做的。

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