std::bind 和右值引用

Question

让我们考虑以下代码：

class Widget{
};

int main(){
Widget w;
auto lambda = bind([](Widget&& ref){ return; }, std::move(w));

return 0;
}

并触发错误

no match for call to ‘(std::_Bind<main()::<lambda(Widget&&)>(Widget)>) ()’
     lambda();

我的问题是：为什么会出现错误？毕竟，我对右值引用进行了显式转换——我的意思是 std::move(w) 并且我通过右值引用进行论证——我的意思是 Widget&& ref.

怎么了？

而且下面的代码是有效的，更让我担心的是：

class Widget{
};

int main(){
Widget w;
auto lambda = bind([](Widget& ref){ return; }, std::move(w));

return 0;
}

Answer 1

如果您写下 std::bind 的示意性作用，可能会更清楚。

// C++14, you'll have to write a lot of boilerplate code for C++11
template <typename FuncT, typename ArgT>
auto
bind(FuncT&& func, ArgT&& arg)
{
  return
    [
      f = std::forward<FuncT>(func),
      a = std::forward<ArgT>(arg)
    ]() mutable { return f(a); };  // NB: a is an lvalue here
}

因为你可以多次调用std::bind给你的函数对象，它不能“用完”捕获的参数，所以它会作为左值引用传递。您将 bind 本身传递给右值这一事实仅意味着在初始化 a 的行上没有进行复制。

如果您尝试使用上面显示的原理图 bind 编译您的示例，您还会从编译器中获得更有用的错误消息。

main.cxx: In instantiation of ‘bind(FuncT&&, ArgT&&)::<lambda()> mutable [with FuncT = main()::<lambda(Widget&&)>; ArgT = Widget]’:
main.cxx:10:33:   required from ‘struct bind(FuncT&&, ArgT&&) [with FuncT = main()::<lambda(Widget&&)>; ArgT = Widget]::<lambda()>’
main.cxx:11:31:   required from ‘auto bind(FuncT&&, ArgT&&) [with FuncT = main()::<lambda(Widget&&)>; ArgT = Widget]’
main.cxx:18:59:   required from here
main.cxx:11:26: error: no match for call to ‘(main()::<lambda(Widget&&)>) (Widget&)’
    ]() mutable { return f(a); };  // NB: a is an lvalue here
                          ^
main.cxx:11:26: note: candidate: void (*)(Widget&&) <conversion>
main.cxx:11:26: note:   conversion of argument 2 would be ill-formed:
main.cxx:11:26: error: cannot bind ‘Widget’ lvalue to ‘Widget&&’
main.cxx:18:33: note: candidate: main()::<lambda(Widget&&)> <near match>
   auto lambda = bind([](Widget&&){ return; }, std::move(w));
                                 ^
main.cxx:18:33: note:   conversion of argument 1 would be ill-formed:
main.cxx:11:26: error: cannot bind ‘Widget’ lvalue to ‘Widget&&’
    ]() mutable { return f(a); };  // NB: a is an lvalue here

Answer 2

要让它工作，你需要这样写：

#include <functional>
#include <iostream>

class Widget{};

int main()
{
    Widget a;
    auto lf = [](Widget&& par){ };

    auto f = std::bind
    (
        lf,
        std::bind
        (
            std::move<Widget&>, a
        )
    );
    f();
    return 0;
}

我的编译器是gcc version 4.9.2 20141101 (Red Hat 4.9.2-1) (GCC)

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