std::bind 和/或 std::forward 的语义

Question

我发现以下代码无法编译非常令人困惑

#include <functional>

class Mountain {
public:
  Mountain() {}
  Mountain(const Mountain&) = delete;
  Mountain(Mountain&&) = delete;
  ~Mountain() {}
};

int main () {
  Mountain everest;
  // shouldn't the follwing rvalues be semantically equivalent?
  int i = ([](const Mountain& c) { return 1; })(everest);
  int j = (std::bind([](const Mountain& c) {return 1;},everest))();
  return 0;
}

编译错误是：

$ g++ -std=c++20 test.cpp -o test
In file included from test.cpp:1:
/usr/bin/../lib/gcc/x86_64-linux-gnu/10/../../../../include/c++/10/functional:486:26: error: no
      matching constructor for initialization of 'tuple<Mountain>'
        : _M_f(std::move(__f)), _M_bound_args(std::forward<_Args>(__args)...)
                                ^             ~~~~~~~~~~~~~~~~~~~~~~~~~~~
/usr/bin/../lib/gcc/x86_64-linux-gnu/10/../../../../include/c++/10/functional:788:14: note: in
      instantiation of function template specialization 'std::_Bind<(lambda at test.cpp:14:22)
      (Mountain)>::_Bind<Mountain &>' requested here
      return typename __helper_type::type(std::forward<_Func>(__f),
             ^
test.cpp:14:17: note: in instantiation of function template specialization 'std::bind<(lambda at
      test.cpp:14:22), Mountain &>' requested here
  int j = (std::bind([](const Mountain& c) {return 1;}, everest))();
                ^
...

因此即使 lambda 只想引用它，也std::bind偷偷地尝试复制everest。我是在摩擦一个没人关心的奇怪的边缘情况（例如，总是可以只用 lambda 捕获对 的引用everest）还是有理由？如果基本原理是 bind 在everest被销毁后保护我免于调用 lambda ，那么是否有不安全的 bind 版本不会这样做？

Answer 1

是的，参数std::bind将被复制（或移动）。

bind 的参数被复制或移动，除非包裹在std::ref或 中，否则永远不会通过引用传递std::cref。

您可以使用std::cref(或std::ref) 代替。例如

int j = (std::bind([](const Mountain& c) {return 1;}, std::cref(everest)))();
//                                                    ^^^^^^^^^^       ^

居住