std::bind 和/或 std::forward

问题描述

我觉得下面的代码无法编译很令人困惑

#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))();
                ^
...

所以 std::bind 偷偷地尝试复制 everest，即使 lambda 只想要对它的引用。我是在反对一个没人关心的奇怪的边缘情况（例如，总是可以通过 lambda 捕获对 everest 的引用）还是有理由？如果基本原理是在 everest 被销毁后 bind 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？

解决方法

暂无找到可以解决该程序问题的有效方法，小编努力寻找整理中！

如果你已经找到好的解决方法，欢迎将解决方案带上本链接一起发送给小编。

小编邮箱:dio#foxmail.com (将#修改为@）