在std :: function中存储不可复制但可移动的对象 [英] Storing noncopyable but movable object in std::function

问题描述

假设我有一个函数s，它是不可复制但可移动的，我如何将它存储在std ::函数？即，如何使下面的代码编译？ （使用gcc 4.6）

Suppose I have a functor s, which is noncopyable but movable, how can I store it in a std::function? i.e, how to make the following code compile? (using gcc 4.6)

#include <functional>
#include <iostream>

struct S
{
  S() = default;
  S(S const&) = delete;
  S& operator=(S const&) = delete;
  S(S&&) { }
  void operator()() { }
};

std::function<void()> func;

void make_func()
{
  S s;
  func = std::bind(std::move(s));  // This won't compile
}

int main()
{
  make_func();
}

推荐答案

标准， std :: function 应该是可复制的。因此，您无法直接达到您想要的效果。

As far as I understand the standard, std::function is supposed to be copyable. Therefore, you cannot directly achieve what you want.

您可以使用一些自定义封装，它会像这样：

You can get away with some custom wrapper I'd guess, though. It would go something like this:

• 使你的包装包含 std :: shared_ptr 到实际函子;


• 当函数从右边构造一个包装器时，将函子移动到动态分配的内存中;


• 复制构造函数包装和析构函数由 shared_ptr copy-ctor / destructor;


• c $ c>为封装器解除引用到真实函数的智能指针，并委托 operator（）。



• make your wrapper contain an std::shared_ptr to the actual functor;
• when a wrapper is constructed from a functor rvalue, move the functor to dynamically allocated memory;
• copy constructor for the wrapper and destructor are simply handled by shared_ptr copy-ctor/destructor;
• operator() for the wrapper dereferences the smart pointer to the real functor and delegates to operator() on it.

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