什么是std :: move和unique_ptr :: reset之间的区别？ [英] what are the differences between std::move and unique_ptr::reset?

问题描述

std :: unique_ptr s p1 和 p2 ， std :: move（）和 std :: unique_ptr :: reset（）？ / p>

For std::unique_ptrs p1 and p2, what are differences between std::move() and std::unique_ptr::reset()?

p1 = std::move(p2);

p1.reset(p2.release());

推荐答案

分配给[unique.ptr.single.assign] / 2：

The answer should be obvious from the standard's specification of move assignment in [unique.ptr.single.assign]/2:

效果：如果通过调用 reset（u.release（）），则 u $ c>后面是 std :: forward< D>（u.get_deleter（））的分配。

Effects: Transfers ownership from u to *this as if by calling reset(u.release()) followed by an assignment from std::forward<D>(u.get_deleter()).

清除移动赋值不等于 reset（u.release（）），因为它有额外的功能。

Clearly move assignment is not the same as reset(u.release()) because it does something additional.

额外的效果很重要，没有它你可以得到未定义的行为与自定义删除：

The additional effect is important, without it you can get undefined behaviour with custom deleters:

#include <cstdlib>
#include <memory>

struct deleter
{
  bool use_free;
  template<typename T>
    void operator()(T* p) const
    {
      if (use_free)
      {
        p->~T();
        std::free(p);
      }
      else
        delete p;
    }
};

int main()
{
  std::unique_ptr<int, deleter> p1((int*)std::malloc(sizeof(int)), deleter{true});
  std::unique_ptr<int, deleter> p2;
  std::unique_ptr<int, deleter> p3;

  p2 = std::move(p1);  // OK

  p3.reset(p2.release());  // UNDEFINED BEHAVIOUR!
}

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