被移动对象是否需要被破坏？ [英] Are moved-from objects required to be destructed?

问题描述

如果我从 b 中移动构建 a ，仍然需要 破坏 b ，或者我可以不这样做吗？

可选< T> 模板。摘录：

 〜可选（）
 {
 if（初始化）
 {
 reinterpret_cast< T *>（data） - >〜T（）; 
} 
} 
 
可选（可选& o）：已初始化（o.initialized）
 {
 if（已初始化）
 {
 new（data）T（std :: move（* o））; // move from o.data 
 o.initialized = false; // o.data不会被破坏了！ 
} 
} 
  

当然，我可以替换bool 初始化使用三值枚举来区分初始化，非初始化和移出。

是的，仍然有必要销毁。

b 。从对象移动是一个有效的，构造的对象。在某些情况下，它甚至可能拥有仍然需要处理的资源。在诸如你显示的通用代码中， T 甚至可能没有移动构造函数。在这种情况下，您可以调用复制构造函数。所以你绝对不能假设〜T（）是一个无操作，可以省略。

If I move-construct a from b, is it still necessary to destruct b, or can I get away without doing so?

This question crossed my mind during the implementation of an optional<T> template. Excerpt:

~optional()
{
    if (initialized)
    {
        reinterpret_cast<T*>(data)->~T();
    }
}

optional(optional&& o) : initialized(o.initialized)
{
    if (initialized)
    {
        new(data) T(std::move(*o));   // move from o.data
        o.initialized = false;        // o.data won't be destructed anymore!
    }
}

Of course, I could just replace the bool initialized with a three-valued enumeration that distinguishes between initialized, non-initialized and moved-from. I just want to know if this is strictly necessary.

解决方案

Yes, it is still necessary to destruct b. A moved from object is a valid, constructed object. In some cases, it may even hold resources that still need to be disposed of. In generic code such as you show, T may not even have a move constructor. You may invoke a copy constructor instead in this case. So you can definitely not assume that ~T() is a no-op and can be elided.

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