什么时候允许c ++ 11中的类型被记忆? [英] When is a type in c++11 allowed to be memcpyed?

问题描述

我的问题如下:

如果我想复制一个类类型，memcpy可以非常快地完成.在某些情况下允许这样做.

If I want to copy a class type, memcpy can do it very fast. This is allowed in some situations.

我们有一些类型特征:

• is_standard_layout.
• is_trivially_copyable.

我想知道的是类型将是按位可复制"的 exact 要求.

What I would like to know is the exact requirements when a type will be "bitwise copyable".

我的结论是，如果is_trivally_copyable和is_standard_layout特征都为真，则类型是按位可复制的:

My conclusion is that a type is bitwise copyable if both of is_trivally_copyable and is_standard_layout traits are true:

1. 正是我需要按位复制吗?
2. 约束是否过多?
3. 约束不足吗?

P.S .:当然，memcpy的结果必须正确.我知道我可以在任何情况下进行memcpy，但是方法不正确.

P.S.: of course, the result of memcpy must be correct. I know I could memcpy in any situation but incorrectly.

推荐答案

当is_trivially_copyable<T>::value为true时，可以使用memcpy复制类型T的对象.该类型不必特别是标准布局类型. 平凡可复制"的定义本质上是可以安全地做到这一点的.

You can copy an object of type T using memcpy when is_trivially_copyable<T>::value is true. There is no particular need for the type to be a standard layout type. The definition of 'trivially copyable' is essentially that it's safe to do this.

可以安全地使用memcpy复制但不是标准布局的类的示例:

An example of a class that is safe to copy with memcpy but which is not standard layout:

struct T {
  int i;
private:
  int j;
};

由于该类对不同的非静态数据成员使用不同的访问控制，因此它不是标准布局，但仍然可以轻松复制.

Because this class uses different access control for different non-static data members it is not standard layout, but it is still trivially copyable.

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