c++11 中的类型什么时候可以被 memcpyed?

When is a type in c++11 allowed to be memcpyed?(c++11 中的类型什么时候可以被 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.这在某些情况下是允许的.

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.

我想知道的是当一个类型可以按位复制"时的确切要求.

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，则类型是可按位复制的:

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

1. 这正是我需要按位复制的内容吗?
2. 是否过度约束?
3. 是否受到限制?

P.S.:当然，memcpy 的结果一定是正确的.我知道我可以在任何情况下进行 memcpy，但不正确.

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::value 是真的.没有特别需要该类型是标准布局类型.可简单复制"的定义本质上是这样做是安全的.

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.

使用 memcpy 可以安全复制但不是标准布局的类示例:

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;
};

因为这个类对不同的非静态数据成员使用不同的访问控制，所以不是标准布局，但它仍然可以简单地复制.

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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