为什么std :: memcpy的行为对于不是TriviallyCopyable的对象是未定义的？

问题描述

从：

在我的工作中，我们使用 std :: memcpy 很长时间不使用TriviallyCopyable的位交换对象：

At my work, we have used std::memcpy for a long time to bitwise swap objects that are not TriviallyCopyable using:

void swapMemory(Entity* ePtr1, Entity* ePtr2)
{
   static const int size = sizeof(Entity); 
   char swapBuffer[size];

   memcpy(swapBuffer, ePtr1, size);
   memcpy(ePtr1, ePtr2, size);
   memcpy(ePtr2, swapBuffer, size);
}

，从未出现任何问题。

我理解，对非TriviallyCopyable对象滥用 std :: memcpy 并导致下游未定义的行为是微不足道的。但是，我的问题：

I understand that it is trivial to abuse std::memcpy with non-TriviallyCopyable objects and cause undefined behavior downstream. However, my question:

为什么 std :: memcpy TriviallyCopyable对象？为什么标准认为有必要指定？

Why would the behavior of std::memcpy itself be undefined when used with non-TriviallyCopyable objects? Why does the standard deem it necessary to specify that?

UPDATE

已被修改，以响应原始帖子和帖子的答案。当前描述为：

It turns out that the standard does not say anything about the behavior of std::memcpy for objects that are not TriviallyCopyable. The contents of http://en.cppreference.com/w/cpp/string/byte/memcpy have been modified in response to the original post and the answers to the post. The current description says:

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