在STL列表中交换两个节点的任何快速方法？ [英] Any fast method to swap two nodes in a STL list?

问题描述

要切换到列表中的节点，更改指针就足够了。但stl swap

方法似乎复制了它们的值，不是吗？

To swap to nodes in list, changing pointers is enough. But the stl swap
method seems to copy their values, doesn''t it?

推荐答案

Atlas <马***** @ gmail.com>在消息中写道

news：11 ********************* @ f14g2000cwb.googlegro ups.com ...

"Atlas" <Ma*****@gmail.com> wrote in message
news:11*********************@f14g2000cwb.googlegro ups.com...

要交换到列表中的节点，更改指针就足够了。但是stl swap
方法似乎复制了它们的值，不是吗？

To swap to nodes in list, changing pointers is enough. But the stl swap
method seems to copy their values, doesn''t it?

所有标准库容器都有复制语义。

为什么这是一个问题？


-Mike

All standard library containers have copy semantics.
Why is this a problem?

-Mike

Atlas写道：

To交换到列表中的节点，更改指针就足够了。但stl swap
方法似乎复制了它们的值，不是吗？

To swap to nodes in list, changing pointers is enough. But the stl swap
method seems to copy their values, doesn''t it?

std :: list有几个叫做''splice''的方法它可以移动

列表中的节点而无需复制。


john

std::list have several methods called ''splice'' which can move nodes in a
list without copying.

john

John Harrison写道：

John Harrison wrote:

Atlas写道：

要交换到列表中的节点，更改指针就足够了。但stl swap
方法似乎复制了它们的值，不是吗？

To swap to nodes in list, changing pointers is enough. But the stl swap
method seems to copy their values, doesn''t it?

std :: list有几个叫做''splice''的方法可以移动节点在没有复制的情况下列表。

std::list have several methods called ''splice'' which can move nodes in a
list without copying.

这是不允许他想做的事情。列表可能不会自行拼接（）

。


Kristo

That''s not allowed for what he wants to do. A list may not splice()
with itself.

Kristo

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