如何使用反向迭代器擦除（）？ [英] How can I erase() using a reverse iterator?

问题描述

嗨


我正在使用反向迭代器查看STL列表容器，但它看起来像
似乎erase（）方法只接受普通的迭代器。是否有一个

类似的方法可以接受反向迭代器，或者是

有没有办法将反向迭代器转换为普通迭代器呢？


感谢您的帮助


B2003

Hi

I''m going through an STL list container using a reverse iterator but it
seems the erase() method only accepts ordinary iterators. Is there a
similar method that will accept reverse iterators or alternatively is
there a way to convert a reverse iterator to a normal one?

Thanks for any help

B2003

推荐答案

Boltar napsal：

Boltar napsal:

嗨


我正在使用反向迭代器查看STL列表容器但它似乎
似乎erase（）方法只接受普通的迭代器。是否有一个

类似的方法可以接受反向迭代器，或者是

有没有办法将反向迭代器转换为普通迭代器呢？


感谢您的帮助


B2003

Hi

I''m going through an STL list container using a reverse iterator but it
seems the erase() method only accepts ordinary iterators. Is there a
similar method that will accept reverse iterators or alternatively is
there a way to convert a reverse iterator to a normal one?

Thanks for any help

B2003

我不知道您使用的是哪种类型的容器。无论如何，我认为你

没有一步擦除所有项目。因此迭代器将变为

无效。你应该使用


erase（container.rbegin（），container.rend（））;


或者simillar它应该可以工作。

I do not know which type of container are you using. Anyway I think you
are not erasing all items in one step. So iterators will become
invalid. You should use

erase(container.rbegin(), container.rend());

Or simillar and it should work.

Boltar写道：

Boltar wrote:

我正在使用反向通过STL列表容器迭代器，但它似乎
似乎erase（）方法只接受普通的迭代器。是否有一个

类似的方法可以接受反向迭代器，或者是

有没有办法将反向迭代器转换为普通迭代器？

I''m going through an STL list container using a reverse iterator but it
seems the erase() method only accepts ordinary iterators. Is there a
similar method that will accept reverse iterators or alternatively is
there a way to convert a reverse iterator to a normal one?

reverse_iterator有一个成员函数base（），它将为您提供底层迭代器的

值。但要注意，底层的

迭代器指向不同的元素。这是一个关闭，精确的

关系船在标准[24.4.1 / 1]中给出：


& * （reverse_iterator（i））==& *（i - 1）

Best


Kai-Uwe Bux

The reverse_iterator has a member function base() that will give you the
value of the underlying iterator. Beware, however, that the underlying
iterator points to a different element. It''s off by one, and the precise
relation ship is given in the standard [24.4.1/1] as:

&*(reverse_iterator(i)) == &*(i - 1)
Best

Kai-Uwe Bux

1月25日，11：39，Kai-Uwe Bux< jkherci ... @ gmx.netwrote：

On 25 Jan, 11:39, Kai-Uwe Bux <[e-mail]:

底层迭代器的值。但要注意，底层的

迭代器指向不同的元素。这是一个关闭，精确的

关系船在标准[24.4.1 / 1]中给出：


& * （reverse_iterator（i））==& *（i - 1）

value of the underlying iterator. Beware, however, that the underlying
iterator points to a different element. It''s off by one, and the precise
relation ship is given in the standard [24.4.1/1] as:

&*(reverse_iterator(i)) == &*(i - 1)

谢谢。这听起来像是委员会决策的另一个胜利。

为什么他们认为一个人做出来有用呢？啊

好​​......


B2003

Thanks. That sounds like another triumph of committee decision making.
Why on earth did they think having it off by one would be useful? Ah
well....

B2003

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