使用常规迭代器向后迭代，或使用reverse_iterator？ [英] Use a regular iterator to iterate backwards, or struggle with reverse_iterator?

问题描述

我最近学到了使用C ++中的反向迭代器的正确方法（特别是当你需要删除一个时）。 （请参见此问题和这一个。）

I recently learned about the right way to work with reverse iterators in C++ (specifically when you need to erase one). (See this question and this one.)

这是你应该做的：

typedef std::vector<int> IV;
for (IV::reverse_iterator rit = iv.rbegin(), rend = iv.rend();
     rit != rend; ++rit)
{
  // Use 'rit' if a reverse_iterator is good enough, e.g.,
  *rit += 10;
  // Use (rit + 1).base() if you need a regular iterator e.g.,
  iv.erase((rit + 1).base());
}

但是我认为 不要这样做，不符合标准，正如MooingDuck指出的）：

But I think thought this is much better (Don't do this, not standards compliant, as MooingDuck points out):

for (IV::iterator it = iv.end(), begin = iv.begin();
     it-- != begin; )
{
  // Use 'it' for anything you want
  *it += 10;
  iv.erase(it);
}

缺点：

• 你告诉我。发生了什么问题？


• 这是不符合标准的，因为MooingDuck点出来。

• You tell me. What's wrong with it?
• It's not standards compliant, as MooingDuck points out. That pretty much overrules any of the possible advantages below.

优点：

• 使用熟悉的惯用法用于反向for-loops


• 不必记住（或解释） +1


• 减少键入


• 适用于std :: list： it = il.erase ;


• 如果您删除元素，则不必调整迭代器。


• 不必重新计算begin迭代器

• Uses a familiar idiom for reverse for-loops
• Don't have to remember (or explain) the +1
• Less typing
• Works for std::list too: it = il.erase(it);
• If you erase an element, you don't have to adjust the iterator
• If you erase, you don't have to recompute the begin iterator

推荐答案

反向迭代器的原因是标准算法不知道如何对集合进行反向迭代。例如：

The reason for reverse iterators is that the standard algorithms do not know how to iterate over a collection backwards. For example:

#include <string>
#include <algorithm>
std::wstring foo(L"This is a test, with two letter a's involved.");
std::find(foo.begin(), foo.end(), L'a'); // Returns an iterator pointing
                                        // to the first a character.
std::find(foo.rbegin(), foo.rend(), L'a').base()-1; //Returns an iterator
                                                 // pointing to the last A.
std::find(foo.end(), foo.begin(), L'a'); //WRONG!! (Buffer overrun)

使用任何迭代器产生更清晰的代码。

Use whichever iterator results in clearer code.

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