blocks|key|1868217|text|经过进一步的研究和测试，我找到了解决方案。显然，根据标准24.4.1/1，i.base()和I之间的关系是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1868218|&*(reverse_iterator(i))+==+&*(i+-+1)|code-block|syntax|javascript|1868219|(从Dr.+Dobbs+article)：|offset|length|1868220|​|1868221|📷|atomic|1868222|1868223|因此，在获取base()时需要应用偏移量。因此，解决方案是：|1868224|m_CursorStack.erase(+--(i.base())+);|1868225|编辑|style|BOLD|1868226|正在更新C%2B%2B11。|1868227|reverse_iterator+i保持不变：|CODE|1868228|m_CursorStack.erase(+std::next(i).base()+);|1868229|reverse_iterator+i是高级的：|1868230|std::advance(i,+1);
m_CursorStack.erase(+i.base()+);|1868231|我发现这一点比我之前的解决方案要清晰得多。使用您需要的任何一个。|1868232|entityMap|0|LINK|mutability|MUTABLE|url|http://www.ddj.com/cpp/184401406|1|IMAGE|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/430eb835cd9d6fe8e520e178d7d60a61.gif|imageAlt^0|0|0|2|H|0|0|0|0|1|1|0|0|0|0|0|2|0|0|H|1|0|0|H|1|0|0|0^^$0|@$1|2|3|4|5|6|7|1P|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1Q|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1R|8|@]|9|@$I|1S|J|1T|1|1U]]|A|$]]|$1|K|3|L|5|6|7|1V|8|@]|9|@]|A|$]]|$1|M|3|N|5|O|7|1W|8|@]|9|@$I|1X|J|1Y|1|1Z]]|A|$]]|$1|P|3|L|5|6|7|20|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|21|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|22|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|23|8|@$I|24|J|25|W|X]]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|26|8|@]|9|@]|A|$]]|$1|10|3|11|5|6|7|27|8|@$I|28|J|29|W|12]]|9|@]|A|$]]|$1|13|3|14|5|D|7|2A|8|@]|9|@]|A|$E|F]]|$1|15|3|16|5|6|7|2B|8|@$I|2C|J|2D|W|12]]|9|@]|A|$]]|$1|17|3|18|5|D|7|2E|8|@]|9|@]|A|$E|F]]|$1|19|3|1A|5|6|7|2F|8|@]|9|@]|A|$]]|$1|1B|3|-4|5|6|7|2G|8|@]|9|@]|A|$]]]|1C|$1D|$5|1E|1F|1G|A|$1H|1I]]|1J|$5|1K|1F|1L|A|$1M|1N|1O|-4]]]]

After some more research and testing I found the solution. Apparently according to the standard [24.4.1/1] the relationship between i.base() and i is:

<pre><code>&amp;*(reverse_iterator(i)) == &amp;*(i - 1)
</code></pre>

(from a <a href="http://www.ddj.com/cpp/184401406" rel="noreferrer">Dr. Dobbs article</a>):

<img src="https://i.cmpnet.com/ddj/cuj/images/cuj0106smeyers/diagram2.gif" alt="alt text">

So you need to apply an offset when getting the base(). Therefore the solution is:

<pre><code>m_CursorStack.erase( --(i.base()) );
</code></pre>

EDIT

Updating for C++11.

reverse_iterator <code>i</code> is unchanged:

<pre><code>m_CursorStack.erase( std::next(i).base() );
</code></pre>

reverse_iterator <code>i</code> is advanced:

<pre><code>std::advance(i, 1);
m_CursorStack.erase( i.base() );
</code></pre>

I find this much clearer than my previous solution. Use whichever you require.

blocks|key|567897|text|虽然使用reverse_iterator的base()方法并递减结果在这里有效，但值得注意的是，reverse_iterators没有被赋予与常规iterators相同的状态。通常，您应该更喜欢常规iterators而不是reverse_iterators+(以及const_iterators和const_reverse_iterators)，原因就像这样。有关原因的深入讨论，请参阅Doctor+Dobbs'+Journal。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|567898|entityMap|0|LINK|mutability|MUTABLE|url|http://www.ddj.com/cpp/184401406^0|4|G|L|6|1C|G|21|8|2R|8|33|G|3O|E|44|M|5D|L|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]|$9|W|A|X|B|C]|$9|Y|A|Z|B|C]|$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@$9|14|A|15|1|16]]|E|$]]|$1|F|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

While using the <code>reverse_iterator</code>'s <code>base()</code> method and decrementing the result works here, it's worth noting that <code>reverse_iterator</code>s are not given the same status as regular <code>iterator</code>s. In general, you should prefer regular <code>iterator</code>s to <code>reverse_iterator</code>s (as well as to <code>const_iterator</code>s and <code>const_reverse_iterator</code>s), for precisely reasons like this. See <a href="http://www.ddj.com/cpp/184401406" rel="nofollow noreferrer">Doctor Dobbs' Journal</a> for an in-depth discussion of why.

blocks|key|603216|text|请注意，如果在for循环中使用m_CursorStack.erase(+(%2B%2Bi).base())可能会有问题(参见原始问题)，因为它会更改i的值。正确的表达式是m_CursorStack.erase((i%2B1).base())|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|603217|entityMap^0|7|3|F|Y|29|X|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|O|8|@]|D|@]|E|$]]]|G|$]]

Please note that <code>m_CursorStack.erase( (++i).base())</code> may be a problem if used in a <code>for</code> loop (see original question) because it changes the value of i. Correct expression is <code>m_CursorStack.erase((i+1).base())</code>

blocks|key|2011041|text|如果你不需要在执行过程中删除所有内容，那么可以使用erase-remove成语来解决问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2011042|m_CursorStack.erase(std::remove(m_CursorStack.begin(),+m_CursorStack.end(),+pCursor),+m_CursorStack.end());|code-block|syntax|javascript|2011043|std::remove将容器中与pCursor匹配的所有项交换到最后，并返回第一个匹配项的迭代器。然后，使用范围的erase将从第一个匹配中擦除，并转到末尾。保留非匹配元素的顺序。|offset|length|style|CODE|2011044|如果你使用的是std::vector，这可能会更快，因为在内容中间进行擦除可能涉及到大量的复制或移动。|2011045|当然，上面解释reverse_iterator::base()用法的答案很有趣，值得了解，为了解决所述的确切问题，我认为std::remove更合适。|2011046|entityMap^0|0|0|0|B|G|7|1L|5|0|7|B|0|7|O|1O|B|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|11|8|@$I|12|J|13|K|L]]|9|@]|A|$]]|$1|O|3|P|5|6|7|14|8|@$I|15|J|16|K|L]|$I|17|J|18|K|L]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|R|$]]

If you don't need to erase everything as you go along, then to solve the problem, you can use the erase-remove idiom:

<pre><code>m_CursorStack.erase(std::remove(m_CursorStack.begin(), m_CursorStack.end(), pCursor), m_CursorStack.end());
</code></pre>

<code>std::remove</code> swaps all the items in the container that match <code>pCursor</code> to the end, and returns an iterator to the first match item. Then, the <code>erase</code> using a range will erase from the first match, and go to the end. The order of the non-matching elements is preserved.

This might work out faster for you if you're using a <code>std::vector</code>, where erasing in the middle of the contents can involve a lot of copying or moving.

Or course, the answers above explaining the use of <code>reverse_iterator::base()</code> are interesting and worth knowing, to solve the exact problem stated, I'd argue that <code>std::remove</code> is a better fit.

blocks|key|1870674|text|typedef+std::map<size_t,+some_class*>+TMap;
TMap+Map;
.......

for(+TMap::const_reverse_iterator+It+=+Map.rbegin(),+end+=+Map.rend();+It+!=+end;+It%2B%2B+)
{
++++TMap::const_iterator+Obsolete+=+It.base();+++//+conversion+into+const_iterator
++++It%2B%2B;
++++Map.erase(+Obsolete+);
++++It--;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1870675|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>typedef std::map&lt;size_t, some_class*&gt; TMap;
TMap Map;
.......

for( TMap::const_reverse_iterator It = Map.rbegin(), end = Map.rend(); It != end; It++ )
{
 TMap::const_iterator Obsolete = It.base(); // conversion into const_iterator
 It++;
 Map.erase( Obsolete );
 It--;
}
</code></pre>

blocks|key|2011113|text|我只想澄清一些事情:在上面的一些评论和答案中，erase的可移植版本被提到为(%2B%2Bi).base()。但是，除非我遗漏了什么，否则正确的语句是(%2B%2Bri).base()，这意味着您可以‘递增’reverse_iterator+(而不是迭代器)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2011114|昨天我遇到了做类似事情的需要，这篇文章很有帮助。谢谢大家。|2011115|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Just wanted to clarify something: In some of the above comments and answers the portable version for erase is mentioned as (++i).base(). However unless I am missing something the correct statement is (++ri).base(), meaning you 'increment' the reverse_iterator (not the iterator).

I ran into a need to do something similar yesterday and this post was helpful. Thanks everyone.

blocks|key|2011140|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2011141|+...或者是从列表中删除此元素的另一种方法？|blockquote|2011142|2011143|2011144|这需要-std=c%2B%2B11标志(对于auto)：|offset|length|style|CODE|2011145|auto+it=vt.end();
while+(it>vt.begin())
{
++++it--;
++++if+(*it+==+pCursor)+//{+delete+*it;
++++++++it+=+vt.erase(it);+//}
}|code-block|syntax|javascript|2011146|entityMap^0|0|0|0|0|3|A|I|4|0|0^^$0|@$1|2|3|-4|4|5|6|S|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|T|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|U|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|V|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|W|7|@$H|X|I|Y|J|K]|$H|Z|I|10|J|K]]|8|@]|9|$]]|$1|L|3|M|4|N|6|11|7|@]|8|@]|9|$O|P]]|$1|Q|3|-4|4|5|6|12|7|@]|8|@]|9|$]]]|R|$]]

<blockquote>
 ... or another way to remove this element from the list?
</blockquote>

This requires the <code>-std=c++11</code> flag (for <code>auto</code>):

<pre><code>auto it=vt.end();
while (it&gt;vt.begin())
{
 it--;
 if (*it == pCursor) //{ delete *it;
 it = vt.erase(it); //}
}
</code></pre>

blocks|key|1868390|text|下面这段代码将erase的结果转换回反向迭代器，以便在反向迭代时擦除容器中的元素。有点奇怪，但它甚至在擦除第一个或最后一个元素时也能正常工作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1868391|std::set<int>+set{1,2,3,4,5};

for+(auto+itr+=+set.rbegin();+itr+!=+set.rend();+)
{++++
++++if+(*itr+==+3)
++++{
++++++++auto+it+=+set.erase(--itr.base());
++++++++itr+=+std::reverse_iterator(it);++++++++++++
++++}
++++else
++++++++%2B%2Bitr;
}|code-block|syntax|javascript|1868392|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

And here is the piece of code to convert the result of erase back to a reverse iterator in order to erase an element in a container while iterating in the reverse. A bit strange, but it works even when erasing the first or last element:

<pre><code>std::set&lt;int&gt; set{1,2,3,4,5};

for (auto itr = set.rbegin(); itr != set.rend(); )
{ 
 if (*itr == 3)
 {
 auto it = set.erase(--itr.base());
 itr = std::reverse_iterator(it); 
 }
 else
 ++itr;
}
</code></pre>

blocks|key|568021|text|有趣的是，这个页面上还没有正确的解决方案。因此，下面是正确的一条：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|568022|在前向迭代器的情况下，解决方案是直接的：|568023|std::list<+int+>::iterator+i+=+myList.begin();
while+(+i+!=+myList.end()+)+{
++if+(+*i+==+to_delete+)+{
++++i+=+myList.erase(+i+);
++}+else+{
++++%2B%2Bi;
++}+
}|code-block|syntax|javascript|568024|在使用反向迭代器的情况下，您需要执行相同的操作：|568025|std::list<+int+>::reverse_iterator+i+=+myList.rbegin();
while+(+i+!=+myList.rend()+)+{
++if+(+*i+==+to_delete+)+{
++++i+=+decltype(i)(myList.erase(+std::next(i).base()+));
++}+else+{
++++%2B%2Bi;
++}+
}|568026|备注：|568027|您可以从迭代器构造reverse_iterator+|offset|length|style|CODE|568028|您可以使用std::list::erase|unordered-list-item|568029|
的返回值|568030|entityMap^0|0|0|0|0|0|0|9|G|0|5|G|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|12|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|13|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|14|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|15|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|16|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|17|8|@$Q|18|R|19|S|T]]|9|@]|A|$]]|$1|U|3|V|5|W|7|1A|8|@$Q|1B|R|1C|S|T]]|9|@]|A|$]]|$1|X|3|Y|5|6|7|1D|8|@]|9|@]|A|$]]|$1|Z|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]]|10|$]]

Funny that there is no correct solution on this page yet. So, the following is the correct one:
In case of the forward iterator the solution is straight forward:
<pre><code>std::list&lt; int &gt;::iterator i = myList.begin();
while ( i != myList.end() ) {
 if ( *i == to_delete ) {
 i = myList.erase( i );
 } else {
 ++i;
 } 
}
</code></pre>
In case of reverse iterator you need to do the same:
<pre><code>std::list&lt; int &gt;::reverse_iterator i = myList.rbegin();
while ( i != myList.rend() ) {
 if ( *i == to_delete ) {
 i = decltype(i)(myList.erase( std::next(i).base() ));
 } else {
 ++i;
 } 
}
</code></pre>
Notes:
<ul>
<li>You can construct a <code>reverse_iterator</code> from an iterator</li>
<li>You can use the return value of <code>std::list::erase</code></li>
</ul>

blocks|key|2011275|text|为了补充其他人的回答，也因为我在搜索std::string时偶然发现了这个问题，所以这里给出了一个使用std::string、std::string::erase和std::reverse_iterator的响应|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2011276|我的问题是从一个完整的文件名字符串中删除图像文件名。它最初是用std::string::find_last_of解决的，但我研究了一种用std::reverse_iterator的替代方法。|2011277|std::string+haystack("\\\\UNC\\complete\\file\\path.exe");
auto&&+it+=+std::find_if(+std::rbegin(haystack),+std::rend(haystack),+[](+char+ch){+return+ch+==+'\\';+}+);
auto&&+it2+=+std::string::iterator(+std::begin(+haystack+)+%2B+std::distance(it,+std::rend(haystack))+);
haystack.erase(it2,+std::end(haystack));
std::cout+<<+haystack;++//////+prints:+'\\UNC\complete\file\'|code-block|syntax|javascript|2011278|这使用了算法、迭代器和字符串头。|2011279|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

To complement other's answers and because I stumbled upon this question whilst searching about std::string without much success, here goes a response with the usage of std::string, std::string::erase and std::reverse_iterator 

My problem was erasing the an image filename from a complete filename string. It was originally solved with std::string::find_last_of, yet I research an alternative way with std::reverse_iterator.

<pre><code>std::string haystack("\\\\UNC\\complete\\file\\path.exe");
auto&amp;&amp; it = std::find_if( std::rbegin(haystack), std::rend(haystack), []( char ch){ return ch == '\\'; } );
auto&amp;&amp; it2 = std::string::iterator( std::begin( haystack ) + std::distance(it, std::rend(haystack)) );
haystack.erase(it2, std::end(haystack));
std::cout &lt;&lt; haystack; ////// prints: '\\UNC\complete\file\'
</code></pre>

This uses algorithm, iterator and string headers.

blocks|key|603518|text|反向迭代器很难使用。所以只使用了通用迭代器。'r‘它从最后一个元素开始。当找到要删除的东西时。擦除它并返回下一个迭代器。例如，当删除第三个元素时，它将指向当前的第四个元素。和新的第三个。所以向左移动应该减1。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|603519|void+remchar(string&+s,char+c)
{++++++
++++auto+r+=+s.end()+-+1;
++++while+(r+>=+s.begin()+&&+*r+==+c)
++++{
++++++++r+=+s.erase(r);
++++++++r+-=+1;
++++}
}|code-block|syntax|javascript|603520|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

reverse iterator is quite hard to use. So just used general iterator. 'r' It start from last element. When find something to erase. erase it and return next iterator. eg when delete 3rd element it will pointing current 4th element. and new 3rd. So it should be decreased 1 to move left

<pre><code>void remchar(string&amp; s,char c)
{ 
 auto r = s.end() - 1;
 while (r &gt;= s.begin() &amp;&amp; *r == c)
 {
 r = s.erase(r);
 r -= 1;
 }
}
</code></pre>

blocks|key|568083|text|如果m_CursorStack是一个向量，你可以通过索引来擦除：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|568084|m_CursorStack.erase(m_CursorStack.begin()+%2B+m_CursorStack.size()+%2B+int(m_CursorStack.rbegin()+-+i)+-+1);|code-block|syntax|javascript|568085|entityMap^0|2|D|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

If <code>m_CursorStack</code> is a vector, you can erase by taking index:
<pre><code>m_CursorStack.erase(m_CursorStack.begin() + m_CursorStack.size() + int(m_CursorStack.rbegin() - i) - 1);
</code></pre>

I am trying to do something like this:

<pre><code>for ( std::list&lt; Cursor::Enum &gt;::reverse_iterator i = m_CursorStack.rbegin(); i != m_CursorStack.rend(); ++i )
{
 if ( *i == pCursor )
 {
 m_CursorStack.erase( i );
 break;
 }
}
</code></pre>

However erase takes an iterator and not a reverse iterator. is there a way to convert a reverse iterator to a regular iterator or another way to remove this element from the list?

How to call erase with a reverse iterator

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我正在尝试这样做：for ( std::list< Cursor::Enum >::reverse_iterator i = m_CursorStack.rbegin(); i != m_CursorStack.rend(); ++i ){ if ( *i == pCursor ) { m_C...

问如何使用反向迭代器调用erase
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何使用反向迭代器调用eraseEN