blocks|key|1751814|text|这个怎么样：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1751815|deque+dq;
//fill+it
for(int+i+=+0;+i+<+dq.size()+-+1;+i%2B%2B)+//notice+the+"-1"+to+avoid+reaching+the+last+element
{
++++if+(+cond()+)+
++++{
++++++++dq.erase(dq.begin()%2Bi);
++++++++i--;
++++}+
}|code-block|syntax|javascript|1751816|您可以使用deque访问operator[]元素，因此dq[i]是deque的元素号i。|offset|length|style|CODE|1751817|entityMap^0|0|0|5|5|C|A|R|5|16|1|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|N|$]]

How about this:

<pre><code>deque dq;
//fill it
for(int i = 0; i &lt; dq.size() - 1; i++) //notice the "-1" to avoid reaching the last element
{
 if ( cond() ) 
 {
 dq.erase(dq.begin()+i);
 i--;
 } 
}
</code></pre>

You can access <code>deque</code> elements with <code>operator[]</code>, so <code>dq[i]</code> is element number <code>i</code> of the deque.

blocks|key|847754|text|您可以使用dq.end()+-+1在结束前获得一个。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|847755|您还可以使用std::remove或std::remove_if来删除您想要的项，而不是自己完成工作。注意，这基本上有点像分区操作--它返回一个迭代器，您想要保存的所有东西都是从范围的开始到那个迭代器。您要删除的所有内容都是在该迭代器之后，到您作为输入提供的范围的末尾。|847756|在这种情况下，您可以按照一般顺序做一些事情：|847757|std::deque<int>+vals+{+1,+2,+3,+4,+5,+6,+7,+8};

//+get+an+iterator+one+before+the+end+of+`vals`
auto+end+=+vals.end()-1;

//+remove+the+even+items+in+the+range+(so+all+except+`8`)
auto+pos+=+std::remove_if(vals.begin(),+end,+
++++[](int+v){+return+v+%252+==+0;}+
);

//+erase+the+items+we+just+"removed"
vals.erase(pos,+end);

//+show+the+result
for+(v+:+vals)
++++std::cout+<<+v+<<+",+";|code-block|syntax|javascript|847758|entityMap^0|5|C|0|6|B|I|E|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|Y|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|Z|8|@]|D|@]|E|$M|N]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

You can just use <code>dq.end() - 1</code> to get one before the end.

You can also use <code>std::remove</code> or <code>std::remove_if</code> to remove the items you want instead of doing the job on your own. Note that this is basically kind of like a partitioning operation--it returns an iterator, and everything you want to keep is from the beginning of the range to that iterator. Everything you want to delete is after that iterator, to the end of the range you provided as input.

In this case, you might do something on this general order:

<pre><code>std::deque&lt;int&gt; vals { 1, 2, 3, 4, 5, 6, 7, 8};

// get an iterator one before the end of `vals`
auto end = vals.end()-1;

// remove the even items in the range (so all except `8`)
auto pos = std::remove_if(vals.begin(), end, 
 [](int v){ return v %2 == 0;} 
);

// erase the items we just "removed"
vals.erase(pos, end);

// show the result
for (v : vals)
 std::cout &lt;&lt; v &lt;&lt; ", ";
</code></pre>

blocks|key|847796|text|来自std::deque::erase：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|CODE|entityRanges|data|847797|所有迭代器和引用都无效，除非擦除的元素位于容器的末尾或开头，在这种情况下，只有迭代器和对擦除元素的引用无效。|blockquote|847798|这意味着，您不能使用迭代器和删除某些元素来迭代deque。相反，将std::remove_if与std::deque::erase结合使用。|847799|(未经测试，但你会有个主意)|847800|std::deque<T>+dq+=+/*+...+*/;
auto+from+=+std::remove_if(dq.begin(),+std::prev(dq.end()),
++++[](decltype(dq)::const_reference,+decltype(dq)::const_reference){+return+cond();+}
);
dq.erase(from,+std::prev(dq.end()));|code-block|syntax|javascript|847801|entityMap^0|2|H|2|H|0|0|X|E|1C|H|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|D]]|E|@]|F|$]]|$1|G|3|H|5|I|7|Z|8|@]|E|@]|F|$]]|$1|J|3|K|5|6|7|10|8|@$9|11|A|12|B|D]|$9|13|A|14|B|D]]|E|@]|F|$]]|$1|L|3|M|5|6|7|15|8|@]|E|@]|F|$]]|$1|N|3|O|5|P|7|16|8|@]|E|@]|F|$Q|R]]|$1|S|3|-4|5|6|7|17|8|@]|E|@]|F|$]]]|T|$]]

From <code>std::deque::erase</code>: 

<blockquote>
 All iterators and references are invalidated, unless the erased elements are at the end or the beginning of the container, in which case only the iterators and references to the erased elements are invalidated.
</blockquote>

Which means, that you cannot iterate over a deque using iterators and delete some elements. Instead, use <code>std::remove_if</code> with <code>std::deque::erase</code>.

(Untested, but you'll get an idea)

<pre><code>std::deque&lt;T&gt; dq = /* ... */;
auto from = std::remove_if(dq.begin(), std::prev(dq.end()),
 [](decltype(dq)::const_reference, decltype(dq)::const_reference){ return cond(); }
);
dq.erase(from, std::prev(dq.end()));
</code></pre>

blocks|key|1845165|text|使用it_end.base()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1845166|#include+<iostream>
#include+<deque>
#include+<iterator>

using+namespace+std;

int+main()
{
++++deque<int>+dq+=+{+0,+1,+2,+3,4,+5+};
++++auto+it_end+=+dq.rbegin();+it_end%2B%2B;

++++for+(auto+it+=+dq.begin();+it+!=+it_end.base();+%2B%2Bit)+{
++++++++cout+<<+*it+<<+"+";
++++}

++++cout+<<+"\n";
++++return+0;
}|code-block|syntax|javascript|1845167|产出：|1845168|0+1+2+3+4|1845169|有关base()函数如何为反向迭代器工作的详细信息，请参阅base()。|1845170|entityMap^0|2|D|0|0|0|0|2|6|T|6|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|R|$]]

Use <code>it_end.base()</code>:

<pre><code>#include &lt;iostream&gt;
#include &lt;deque&gt;
#include &lt;iterator&gt;

using namespace std;

int main()
{
 deque&lt;int&gt; dq = { 0, 1, 2, 3,4, 5 };
 auto it_end = dq.rbegin(); it_end++;

 for (auto it = dq.begin(); it != it_end.base(); ++it) {
 cout &lt;&lt; *it &lt;&lt; " ";
 }

 cout &lt;&lt; "\n";
 return 0;
}
</code></pre>

Outputs:

<pre><code>0 1 2 3 4
</code></pre>

See <a href="https://stackoverflow.com/a/10373912/12711">https://stackoverflow.com/a/10373912/12711</a> for some details on how the <code>base()</code> function works for reverse iterators.

I am trying to iterate through a deque up to excluding the last entry. Ideally I would avoid counting and comparing to the length, so I tried

<pre><code> auto it_end = dq.rbegin(); it_end++;
 for ( auto it = dq.begin(); it !=it_end; ) {
 if ( cond() ) {
 it = dq.erase( it );
 } else {
 it++;
 }
 }
</code></pre>

But the compiler complains there is no match operand which seems understandable since I have a reverse iterator and a regular iterator. Is there an elegant way to stop before the last element that avoids counting? Like, an offset?
I also tried usign back, but that turned out to be a reference not an iterator, so != is not happy either.

C++ deque iterate to second last element

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。

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我正试图迭代一遍，直到排除最后一项。理想情况下，我会避免计算和比较长度，所以我尝试    auto it_end = dq.rbegin(); it_end++;    for ( auto it = dq.begin(); it !=it_end; ) {        if ( cond() ) {        ...

问迭代到第二个最后元素的C++方法
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问迭代到第二个最后元素的C++方法EN