blocks|key|826874|text|res.erase(it)总是返回下一个有效的迭代器，如果删除最后一个元素，它将指向.end()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|826875|在循环结束时，总是会调用%2B%2Bit，因此您会递增.end()，这是不允许的。|826876|不过，简单地检查.end()仍然会留下一个bug，因为您总是在每次迭代中跳过一个元素(it通过.erase()的返回“递增”，然后再通过循环递增)|826877|你可能想要这样的东西：|826878|+while+(it+!=+res.end())+{
++++++++it+=+res.erase(it);++++
+}|code-block|syntax|javascript|826879|删除每个元素的步骤|826880|(为了完整性:我假设这是一个简化的示例，如果您只想删除每个元素，而不必对其执行操作(例如，删除)，则应该简单地调用res.clear())。|826881|当您只有条件地擦除元素时，您可能需要类似这样的东西|826882|for+(+;+it+!=+res.end();+)+{
++if+(condition)+{
++++it+=+res.erase(it);
++}+else+{
++++%2B%2Bit;
++}
}|826883|entityMap^0|0|D|16|6|0|C|4|N|6|0|8|6|17|2|1B|8|0|0|0|0|1L|B|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|15|8|@$9|16|A|17|B|C]|$9|18|A|19|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|1A|8|@$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|1H|8|@]|D|@]|E|$]]|$1|L|3|M|5|N|7|1I|8|@]|D|@]|E|$O|P]]|$1|Q|3|R|5|6|7|1J|8|@]|D|@]|E|$]]|$1|S|3|T|5|6|7|1K|8|@$9|1L|A|1M|B|C]]|D|@]|E|$]]|$1|U|3|V|5|6|7|1N|8|@]|D|@]|E|$]]|$1|W|3|X|5|N|7|1O|8|@]|D|@]|E|$O|P]]|$1|Y|3|-4|5|6|7|1P|8|@]|D|@]|E|$]]]|Z|$]]

<code>res.erase(it)</code> always returns the next valid iterator, if you erase the last element it will point to <code>.end()</code>

At the end of the loop <code>++it</code> is always called, so you increment <code>.end()</code> which is not allowed. 

Simply checking for <code>.end()</code> still leaves a bug though, as you always skip an element on every iteration (<code>it</code> gets 'incremented' by the return from <code>.erase()</code>, and then again by the loop)

You probably want something like:

<pre><code> while (it != res.end()) {
 it = res.erase(it); 
 }
</code></pre>

to erase each element

(for completeness: I assume this is a simplified example, if you simply want every element gone without having to perform an operation on it (e.g. delete) you should simply call <code>res.clear()</code>)

When you only conditionally erase elements, you probably want something like

<pre><code>for ( ; it != res.end(); ) {
 if (condition) {
 it = res.erase(it);
 } else {
 ++it;
 }
}
</code></pre>

blocks|key|826988|text|不要擦除然后递增迭代器。不需要递增，如果你的向量有奇数(或者偶数，我不知道)的元素，你会错过向量的结尾。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|826989|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Do not erase and then increment the iterator. No need to increment, if your vector has an odd (or even, I don't know) number of elements you will miss the end of the vector.

blocks|key|1041868|text|it%2B%2B指令在块的末尾完成。因此，如果要擦除最后一个元素，则尝试递增指向空集合的迭代器。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1041869|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

The it++ instruction is done at the end of the block. So if your are erasing the last element, then you try to increment the iterator that is pointing to an empty collection.

blocks|key|827125|text|在for循环的循环表达式中，将it递增到(空)容器的末尾。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|827126|entityMap^0|F|2|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

You increment <code>it</code> past the end of the (empty) container in the for loop's loop expression.

blocks|key|827180|text|for(+;+it+!=+res.end();)
{
++++it+=+res.erase(it);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|827181|或者，更一般地说：|unstyled|827182|for(+;+it+!=+res.end();)
{
++++if+(smth)
++++++++it+=+res.erase(it);
++++else
++++++++%2B%2Bit;
}|827183|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>for( ; it != res.end();)
{
 it = res.erase(it);
}
</code></pre>

or, more general:

<pre><code>for( ; it != res.end();)
{
 if (smth)
 it = res.erase(it);
 else
 ++it;
}
</code></pre>

blocks|key|1042040|text|下面的方法似乎也很有效：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1042041|for+(vector<int>::iterator+it+=+res.begin();+it+!=+res.end();+it%2B%2B)
{
++res.erase(it--);
}|code-block|syntax|javascript|1042042|不确定这个版本中是否有任何缺陷？|1042043|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

The following also seems to work :

<pre><code>for (vector&lt;int&gt;::iterator it = res.begin(); it != res.end(); it++)
{
 res.erase(it--);
}
</code></pre>

Not sure if there's any flaw in this ?

blocks|key|1042057|text|if(allPlayers.empty()+==+false)+{
++++for(int+i+=+allPlayers.size()+-+1;+i+>=+0;+i--)
++++{
++++++++if(allPlayers.at(i).getpMoney()+<=+0)+
++++++++++++allPlayers.erase(allPlayers.at(i));
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1042058|这对我很有效。而且不需要考虑索引已经被擦除了。|unstyled|1042059|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>if(allPlayers.empty() == false) {
 for(int i = allPlayers.size() - 1; i &gt;= 0; i--)
 {
 if(allPlayers.at(i).getpMoney() &lt;= 0) 
 allPlayers.erase(allPlayers.at(i));
 }
}
</code></pre>

This works for me. And Don't need to think about indexes have already erased.

blocks|key|1042119|text|作为对crazylammer答案的修改，我经常使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1042120|your_vector_type::iterator+it;
for(+it+=+res.start();+it+!=+res.end();)
{
++++your_vector_type::iterator+curr+=+it%2B%2B;
++++if+(something)
++++++++res.erase(curr);
}|code-block|syntax|javascript|1042121|这样做的好处是，你不必担心忘记递增你的迭代器，当你有复杂的逻辑时，它不太容易出错。在循环中，curr永远不会等于res.end()，并且无论您是否将其从向量中删除，它都将位于下一个元素。|1042122|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

As a modification to crazylammer's answer, I often use:

<pre><code>your_vector_type::iterator it;
for( it = res.start(); it != res.end();)
{
 your_vector_type::iterator curr = it++;
 if (something)
 res.erase(curr);
}
</code></pre>

The advantage of this is that you don't have to worry about forgetting to increment your iterator, making it less bug prone when you have complex logic. Inside the loop, curr will never be equal to res.end(), and it will be at the next element regardless of if you erase it from your vector.

blocks|key|827390|text|因为向量中的erase方法返回所传递迭代器的下一个迭代器。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|827391|我将举例说明如何在迭代时删除向量中的元素。|827392|void+test_del_vector(){
++++std::vector<int>+vecInt{0,+1,+2,+3,+4,+5};

++++//method+1
++++for(auto+it+=+vecInt.begin();it+!=+vecInt.end();){
++++++++if(*it+%25+2){//+remove+all+the+odds
++++++++++++it+=+vecInt.erase(it);+//+note+it+will+=+next(it)+after+erase
++++++++}+else{
++++++++++++%2B%2Bit;
++++++++}
++++}

++++//+output+all+the+remaining+elements
++++for(auto+const&+it:vecInt)std::cout<<it;
++++std::cout<<std::endl;

++++//+recreate+vecInt,+and+use+method+2
++++vecInt+=+{0,+1,+2,+3,+4,+5};
++++//method+2
++++for(auto+it=std::begin(vecInt);it!=std::end(vecInt);){
++++++++if+(*it+%25+2){
++++++++++++it+=+vecInt.erase(it);
++++++++}else{
++++++++++++%2B%2Bit;
++++++++}
++++}

++++//+output+all+the+remaining+elements
++++for(auto+const&+it:vecInt)std::cout<<it;
++++std::cout<<std::endl;

++++//+recreate+vecInt,+and+use+method+3
++++vecInt+=+{0,+1,+2,+3,+4,+5};
++++//method+3
++++vecInt.erase(std::remove_if(vecInt.begin(),+vecInt.end(),
+++++++++++++++++[](const+int+a){return+a+%25+2;}),
+++++++++++++++++vecInt.end());

++++//+output+all+the+remaining+elements
++++for(auto+const&+it:vecInt)std::cout<<it;
++++std::cout<<std::endl;

}|code-block|syntax|javascript|827393|输出aw如下：|827394|024
024
024|827395|一种更具生成性的方法：|827396|template<class+Container,+class+F>
void+erase_where(Container&+c,+F&&+f)
{
++++c.erase(std::remove_if(c.begin(),+c.end(),std::forward<F>(f)),
++++++++++++c.end());
}

void+test_del_vector(){
++++std::vector<int>+vecInt{0,+1,+2,+3,+4,+5};
++++//method+4
++++auto+is_odd+=+[](int+x){return+x+%25+2;};
++++erase_where(vecInt,+is_odd);

++++//+output+all+the+remaining+elements
++++for(auto+const&+it:vecInt)std::cout<<it;
++++std::cout<<std::endl;++++
}|827397|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Because the method erase in vector return the next iterator of the passed iterator.

I will give example of how to remove element in vector when iterating.

<pre><code>void test_del_vector(){
 std::vector&lt;int&gt; vecInt{0, 1, 2, 3, 4, 5};

 //method 1
 for(auto it = vecInt.begin();it != vecInt.end();){
 if(*it % 2){// remove all the odds
 it = vecInt.erase(it); // note it will = next(it) after erase
 } else{
 ++it;
 }
 }

 // output all the remaining elements
 for(auto const&amp; it:vecInt)std::cout&lt;&lt;it;
 std::cout&lt;&lt;std::endl;

 // recreate vecInt, and use method 2
 vecInt = {0, 1, 2, 3, 4, 5};
 //method 2
 for(auto it=std::begin(vecInt);it!=std::end(vecInt);){
 if (*it % 2){
 it = vecInt.erase(it);
 }else{
 ++it;
 }
 }

 // output all the remaining elements
 for(auto const&amp; it:vecInt)std::cout&lt;&lt;it;
 std::cout&lt;&lt;std::endl;

 // recreate vecInt, and use method 3
 vecInt = {0, 1, 2, 3, 4, 5};
 //method 3
 vecInt.erase(std::remove_if(vecInt.begin(), vecInt.end(),
 [](const int a){return a % 2;}),
 vecInt.end());

 // output all the remaining elements
 for(auto const&amp; it:vecInt)std::cout&lt;&lt;it;
 std::cout&lt;&lt;std::endl;

}
</code></pre>

output aw below:

<pre><code>024
024
024
</code></pre>

A more generate method:

<pre><code>template&lt;class Container, class F&gt;
void erase_where(Container&amp; c, F&amp;&amp; f)
{
 c.erase(std::remove_if(c.begin(), c.end(),std::forward&lt;F&gt;(f)),
 c.end());
}

void test_del_vector(){
 std::vector&lt;int&gt; vecInt{0, 1, 2, 3, 4, 5};
 //method 4
 auto is_odd = [](int x){return x % 2;};
 erase_where(vecInt, is_odd);

 // output all the remaining elements
 for(auto const&amp; it:vecInt)std::cout&lt;&lt;it;
 std::cout&lt;&lt;std::endl; 
}
</code></pre>

blocks|key|1042280|text|在现代C%2B%2B中，你可以使用"std::remove_if“和lambda表达式；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1042281|此代码将删除向量中的"3“|1042282|vector<int>+vec+{1,2,3,4,5,6};

vec.erase(std::remove_if(begin(vec),end(vec),[](int+elem){return+(elem+==+3);}),+end(vec));|code-block|syntax|javascript|1042283|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Something that you can do with modern C++ is using "std::remove_if" and lambda expression;

This code will remove "3" of the vector

<pre><code>vector&lt;int&gt; vec {1,2,3,4,5,6};

vec.erase(std::remove_if(begin(vec),end(vec),[](int elem){return (elem == 3);}), end(vec));
</code></pre>

I have this code:

<pre><code>int main()
{
 vector&lt;int&gt; res;
 res.push_back(1);
 vector&lt;int&gt;::iterator it = res.begin();
 for( ; it != res.end(); it++)
 {
 it = res.erase(it);
 //if(it == res.end())
 // return 0;
 }
}
</code></pre>

"A random access iterator pointing to the new location of the element that followed the last element erased by the function call, which is the vector end if the operation erased the last element in the sequence."

This code crashes, but if I use the <code>if(it == res.end())</code> portion and then return, it works. How come? Does the for loop cache the <code>res.end()</code> so the not equal operator fails?

Vector erase iterator

我有这样的代码：int main(){ vector<int> res; res.push_back(1); vector<int>::iterator it = res.begin(); for( ; it != res.end(); it++) { it = res.er...

问向量擦除迭代器
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