如果你在从开始到结束迭代时在地图元素上调用erase()会发生什么? [英] What happens if you call erase() on a map element while iterating from begin to end?
问题描述
在下面的代码中,我通过一个映射循环,并测试一个元素是否需要被删除。是否可以擦除元素并保持迭代,或者我需要收集另一个容器中的键,并执行第二个循环调用erase()?
In the following code I loop through a map and test if an element needs to be erased. Is it safe to erase the element and keep iterating or do I need to collect the keys in another container and do a second loop to call the erase()?
map<string, SerialdMsg::SerialFunction_t>::iterator pm_it;
for (pm_it = port_map.begin(); pm_it != port_map.end(); pm_it++)
{
if (pm_it->second == delete_this_id) {
port_map.erase(pm_it->first);
}
}
更新:当然, http://stackoverflow.com/questions/52714/stl-vector-vs-map-erase\">读这个问题,我不认为会相关,但回答我的问题。
UPDATE: Of course, I then read this question which I didn't think would be related but answers my question.
推荐答案
C ++ 11
擦除已在所有容器类型中得到改进/保持一致。)
擦除方法现在返回下一个迭代器。
C++11
This has been fixed in C++11 (or erase has been improved/made consistent across all container types).
The erase method now returns the next iterator.
auto pm_it = port_map.begin();
while(pm_it != port_map.end())
{
if (pm_it->second == delete_this_id)
{
pm_it = port_map.erase(pm_it);
}
else
{
++pm_it;
}
}
C ++ 03
删除地图中的元素不会使任何迭代器无效。
(除了被删除的元素上的迭代器)
C++03
Erasing elements in a map does not invalidate any iterators.
(apart from iterators on the element that was deleted)
实际插入或删除不会使任何迭代器无效:
Actually inserting or deleting does not invalidate any of the iterators:
也看到这个答案:
Mark Ransom技术
但您需要更新代码:
你的代码在调用erase后增加pm_it。
But you do need to update your code:
In your code you increment pm_it after calling erase. At this point it is too late and is already invalidated.
map<string, SerialdMsg::SerialFunction_t>::iterator pm_it = port_map.begin();
while(pm_it != port_map.end())
{
if (pm_it->second == delete_this_id)
{
port_map.erase(pm_it++); // Use iterator.
// Note the post increment.
// Increments the iterator but returns the
// original value for use by erase
}
else
{
++pm_it; // Can use pre-increment in this case
// To make sure you have the efficient version
}
}
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