从std :: vector中删除,同时为每个? [英] Erasing from a std::vector while doing a for each?
问题描述
迭代的正确方法是使用迭代器。然而,我认为通过擦除,迭代器是无效的。
基本上我想做的是:
for(iterator it = begin; it!= end; ++ it)
{
if(it-> somecondition())
{
erase it
}
}
感谢
struct RemoveTimedEvent
{
bool operator()(const AguiTimedEvent& pX,AguiWidgetBase * widget)const
{
return pX.getCaller()== widget;
}
};
void AguiWidgetContainer :: clearTimedEvents(AguiWidgetBase * widget)
{
std :: vector< AguiTimedEvent> :: iterator it = std :: remove_if(timedEvents.begin
timedEvents.end(),RemoveTimedEvent());
timedEvents.erase(it,timedEvents.end());
}
erase()
返回一个新的迭代器:
for(iterator it = begin; it!= end(container)/ * !!! * /;)
{
if(it-> somecondition())
{
it = vec.erase它); //返回新的迭代器继续。
}
else
{
++ it;
}
}
请注意,我们无法再将其与预先计算的结束,因为我们可能会擦除它,因此使它无效。
一个更好的方法可能是结合 std :: remove_if
code> erase()。您从O(N 2 )变为O(N):
iterator it = std :: remove_if(begin,end,pred);
vec.erase(it,vec.end());
其中 pred
是您的移除谓词as:
struct predicate //做一个更好的名字
{
bool operator T& pX)const //用你的类型替换T
{
return pX.shouldIBeRemoved();
}
};
iterator it = std :: remove_if(begin,end,predicate());
vec.erase(it,vec.end());
在你的情况下,你可以使它一般:
class remove_by_caller
{
public:
remove_by_caller(AguiWidgetBase * pWidget):
mWidget(pWidget)
{}
//如果每一个有getCaller的东西都有一个基础,那么使用
template< typename T> //现在为模板
bool operator()(const T& pX)const
{
return pX.getCaller()== mWidget;
}
private:
AguiWidgetBase * mWidget;
};
std :: vector< AguiTimedEvent> :: iterator it =
std :: remove_if(timedEvents.begin(),timedEvents.end(),remove_by_caller(widget));
timedEvents.erase(it,timedEvents.end());
注意lambda存在以简化Boost和C ++ 11中的这个过程。
The proper way to iterate is to use iterators. However, I think by erasing, the iterator is invalidated.
Basically what I want to do is:
for(iterator it = begin; it != end; ++it)
{
if(it->somecondition() )
{
erase it
}
}
How could I do this without v[i] method?
Thanks
struct RemoveTimedEvent
{
bool operator()(const AguiTimedEvent& pX, AguiWidgetBase* widget) const
{
return pX.getCaller() == widget;
}
};
void AguiWidgetContainer::clearTimedEvents( AguiWidgetBase* widget )
{
std::vector<AguiTimedEvent>::iterator it = std::remove_if(timedEvents.begin(),
timedEvents.end(), RemoveTimedEvent());
timedEvents.erase(it, timedEvents.end());
}
erase()
returns a new iterator:
for(iterator it = begin; it != end(container) /* !!! */;)
{
if (it->somecondition())
{
it = vec.erase(it); // Returns the new iterator to continue from.
}
else
{
++it;
}
}
Note that we can no longer compare it against a precalculated end, because we may erase it and therefore invalidate it. We must get the end explicitly each time.
A better method might be to combine std::remove_if
and erase()
. You change from being O(N2) (every element gets erased and shifted as you go) to O(N):
iterator it = std::remove_if(begin, end, pred);
vec.erase(it, vec.end());
Where pred
is your removal predicate, such as:
struct predicate // do choose a better name
{
bool operator()(const T& pX) const // replace T with your type
{
return pX.shouldIBeRemoved();
}
};
iterator it = std::remove_if(begin, end, predicate());
vec.erase(it, vec.end());
In your case, you can make it pretty general:
class remove_by_caller
{
public:
remove_by_caller(AguiWidgetBase* pWidget) :
mWidget(pWidget)
{}
// if every thing that has getCaller has a base, use that instead
template <typename T> // for now a template
bool operator()(const T& pX) const
{
return pX.getCaller() == mWidget;
}
private:
AguiWidgetBase* mWidget;
};
std::vector<AguiTimedEvent>::iterator it =
std::remove_if(timedEvents.begin(), timedEvents.end(), remove_by_caller(widget));
timedEvents.erase(it, timedEvents.end());
Note lambda's exist to simplify this process, both in Boost and C++11.
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