从std :: vector中删除,同时为每个执行a？

Question

迭代的正确方法是使用迭代器.但是,我认为通过擦除,迭代器无效.

基本上我想做的是:

for(iterator it = begin; it != end; ++it)
{
    if(it->somecondition() )
    {
     erase it
    }

}

没有v [i]方法我怎么能这样做？

谢谢

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());

}

Answer 1

erase() 返回一个新的迭代器:

for(iterator it = begin; it != end(container) /* !!! */;)
{
    if (it->somecondition())
    {
        it = vec.erase(it);  // Returns the new iterator to continue from.
    }
    else
    {
        ++it;
    }
}

请注意,我们无法再将其与预先计算的结果进行比较,因为我们可能会将其删除,从而使其无效.我们必须每次都明确地结束.

一个更好的方法可能是结合std::remove_if和erase().你从O(N ²)(每个元素被擦除并随着你移动)变为O(N):

iterator it = std::remove_if(begin, end, pred);
vec.erase(it, vec.end());

pred你的删除谓词在哪里,例如:

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());

在你的情况下,你可以使它非常一般:

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());

注意在Boost和C++ 11中都存在lambda来简化这个过程.