使用swap-and-pop迭代时擦除向量中的元素 [英] Erasing element in a vector while iterating using swap-and-pop

问题描述

我想在迭代向量时删除某些元素，但以下代码导致向量迭代器不可递增"断言失败.

I want to erase certain elements while iterating a vector, but the below code results in a "vector iterator not incrementable" assertion failure.

for(auto iter=vec.begin(); iter!=vec.end(); ++iter) 
{     
    if((*iter).isDead())
    {        
        std::swap(*iter, vec.back());//swap with the back       
        vec.pop_back();     //erase the element
    } 
} 

这段代码有什么问题?

推荐答案

您正在增加与当前元素交换的元素；如果那是最后一个元素，那么您刚刚删除了它并使您的迭代器无效.如果你没有擦除，你只想增加，使用类似的东西:

You are incrementing past the element you've swapped with the current element; and if that was the last element, then you have just erased it and invalidated your iterator. You only want to increment if you didn't erase, using something like:

for(auto iter=vec.begin(); iter!=vec.end();) 
{     
    if((*iter).isDead())
    {        
        std::swap(*iter, vec.back());//swap with the back       
        vec.pop_back();     //erase the element
    } else {
        ++iter;
    }
} 

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