remove_if 等效于 std::map [英] remove_if equivalent for std::map

问题描述

我试图根据特定条件从地图中删除一系列元素.我如何使用 STL 算法做到这一点?

I was trying to erase a range of elements from map based on particular condition. How do I do it using STL algorithms?

最初我想使用 remove_if 但这是不可能的，因为 remove_if 不适用于关联容器.

Initially I thought of using remove_if but it is not possible as remove_if does not work for associative container.

是否有任何适用于地图的remove_if"等效算法?

Is there any "remove_if" equivalent algorithm which works for map ?

作为一个简单的选择，我想到了遍历地图并擦除.但是循环遍历地图并擦除安全选项吗?(因为擦除后迭代器无效)

As a simple option, I thought of looping through the map and erase. But is looping through the map and erasing a safe option?(as iterators get invalid after erase)

我使用了以下示例:

bool predicate(const std::pair<int,std::string>& x)
{
    return x.first > 2;
}

int main(void) 
{

    std::map<int, std::string> aMap;

    aMap[2] = "two";
    aMap[3] = "three";
    aMap[4] = "four";
    aMap[5] = "five";
    aMap[6] = "six";

//      does not work, an error
//  std::remove_if(aMap.begin(), aMap.end(), predicate);

    std::map<int, std::string>::iterator iter = aMap.begin();
    std::map<int, std::string>::iterator endIter = aMap.end();

    for(; iter != endIter; ++iter)
    {
            if(Some Condition)
            {
                            // is it safe ?
                aMap.erase(iter++);
            }
    }

    return 0;
}

推荐答案

差不多.

for(; iter != endIter; ) {
     if (Some Condition) {
          iter = aMap.erase(iter);
     } else {
          ++iter;
     }
}

如果您确实从中删除了一个元素，那么您最初拥有的会增加迭代器两次；您可能会跳过需要删除的元素.

What you had originally would increment the iterator twice if you did erase an element from it; you could potentially skip over elements that needed to be erased.

这是我在很多地方看到使用和记录的常用算法.

This is a common algorithm I've seen used and documented in many places.

您是正确的，迭代器在擦除后失效，但只有引用被擦除元素的迭代器，其他迭代器仍然有效.因此在 erase() 调用中使用 iter++.

You are correct that iterators are invalidated after an erase, but only iterators referencing the element that is erased, other iterators are still valid. Hence using iter++ in the erase() call.

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