pop_back()是否真的会使std :: vector上的所有迭代器无效? [英] Does pop_back() really invalidate *all* iterators on an std::vector?

问题描述

std::vector<int> ints;

// ... fill ints with random values

for(std::vector<int>::iterator it = ints.begin(); it != ints.end(); )
{
    if(*it < 10)
    {
        *it = ints.back();
        ints.pop_back();
        continue;
    }
    it++;
}

此代码不起作用，因为调用 pop_back()时， it 无效.但是我在 std :: vector :: pop_back()中找不到任何有关迭代器无效的文档.

This code is not working because when pop_back() is called, it is invalidated. But I don't find any doc talking about invalidation of iterators in std::vector::pop_back().

您是否有与此相关的链接?

Do you have some links about that?

推荐答案

对 pop_back() 删除向量中的最后一个元素，因此对该元素的迭代器无效. pop_back()调用不会使最后一个元素之前的项目的迭代器无效，只有重新分配才能做到.摘自Josuttis的"C ++标准库参考":

The call to pop_back() removes the last element in the vector and so the iterator to that element is invalidated. The pop_back() call does not invalidate iterators to items before the last element, only reallocation will do that. From Josuttis' "C++ Standard Library Reference":

插入或删除元素使引用，指针无效，并且引用以下内容的迭代器元素.如果插入导致重新分配，它会使所有引用，迭代器和指针.

Inserting or removing elements invalidates references, pointers, and iterators that refer to the following element. If an insertion causes reallocation, it invalidates all references, iterators, and pointers.

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