c++ - 为什么这个 vector 迭代器不会失效？

Question

我读过一些关于迭代器失效的帖子，似乎需要 vector 重新分配的插入会使迭代器失效。 vector 中间的删除也不应该导致失效吗？

我对此没有清楚的理解，不确定为什么在调整大小并从开始、中间和结束处删除后使用这些迭代器不会破坏它们:

#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main(int argc, char** argv) {

    vector<int> v;
    v.reserve(10);

    for (int i = 0; i < 10; i++)
        v.push_back(i);

    for (auto x = v.begin(); x != v.end(); x++) {
        cout << *x << endl;
    }

    cout << endl << "RESIZE" << endl << endl;

    for (int i = 10; i < 20; i++)
        v.push_back(i);

    for (auto x = v.begin(); x != v.end(); x++) {
        cout << *x << endl;
    }

    cout << endl << "RESIZE 2" << endl << endl;

    for (int i = 20; i < 200; i++)
        v.push_back(i);

    for (auto x = v.begin(); x != v.end(); x++) {
        cout << *x << endl;
    }

    cout << endl << "REMOVES" << endl << endl;

    v.erase(v.begin());
    v.pop_back();
    v.erase(v.begin() + 17);

    for (auto x = v.begin(); x != v.end(); x++) {
        cout << *x << endl;
    }

    return 0;
}

Answer 1

请注意，调用 begin() 或 end() 将始终提供一个合理的迭代器

不过是这样的:

std:vector<int> v;
....

std::vector<int>::iterator i=v.begin();
v.erase(i);
std::cout << *i << std::endl;  // i iterator was invalidated by erasing it
                               // trying to access it or increment it is undefined behaviour.
std::cout << *v.begin() << std::endl;  // begin() provides always a sane iterator.