C++ move () : what's left in the vector?

Question

我有一段代码，在 vector 中, 元素是成对的 int和 string .然后我想从 vector move 所有元素进入 unordered_map<int, string> :

#include <algorithm>
#include <iostream>
#include <iterator>
#include <unordered_map>
#include <vector>
using namespace std;

template <typename C>
void print(const C& container) {
    for (const auto& ele : container) {
        cout << "(" << ele.first << ", " << ele.second << "), ";
    }
    cout << endl;
}

int main() {
    vector<pair<int, string>> v {
        {1, "one"},
        {2, "two"},
        {3, "three"},
        {4, "four"},
        {5, "five"}
    };
    unordered_map<int, string> uMap;

    move(begin(v), end(v), inserter(uMap, begin(uMap)));

    cout << "In unordered_map:" << endl;
    print(uMap);
    cout << endl << "In vector:" << endl;
    print(v);

    return 0;
}

我不明白的是结果:

In unordered_map:
(5, five), (4, four), (3, three), (2, two), (1, one), 
In vector:
(1, ), (2, ), (3, ), (4, ), (5, ),

为什么那些整数留在vector中？我以为move()函数将从 vector move 所有元素进入unordered_map , 这样 vector 中就不会留下任何东西了？

Answer 1

来自 cppreference :

Moves the elements in the range [first, last), to another range beginning at d_first, starting from first and proceeding to last - 1. After this operation the elements in the moved-from range will still contain valid values of the appropriate type, but not necessarily the same values as before the move.

因为该算法在迭代器上运行，所以它实际上无法从 vector 中删除元素。它们仍然存在，并且可以具有与以前相同的值(但不一定)。

此外，字符串仍然存在于 vector 中。从 std::string move 后，它处于有效但未指定的状态。在这种情况下，字符串为空。