c++ - 使用 vector 时，pop_back 是否会连同元素一起删除值？

Question

为什么这样说:

cout << values[0] << " " << values[1] << " " << values[2] << endl;

显示

1 2 3

尽管使用了 pop back 从 vector 中删除最后一个元素。这些值不应该也被删除吗？还是即使删除了元素， vector 也会调整大小？

示例代码如下:

// This program demonstrates the vector pop_back member function.
#include <iostream>
#include <vector>
using namespace std;

int main()
{
    vector<int> values;

    // Store values in the vector.
    values.push_back(1); // Last element in values is 1
    values.push_back(2); // Now elements in values are 1,2
    values.push_back(3); // Now elements in values are 1,2,3

    cout << "The size of values is " << values.size() << endl; // values has 3 elements

    // Remove a value from the vector.
    cout << "Popping a value from the vector...\n";
    values.pop_back();
    cout << "The size of values is now " << values.size() << endl; // 1 is Removed thus size is 2
    cout << values[0] << " " << values[1] << " " << values[2] << endl;

    // Now remove another value from the vector.
    cout << "Popping a value from the vector...\n";
    values.pop_back();
    cout << "The size of values is now " << values.size() << endl;
    cout << values[0] << " " << values[1] << " " << values[2] << endl;


    // Remove the last value from the vector.
    cout << "Popping a value from the vector...\n";
    values.pop_back();
    cout << "The size of values is now " << values.size() << endl;
    cout << values[0] << " " << values[1] << " " << values[2] << endl;

    return 0;

}

Answer 1

“值”和“元素”是一回事。 pop_back() 如果 vector 不为空，则从 vector 中删除最后一个值。

vector 的设计使得程序员有责任不越界访问 vector。如果一个 vector 有 2 个元素并且您尝试通过 at() 以外的任何方法访问第三个元素，您会导致 undefined behaviour .

要进行边界检查，请使用 values.at(0) 而不是 values[0] 等，并包含一个 try ...catch block 以捕获生成的异常。