vector<bool>性能测试

cpp 复制代码
#include <iostream>
#include <vector>
#include <sstream>
#include <numeric>
#include <string>
#include <unordered_map>
#include <queue>
#include <map>
using namespace std;


int main()
{
    clock_t t1, t2, t3;
    vector<bool> vb;
    vector<int> vi;
    vector<char> vc;

    t1 = clock();
    for (int i = 0; i < 1000; ++i)
    {
        for (int j = 0; j < 1000; ++j)
        {
            vb.push_back(true);
        }
    }
    t1 = clock() - t1;

    t2 = clock();
    for (int i = 0; i < 1000; ++i)
    {
        for (int j = 0; j < 1000; ++j)
        {
            vi.push_back(1);
        }
    }
    t2 = clock() - t2;

    t3 = clock();
    for (int i = 0; i < 1000; ++i)
    {
        for (int j = 0; j < 1000; ++j)
        {
            vc.push_back('a');
        }
    }
    t3 = clock() - t3;

    cout << "'vector<bool>::push_back(true)' 1000000 times cost: " << t1 << endl;
    cout << "'vector<int>::push_back(1)' 1000000 times cost: " << t2 << endl;
    cout << "'vector<char>::push_back('a')' 1000000 times cost: " << t3 << endl;

    t1 = clock();
    for (int i = 0; i < 1000; ++i)
    {
        for (int j = 0; j < 1000; ++j)
        {
            bool b = vb[j + 1000 * i];
        }
    }
    t1 = clock() - t1;

    t2 = clock();
    for (int i = 0; i < 1000; ++i)
    {
        for (int j = 0; j < 1000; ++j)
        {
            int b = vi[j + 1000 * i];
        }
    }
    t2 = clock() - t2;

    t3 = clock();
    for (int i = 0; i < 1000; ++i)
    {
        for (int j = 0; j < 1000; ++j)
        {
            char b = vc[j + 1000 * i];
        }
    }
    t3 = clock() - t3;

    cout << "'vector<bool>::operator[]' 1000000 times cost: " << t1 << endl;
    cout << "'vector<int>::operator[]' 1000000 times cost: " << t2 << endl;
    cout << "'vector<char>::operator[]' 1000000 times cost: " << t3 << endl;

    return 0;
}

测试结果

bash 复制代码
'vector<bool>::push_back(true)' 1000000 times cost: 4824
'vector<int>::push_back(1)' 1000000 times cost: 62
'vector<char>::push_back('a')' 1000000 times cost: 59
'vector<bool>::operator[]' 1000000 times cost: 312
'vector<int>::operator[]' 1000000 times cost: 5
'vector<char>::operator[]' 1000000 times cost: 6

原理:

vector<bool> 源码分析

相关推荐
YuTaoShao13 分钟前
【LeetCode 热题 100】73. 矩阵置零——(解法二)空间复杂度 O(1)
java·算法·leetcode·矩阵
Heartoxx14 分钟前
c语言-指针(数组)练习2
c语言·数据结构·算法
zzywxc78716 分钟前
AI 正在深度重构软件开发的底层逻辑和全生命周期,从技术演进、流程重构和未来趋势三个维度进行系统性分析
java·大数据·开发语言·人工智能·spring
大熊背28 分钟前
图像处理专业书籍以及网络资源总结
人工智能·算法·microsoft
满分观察网友z32 分钟前
别怕树!一层一层剥开它的心:用BFS/DFS优雅计算层平均值(637. 二叉树的层平均值)
算法
江理不变情35 分钟前
图像质量对比感悟
c++·人工智能
灵性花火36 分钟前
Qt的前端和后端过于耦合(0/7)
开发语言·前端·qt
DES 仿真实践家1 小时前
【Day 11-N22】Python类(3)——Python的继承性、多继承、方法重写
开发语言·笔记·python
apocelipes2 小时前
记一次ADL导致的C++代码编译错误
c++·开发工具和环境
杰克尼2 小时前
1. 两数之和 (leetcode)
数据结构·算法·leetcode