更快，用vector :: iterator或at（）迭代一个STL向量？ [英] What's faster, iterating an STL vector with vector::iterator or with at()?

问题描述

在性能方面，什么会更快？有区别吗？它是平台依赖吗？

In terms of performance, what would work faster? Is there a difference? Is it platform dependent?

//1. Using vector<string>::iterator:
vector<string> vs = GetVector();

for(vector<string>::iterator it = vs.begin(); it != vs.end(); ++it)
{
   *it = "Am I faster?";
}

//2. Using size_t index:
for(size_t i = 0; i < vs.size(); ++i)
{
   //One option:
   vs.at(i) = "Am I faster?";
   //Another option:
   vs[i] = "Am I faster?";
}

推荐答案

找出？

编辑：我的错误 - 我以为我在计时优化版本，但没有。在我的机器上，使用g ++ -O2编译，迭代器版本比operator []版本稍微慢，但可能不是那么明显。

My bad - I thought I was timing the optimised version but wasn't. On my machine, compiled with g++ -O2, the iterator version is slightly slower than the operator[] version, but probably not significantly so.

#include <vector>
#include <iostream>
#include <ctime>
using namespace std;

int main() {
    const int BIG = 20000000;
    vector <int> v;
    for ( int i = 0; i < BIG; i++ ) {
        v.push_back( i );
    }

    int now = time(0);
    cout << "start" << endl;
    int n = 0;
    for(vector<int>::iterator it = v.begin(); it != v.end(); ++it) {
        n += *it;
    }

    cout << time(0) - now << endl;
    now = time(0);
    for(size_t i = 0; i < v.size(); ++i) {
        n += v[i];
    }
    cout << time(0) - now << endl;

    return n != 0;
}

这篇关于更快，用vector :: iterator或at（）迭代一个STL向量？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！