【c++】遍历容器，哪一种方法速度最快？

终于有一个简单的每日一题！写完的时候甚至代码还没有编译结束！刚好借今天的每日一题探究一下一直以来的一些疑惑：容器的遍历。
题目大概是这样的：

我们一眼就看到了容器的遍历！！那么众所周知，容器的遍历最常用的有三种：

1. 基础for，也就是for(int i = 0; i < 10; i++)这种格式
2. c++11 的for
3. algorithm中的for_each

下面蘸着这个题写三种遍历，看看执行时间：

#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

int countSeniors(vector<string> &details)
{
  int cnt = 0;
  for_each(details.begin(), details.end(), [&cnt](const string &detail)
           {
            if (detail[11] > '6' || (detail[11] == '6' && detail[12] != '0')) {
                cnt++;
            } });

  return cnt;
}

int countSeniors2(vector<string> &details)
{
  int cnt = 0;
  for (const auto &detail : details)
  {
    if (detail[11] > '6' || (detail[11] == '6' && detail[12] != '0'))
    {
      cnt++;
    }
  }
  return cnt;
}

int countSeniors3(vector<string> &details)
{
  int cnt = 0;
  for (int i = 0; i < static_cast<int>(details.size()); i++)
  {
    if (details[i][11] > '6' || (details[i][11] == '6' && details[i][12] != '0'))
    {
      cnt++;
    }
  }
  return cnt;
}

int main()
{
  vector<string> details = {"7868190130M7522", "5303914400F9211", "9273338290F4010"};
  cout << setprecision(10);
  {
    clock_t startT, finishT;
    startT = clock();
    countSeniors(details);
    finishT = clock();
    double duration = static_cast<double>(finishT - startT) / CLOCKS_PER_SEC;
    cout << "匿名函数： " << duration << endl;
  }
  {
    clock_t startT, finishT;
    startT = clock();
    countSeniors2(details);
    finishT = clock();
    double duration = static_cast<double>(finishT - startT) / CLOCKS_PER_SEC;
    cout << "c++11 for: " << duration << endl;
  }
  {
    clock_t startT, finishT;
    startT = clock();
    countSeniors3(details);
    finishT = clock();
    double duration = static_cast<double>(finishT - startT) / CLOCKS_PER_SEC;
    cout << "基础for： " << duration << endl;
  }
  {
    clock_t startT, finishT;
    startT = clock();
    for (int i = 0; i < 100; i++)
    {
      countSeniors(details);
    }
    finishT = clock();
    double duration = static_cast<double>(finishT - startT) / CLOCKS_PER_SEC;
    cout << "100 次：匿名函数： " << duration << endl;
  }
  {
    clock_t startT, finishT;
    startT = clock();
    for (int i = 0; i < 100; i++)
    {
      countSeniors2(details);
    }
    finishT = clock();
    double duration = static_cast<double>(finishT - startT) / CLOCKS_PER_SEC;
    cout << "100 次：c++11 for： " << duration << endl;
  }
  {
    clock_t startT, finishT;
    startT = clock();
    for (int i = 0; i < 100; i++)
    {
      countSeniors3(details);
    }
    finishT = clock();
    double duration = static_cast<double>(finishT - startT) / CLOCKS_PER_SEC;
    cout << "100 次：基础for： " << duration << endl;
  }
  return 0;
}

运行一遍，得到了这样的输出：

Amazing！可以看到这场比赛，匿名函数输得一塌糊涂，反而是朴实无华的for夺魁！
这到底是为什么呢？

–没写完，代码跑完了，一会儿干完活再写