printf比std :: cout快5倍以上？

问题描述

  #include< iostream> 
 #include< cstdlib> 
 #include< cstdio> 
 #include< ctime> 
 
 int main（int argc，char * argv []）
 {
 std :: clock_t start; 
 double duration; 
 
 std :: cout<< 启动std :: cout测试。 << std :: endl; 
 start = std :: clock（）; 
 
 for（int i = 0; i  {
 std :: cout< Hello，World！（<< i<<）< std :: endl; 
} 
 
 duration =（std :: clock（） -  start）/（double）CLOCKS_PER_SEC; 
 
 std :: cout<< 结束std :: cout测试。 << std :: endl; 
 std :: cout<< 所用时间：<持续时间< std :: endl; 
 
 std :: system（pause）; 
 
 std :: cout<< 启动std :: printf测试。 << std :: endl; 
 start = std :: clock（）; 
 
 for（int i = 0; i  {
 std :: printf（Hello，World！（％i）\\\
一世）; 
 std :: fflush（stdout）; 
} 
 
 duration =（std :: clock（） -  start）/（double）CLOCKS_PER_SEC; 
 
 std :: cout<< Ending std :: printf test。 << std :: endl; 
 std :: cout<< 所用时间：<持续时间< std :: endl; 
 
系统（pause）; 
 
 return 0; 
} 
  

现在，这里是前五次运行的时间：

• std :: cout test： 1.125 s; printf测试： 0.195 s


• std :: cout test： 1.154 printf测试： 0.230 s


• std :: cout test： 1.142 printf测试： 0.216 s


• std :: cout test： 1.322 printf测试： 0.221 s


• std :: cout test： 1.108 printf test： 0.232 s

正如你所看到的，使用 printf 然后 fflush ing比使用 std :: cout 大约少5倍的时间。

虽然我希望使用 std :: cout 的< / code>运算符可能有点慢（几乎最小），我没有准备好这个巨大的差异。我做一个公平的测试？如果是这样，那么什么使第一个测试比第二个测试慢得多，如果他们基本上做同样的事情？

解决方案

对于真正的苹果对苹果比较，重写你的测试，以便在测试用例之间改变的只是使用的打印函数：

  int main（int argc，char * argv []）
 {
 const char * teststring =Test output string\ 
 std :: clock_t start; 
 double duration; 
 
 std :: cout<< 启动std :: cout测试。 << std :: endl; 
 start = std :: clock（）; 
 
 for（int i = 0; i  std :: cout<测试串
 / *显示时间结果，代码为简洁修剪* / 
 
 for（int i = 0; i  std :: printf（teststring） ; 
 std :: fflush（stdout）; 
} 
 / *显示时序结果，代码为了简洁* / 
 return 0; 
} 
  

这样，你将只测试 printf 和 cout 函数调用。您不会因多个<< 电话等造成任何差异。如果您尝试这样做，我怀疑您会得到一个不同的结果。 p>


#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <ctime>

int main(int argc, char* argv[])
{
    std::clock_t start;
    double duration;

    std::cout << "Starting std::cout test." << std::endl;
    start = std::clock();

    for (int i = 0; i < 1000; i++)
    {
        std::cout << "Hello, World! (" << i << ")" << std::endl;
    }

    duration = (std::clock() - start) / (double) CLOCKS_PER_SEC;

    std::cout << "Ending std::cout test." << std::endl;
    std::cout << "Time taken: " << duration << std::endl;

    std::system("pause");

    std::cout << "Starting std::printf test." << std::endl;
    start = std::clock();

    for (int i = 0; i < 1000; i++)
    {
        std::printf("Hello, World! (%i)\n", i);
        std::fflush(stdout);
    }

    duration = (std::clock() - start) / (double) CLOCKS_PER_SEC;

    std::cout << "Ending std::printf test." << std::endl;
    std::cout << "Time taken: " << duration << std::endl;

    system("pause");

    return 0;
}

Now, here are the times for the first five runs:

• std::cout test: 1.125 s ; printf test: 0.195 s
• std::cout test: 1.154 s ; printf test: 0.230 s
• std::cout test: 1.142 s ; printf test: 0.216 s
• std::cout test: 1.322 s ; printf test: 0.221 s
• std::cout test: 1.108 s ; printf test: 0.232 s

As you can see, using printf and then fflushing takes about 5 times less time than using std::cout.

Although I did expect using std::cout's << operator to be perhaps a little slower (almost minimal) , I wasn't prepared for this huge difference. Am I making a fair test? If so, then what makes the first test so much slower than the second one, if they essentially do the exact same thing?

解决方案

For a true apples-to-apples comparison, re-write your test so that the only thing changing between the test cases is the print function being used:

int main(int argc, char* argv[])
{
    const char* teststring = "Test output string\n";
    std::clock_t start;
    double duration;

    std::cout << "Starting std::cout test." << std::endl;
    start = std::clock();

    for (int i = 0; i < 1000; i++)
        std::cout << teststring;
    /* Display timing results, code trimmed for brevity */

    for (int i = 0; i < 1000; i++) {
        std::printf(teststring);
        std::fflush(stdout);
    }
    /* Display timing results, code trimmed for brevity */
    return 0;
}

With that, you will be testing nothing but the differences between the printf and cout function calls. You won't incur any differences due to multiple << calls, etc. If you try this, I suspect that you'll get a much different result.

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