printf比std :: cout快5倍多？

Question

#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;
}

现在,这是前五次运行的时间:

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

如您所见,使用printf然后使用fflush比使用时间少5倍std::cout.

虽然我确实期望使用std::cout's <<运算符可能稍慢(几乎是最小),但我并没有为这个巨大的差异做好准备.我做了一个公平的测试吗？如果是这样,那么是什么让第一次测试比第二次测试慢得多,如果他们基本上做同样的事情呢？

Answer 1

试试这个:

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

int main(int argc, char* argv[])
{
#if defined(NOSYNC)
    std::cout.sync_with_stdio(false);
#endif

    std::cout << "Starting std::cout test." << std::endl;

    std::clock_t start = std::clock();

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

    clock_t mid = std::clock();

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

    std::clock_t end = std::clock();

    std::cout << "Time taken: P1 " << ((mid-start)*1.0/CLOCKS_PER_SEC) << std::endl;

    std::cout << "Time taken: P2 " << ((end-mid)*1.0/CLOCKS_PER_SEC) << std::endl;


    return 0;
}

然后我得到:

> g++ -O3 t13.cpp
> ./a.out
# lots of lines deleted
Time taken: P1 0.002517
Time taken: P2 0.001872

> g++ -O3 t13.cpp -DNOSYNC   
> ./a.out
# lots of lines deleted
Time taken: P1 0.002398
Time taken: P2 0.001878

所以P2时间不会改变.
但是你使用了P1时间(即std :: cout)的改进std::cout.sync_with_stdio(false);.因为代码不再试图保持两个流(std :: cout stdout)同步.如果您正在编写纯C++并且仅使用std :: cout而不是问题.

Answer 2

对于真正的苹果对苹果比较,重新编写测试,以便测试用例之间唯一的变化是使用的打印功能:

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;
}

有了它,你将只测试printf和cout函数调用之间的差异.由于多次<<通话等原因,您不会产生任何差异.如果您尝试这样做,我怀疑您会得到一个非常不同的结果.