c – 每微秒1,000,000,000次计算？

好的,我一直在和一个朋友谈论编译器和优化程序,他建议n * 0.5比n / 2要快.我说编译器自动做这种优化,所以我写了一个小程序来看如果n / 2和n * 0.5之间有差异：

师：

#include <stdio.h>
#include <time.h>

int main(int argc,const char * argv[]) {
    int i,m;
    float n,s;
    clock_t t;

    m = 1000000000;
    t = clock();
    for(i = 0; i < m; i++) {
        n = i / 2;
    }
    s = (float)(clock() - t) / CLOCKS_PER_SEC;

    printf("n = i / 2: %d calculations took %f seconds (last calculation = %f)\n",m,s,n);

    return 0;
}

乘法：

#include <stdio.h>
#include <time.h>

int main(int argc,s;
    clock_t t;

    m = 1000000000;
    t = clock();
    for(i = 0; i < m; i++) {
        n = i * 0.5;
    }
    s = (float)(clock() - t) / CLOCKS_PER_SEC;

    printf("n = i * 0.5: %d calculations took %f seconds (last calculation = %f)\n",n);

    return 0;
}

而对于这两个版本,我得到了0.000002平均.当编译时使用clang main.c -O1.他说时间测量一定有问题.于是他写了一个程序：

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

using namespace std;

int main()
{
    clock_t ts,te;
    double  dT;

    int i,m;
    double n,o,p,q,r,s;
    m = 1000000000;

    cout << "Independent calculations:\n";
    ts = clock();
    for (i = 0; i < m; i++)
    {
        //  make it a trivial pure float calculation with no int casting to float
        n = 11.1 / 2.3;
        o = 22.2 / 2.3;
        p = 33.3 / 2.3;
        q = 44.4 / 2.3;
        r = 55.5 / 2.3;
        s = 66.6 / 2.3;
    }

    te = clock();
    dT = ((float)(te - ts)) / CLOCKS_PER_SEC;   //  make initial call to get the elapsed time to run the loop
    ts = clock();

    printf("Division: %d calculations took %f seconds\n",dT);

    for (i = 0; i < m; i++)
    {
        //  make it a trivial pure float calculation with no int casting to float
        n = 11.1 * 0.53;
        o = 22.2 * 0.53;
        p = 33.3 * 0.53;
        q = 44.4 * 0.53;
        r = 55.5 * 0.53;
        s = 66.6 * 0.53;
    }

    te = clock();
    dT = ((float)(te - ts)) / CLOCKS_PER_SEC;   //  make initial call to get the elapsed time to run the loop
    ts = clock();

    printf("Multiplication: %d calculations took %f seconds\n",dT);

    cout << "\nDependent calculations:\n";
    for (i = 0; i < m; i++)
    {
        //  make it a trivial pure float calculation with no int casting to float
        n = 11.1 / 2.3;
        o = n / 2.3;
        p = o / 2.3;
        q = p / 2.3;
        r = q / 2.3;
        s = r / 2.3;
    }


    te = clock();
    dT = ((float)(te - ts)) / CLOCKS_PER_SEC;   //  make initial call to get the elapsed time to run the loop
    ts = clock();

    printf("Division: %d calculations took %f seconds\n",dT);

    for (i = 0; i < m; i++)
    {
        //  make it a trivial pure float calculation with no int casting to float
        n = 11.1 * 0.53;
        o = n * 0.53;
        p = o * 0.53;
        q = p * 0.53;
        r = q * 0.53;
        s = r * 0.53;
    }


    te = clock();
    dT = ((float)(te - ts)) / CLOCKS_PER_SEC;   //  make initial call to get the elapsed time to run the loop
    ts = clock();

    printf("Multiplication: %d calculations took %f seconds\n",dT);

    return 0;
}

他得到了

1.869570s
1.868254s
25.674016s
3.497555s

…以该顺序.

所以我运行的程序在我的机器编译与clang main.cpp -O1,我得到类似的结果与以前：0.000002到0.000011.

但是,当我没有优化编译程序时,在他的第一次测试中,我得到了类似的结果.所以我的问题是,任何数量的优化如何使程序更快？

解决方法

由于代码在循环的每次迭代中都不会做任何不同的操作,所以编译器可以自由地将循环中的代码移动到外部(结果将完全相同),并且完全删除循环,让您几乎为0运行时,如你所见.

原文地址：https://www.jb51.cc/c/112968.html