好的，我一直在和 friend 讨论编译器和程序优化，他建议n * 0.5 比n/2 快。我说编译器会自动做那种优化，所以我写了一个小程序，看看n/2和n * 0.5有没有区别:

部门:

#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, const char * argv[]) {
    int i, m;
    float n, 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", m, s, n);

    return 0;
}

对于这两个版本，我得到了 0.000002s avg。使用 clang main.c -O1 编译时。他说时间测量肯定有问题。于是他又写了一个程序:

#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", m, 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", m, 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", m, 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", m, dT);

    return 0;
}

为此他得到了……

1.869570s
1.868254s
25.674016s
3.497555s

...按这个顺序。

所以我在使用 clang++ main.cpp -O1 编译的机器上运行程序，得到的结果与以前相似:0.000002 到 0.000011。

但是，当我在没有优化的情况下编译程序时，我在他的第一次测试中得到了与他相似的结果。所以我的问题是，任何程度的优化如何使程序 更快？

最佳答案

由于代码在循环的每次迭代中没有做任何不同的事情，编译器可以自由地将循环内的代码移到外面(结果将完全相同)，并完全删除循环，留下你如您所见，运行时间几乎为 0。

关于c++ - 每微秒 1,000,000,000 次计算？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/28592721/