c - 在 C 问题中使用 time() 和 clock()

Question

我正在做一项编程作业，但我得到了奇怪的结果。这个想法是计算处理器滴答数和运行算法所花费的时间。

通常代码运行得如此之快以至于花费的时间为 0 秒，但我注意到处理器滴答数在开始和结束时为 0，导致处理器滴答数为 0。

我使用 usleep 添加了一个延迟，这样花费的时间不为零，但处理器滴答仍然为零，时间戳之间的计算仍然为零。

几天来我一直在努力解决这个问题，但无法解决这个问题，非常欢迎任何建议。我的代码如下:

/* This program takes an input "n". If n is even it divides n by 2
* If n is odd, it multiples n by 3 and adds 1. Each time through the loop
* it iterates a counter.
* It continues until n is 1
*
* This program will compute the time taken to perform the above algorithm
*/
#include <stdio.h>
#include <time.h>

void delay(int);

int main(void) {
    int n, i = 0;
    time_t start, finish, duration;
    clock_t startTicks, finishTicks, diffTicks;
    printf("Clocks per sec = %d\n", CLOCKS_PER_SEC);
    printf("Enter an integer: ");
    scanf("%d", &n);    // read value from keyboard
    time(&start);       // record start time in ticks
    startTicks = clock();
    printf("Start Clock = %s\n", ctime(&start));
    printf("Start Processor Ticks = %d\n", startTicks);

    while (n != 1) {    // continues until n=1
        i++;    // increment counter
        printf("iterations =%d\t", i);  // display counter iterations
        if (n % 2) {            // if n is odd, n=3n+1
            printf("Input n is odd!\t\t");
            n = (n * 3) + 1;
            printf("Output n = %d\n", n);
            delay(1000000);
        } else {                //if n is even, n=n/2
            printf("Input n is even!\t");
            n = n / 2;
            printf("Output n = %d\n", n);
            delay(1000000);
        }
    }
    printf("n=%d\n", n);
    time(&finish);      // record finish time in ticks
    finishTicks = clock();
    printf("Stop time = %s\n", ctime(&finish));
    printf("Stop Processor Ticks = %d\n", finishTicks);
    duration = difftime(finish, start); // compute difference in time
    diffTicks = finishTicks - startTicks;
    printf("Time elapsed = %2.4f seconds\n", duration);
    printf("Processor ticks elapsed = %d\n", diffTicks);
    return (n);
}

void delay(int us) {
    usleep(us);
}

EDIT: 所以经过进一步研究，我发现 usleep() 不会影响程序运行时间，所以我在 asm 中写了一个延迟函数。现在我得到了处理器滴答的值，但我仍然得到零秒来运行算法。

void delay(int us) {
    for (int i = 0; i < us; i++) {
        __asm__("nop");
    }
}

Answer 1

您可以使用以下公式计算耗时。

double timeDiff  = (double)(EndTime - StartTime) / CLOCKS_PER_SEC.

这是伪代码。

void CalculateTime(clock_t startTime, clock_t endTime)
{
   clock_t diffTime = endTime - startTime;
   printf("Processor time elapsed = %lf\n", (double)diffTime /CLOCKS_PER_SEC);
}

希望这对您有所帮助。