每10秒循环 [英] Loop every 10 second

问题描述

如何每10秒加载一个循环，并为计数加+1，并打印？

How to load a loop every 10 second and add +1 to the count and print it?

如：

int count;
    while(true)
    {
       count +=1;
       cout << count << endl; // print every 10 second 
    }

print：

1
2
3
4
5
ect...

我不知道怎么样，请帮助我。

i dont know how, please help me out guys

推荐答案

我的尝试。 （几乎）完美的POSIX。也可以在POSIX和MSVC / Win32上使用。

My try. (Almost) perfectly POSIX. Works on both POSIX and MSVC/Win32 also.

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

const int NUM_SECONDS = 10;

int main()
{
    int count = 1;

    double time_counter = 0;

    clock_t this_time = clock();
    clock_t last_time = this_time;

    printf("Gran = %ld\n", NUM_SECONDS * CLOCKS_PER_SEC);

    while(true)
    {
        this_time = clock();

        time_counter += (double)(this_time - last_time);

        last_time = this_time;

        if(time_counter > (double)(NUM_SECONDS * CLOCKS_PER_SEC))
        {
            time_counter -= (double)(NUM_SECONDS * CLOCKS_PER_SEC);
            printf("%d\n", count);
            count++;
        }

        printf("DebugTime = %f\n", time_counter);
    }

    return 0;
}

这种方式你也可以控制每次迭代， ）的方法。

This way you can also have the control on each iteration, unlike the sleep()-based approach.

此解决方案（或基于高精度计时器的同样）也确保时序中没有错误累积。

This solution (or the same based on high-precision timer) also ensures that there is no error accumulation in timing.

编辑：OSX的东西，如果所有的失败

OSX stuff, if all else fails

#include <unistd.h>
#include <stdio.h>

const int NUM_SECONDS = 10;

int main()
{
    int i;
    int count = 1;
    for(;;)
    {
        // delay for 10 seconds
        for(i = 0 ; i < NUM_SECONDS ; i++) { usleep(1000 * 1000); }
        // print
        printf("%d\n", count++);
    }
    return 0;
}

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