c - c程序计算执行时间出错？

Question

谁能告诉我如果我在下面的几行中错过双倍会发生什么？

time_spent = (double)(end - begin) / CLOCKS_PER_SEC;

我试图计算我的排序问题的运行时间，却忘记了类型转换为加倍。代码运行了将近 90 分钟，但打印的输出时间为“270.000006”。谁能帮我弄清楚这 270 意味着什么？

它有时也会显示“-ve”值。这个问题的任何解决方案。

编辑 - 我正在对 10^9 和 10^10 数字进行排序，因此代码将运行数小时。

Answer 1

如果代码运行了将近 90 分钟，您将在 72 分钟内获得 32 位架构上的 clock_t 类型溢出。我相信这是你的情况。

Answer 2

Correct usage is:

    clock_t begin, end;
    double time_spent;

    begin = clock();
    /* actual task that needs to be monitored */
    end = clock();
    time_spent = (double)(end - begin) / CLOCKS_PER_SEC; // in seconds

You are measuring CPU time here, not the elapsed time which includes I/O time as well. CLOCKS_PER_SEC is a constant which is declared in . The computation needs to be done in floating point arithmetic. Another alternative to compute timing is to use the time command.

Read How to log the time taken for a unix command?