简单的信号 - C编程及报警功能 [英] Simple Signals - C programming and alarm function

问题描述

#include  <stdio.h>
#include  <signal.h>


void  ALARMhandler(int sig)
{
  signal(SIGALRM, SIG_IGN);          /* ignore this signal       */
  printf("Hello");
  signal(SIGALRM, ALARMhandler);     /* reinstall the handler    */
}

int main(int argc, char *argv[])
{
  alarm(2);                     /* set alarm clock          */
  while (1)
    ;
  printf("All done");
}

我希望该计划到2秒后打印你好，而是输出的zsh：报警./a.out

I expect the program to print "hello" after 2 seconds, but instead the output is "zsh: alarm ./a.out"

任何想法是怎么回事？

推荐答案

你忘了最初设定的报警处理程序。改变的开始的main（）这样的：

You're forgetting to set the alarm handler initially. Change the start of main() like:

int main(int argc, char *argv[])
{
   signal(SIGALRM, ALARMhandler);
   ...

此外，信号处理程序可能会打印什么。这是因为C库高速缓存输出，直到它看到一行结束。所以：

Also, the signal handler will probably print nothing. That's because the C library caches output until it sees an end of line. So:

void  ALARMhandler(int sig)
{
  signal(SIGALRM, SIG_IGN);          /* ignore this signal       */
  printf("Hello\n");
  signal(SIGALRM, ALARMhandler);     /* reinstall the handler    */
}

有关现实世界的计划，从信号处理印刷也不是很安全。一个信号处理器应该做的少，因为它可以，preferably只有在这里或那里设置一个标志。和国旗应该声明挥发性。

For a real-world program, printing from a signal handler is not very safe. A signal handler should do as little as it can, preferably only setting a flag here or there. And the flag should be declared volatile.

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