处理多个SIGCHLD

2 回答

• 19

  在Linux上，在您使用 signalfd() 读取 SIGCHLD 之前终止的多个子节点将被压缩为单个 SIGCHLD . 这意味着当您阅读 SIGCHLD 信号时，您必须在所有已终止的孩子之后进行清理：

  // Do this after you've read() a SIGCHLD from the signalfd file descriptor:
  while (1) {
      int status;
      pid_t pid = waitpid(-1, &status, WNOHANG);
      if (pid <= 0) {
          break;
      }
      // something happened with child 'pid', do something about it...
      // Details are in 'status', see waitpid() manpage
  }
  

  我应该注意到，当两个子进程同时终止时，我实际上看到了这种信号压缩 . 如果我只做了一个 waitpid() ，其中一个终止的孩子没有被处理;并且上面的循环修复了它 .

  相应文件：

  • http://man7.org/linux/man-pages/man7/signal.7.html "By contrast, if multiple instances of a standard signal are delivered while that signal is currently blocked, then only one instance is queued"

  • http://man7.org/linux/man-pages/man3/sigwait.3p.html "If prior to the call to sigwait() there are multiple pending instances of a single signal number, it is implementation-defined whether upon successful return there are any remaining pending signals for that signal number."

  回复于 2024-04-25T13:44:46+08:00
• 0

  实际上，无障碍的方式是 waitfd 功能，允许您向poll（）/ epoll（）添加特定的pid . 不幸的是，多年前它被提出时并未被Linux接受 .

  回复于 2024-04-25T13:44:46+08:00