APUE 说
由于进程组在父进程终止时是孤立的,因此 POSIX.1 要求向新孤立进程组中停止的每个进程(就像我们的子进程一样)发送挂断信号 (SIGHUP),然后发送继续信号 (SIGCONT)
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <signal.h>
#include <errno.h>
#define errexit(msg) do{ perror(msg); exit(EXIT_FAILURE); } while(0)
static void sig_hup(int signo)
{
printf("SIGHUP received, pid = %d\n", getpid());
}
static void sig_cont(int signo)
{
printf("SIGCONT received, pid = %d\n", getpid());
}
static void sig_ttin(int signo)
{
printf("SIGTTIN received, pid = %d\n", getpid());
}
static void pr_ids(char *name)
{
printf("%s: pid = %d, ppid = %d, pgrp = %d, tpgrp = %d\n",
name, getpid(), getppid(), getpgrp(), tcgetpgrp(STDIN_FILENO));
}
int main(int argc, char *argv[])
{
char c;
pid_t pid;
setbuf(stdout, NULL);
pr_ids("parent");
if ((pid = fork()) < 0) {
errexit("fork error");
} else if (pid > 0) { /* parent */
sleep(5);
printf("parent exit\n");
exit(0);
} else { /* child */
pr_ids("child...1");
signal(SIGCONT, sig_cont);
signal(SIGHUP, sig_hup);
signal(SIGTTIN, sig_ttin);
kill(getpid(), SIGTSTP);
//sleep(10);
pr_ids("child...2");
if (read(STDIN_FILENO, &c, 1) != 1) {
printf("read error from controlling TTY, errno = %d\n",
errno);
}
printf("child exit\n");
}
exit(0);
}
程序输出:
parent: pid = 2036, ppid = 1959, pgrp = 2036, tpgrp = 2036
child...1: pid = 2037, ppid = 2036, pgrp = 2036, tpgrp = 2036
parent exit
xiejingfeng@xiejingfeng-desktop:/codes/apue $ SIGCONT 收到,pid = 2037
SIGHUP 收到,pid = 2037
child...2: pid = 2037, ppid = 1, pgrp = 2036, tpgrp = 1959
从控制 TTY 读取错误,errno = 5
child exit
输出不像书中所说的那样,因为程序首先接收 SIGCONT 然后 SIGHUP,这让我很困惑,你们能帮帮我吗?
提前致谢。