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原帖由 无声无息 于 2008-10-24 16:41 发表 ![]()
平时需要在后台执行任务的时候，常使用nohup或者screen之类的东东，确实好用。
不过今天有一个情况有点不解，一直都这么认为
1、nohup command 登录后执行command,logout后,此command不受影响的继续运行
2 ...
调用两次fork()就会出现你说的情况
APUE上的一段
- If we want to write a process so that it forks a child but we don't want to wait for the child to complete and we don't want the child to become a zombie until we terminate, the trick is to call fork twice. The program in Figure 8.8 does this.
- We call sleep in the second child to ensure that the first child terminates before printing the parent process ID. After a fork, either the parent or the child can continue executing; we never know which will resume execution first. If we didn't put the second child to sleep, and if it resumed execution after the fork before its parent, the parent process ID that it printed would be that of its parent, not process ID 1.
- Executing the program in Figure 8.8 gives us
- $ ./a.out
- $ second child, parent pid = 1
- Note that the shell prints its prompt when the original process terminates, which is before the second child prints its parent process ID.
- Figure 8.8. Avoid zombie processes by calling fork twice
- #include "apue.h"
- #include <sys/wait.h>
- int
- main(void)
- {
- pid_t pid;
- if ((pid = fork()) < 0) {
- err_sys("fork error");
- } else if (pid == 0) { /* first child */
- if ((pid = fork()) < 0)
- err_sys("fork error");
- else if (pid > 0)
- exit(0); /* parent from second fork == first child */
- /*
- * We're the second child; our parent becomes init as soon
- * as our real parent calls exit() in the statement above.
- * Here's where we'd continue executing, knowing that when
- * we're done, init will reap our status.
- */
- sleep(2);
- printf("second child, parent pid = %d\n", getppid());
- exit(0);
- }
-
- if (waitpid(pid, NULL, 0) != pid) /* wait for first child */
- err_sys("waitpid error");
- /*
- * We're the parent (the original process); we continue executing,
- * knowing that we're not the parent of the second child.
- */
- exit(0);
- }
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发表于 2008-10-24 16:41
