fork两次避免产生僵尸进程[c/c++]

(APUE)Figure 8.8. Avoid zombie processes by calling fork twice

#include <stdio.h>
#include <sys/wait.h>
#include <sys/types.h>

int main(void)
{
   pid_t pid;

    if ((pid = fork()) < 0)
    {
        fprintf(stderr,"Fork error!\n");
        exit(-1);
    }
    else if (pid == 0) /* first child */
    {
        if ((pid = fork()) < 0)
        {
            fprintf(stderr,"Fork error!\n");
            exit(-1);
        }
        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 */
    {
        fprintf(stderr,"Waitpid error!\n");
        exit(-1);
    }

    /*
     * We're the parent (the original process); we continue executing,
     * knowing that we're not the parent of the second child.
     */
    exit(0);
}