Question

I am new to UNIX programming and i was reading about zombie processes and how to avoid them using fork() twice. I read the code from a book and tried to run it on my system.. I am using ubuntu 12.04. I ran the following code:

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

int main()
{
    pid_t pid;
    if(pid = fork() < 0)
        printf("Fork Error.!!!\n");
    else 
    if(pid == 0)
    {
        if((pid = fork()) < 0)
             printf("Fork2 Error.!!!\n");
        else 
             if(pid > 0)
                   exit(0);

        sleep(2);
        printf("Second Child, parent id: %d\n", getppid());
        exit(0);
    }
    if(waitpid(pid, NULL, 0) != pid)
        printf("Waitpid Error.!!!\n");

    exit(0);
}

The output that i get is as follows:

Second Child, parent id: 1
Second Child, parent id: 1

The book says this should be printed only once and that is also what i feel should happen when i see what’s happening in the code. I dont understand why its getting printed twice. I found this code at many places on the net but could not get something that explains this. Any help is welcome. Thanks.!!

Answer 1

second line of the code in main – missing brackets around pid = fork()

if( ( pid = fork() ) < 0){