fork() problem

linuxanswer · 04-02-2004, 09:44 AM

watch in this: I want to code a stupid program tha fork 1 child, in this i fork other 2 child that execute pwd and id, and in the end after i wait the 3 child and subchild print the pid of th PARENTS.But i have an error that not understand, why at last i receive also the pid of the first child if i ask only the ppid.Tnx in advance : )

#define PWD "/bin/pwd"
#define ID "/usr/bin/id"

int
main(void)
{
pid_t pid_1, pid_2, pid_3;
int status_1, status_2, status_3;
char *en[1];
char *in[2];

pid_1 = fork();
if (pid_1 == 0) {
fprintf(stdout,"First child with pid :: %i", getpid());

en[0] = ID;
pid_2 = fork();
if (pid_2 == 0) {
fprintf("Second child of child with pid :: %i, and parent pid :: %i", getpid(), getppid());
execv(en[0], en);
exit(0);
}

in[0] = PWD;
pid_2 = fork();
if (pid_2 == 0) {
fprintf("Third child of child with pid :: %i, and parent pid :: %i", getpid(), getppid());
execv(en[0], en);
exit(0);
}
wait(&status_3);
wait(&status_2);
}
wait(&_status1);
fprintf(stdout,"We are in parent with pid :: %i", getpid());
exit( EXIT_SUCCESS );
}

0k n0w if i run obatin this ::

First child with pid :: 1573 //ok
Second child of child with pid :: 1574 and parent pid :: 1573 //ok
/root/System //ok
Second child of child with pid :: 1575 and parent pid :: 1573 //ok
uid=0(root)...........groups=0(root) //ok
We are in parent with pid :: 1573 <- ouCh :/ why i obtain this ???
We are in parent with pid :: 1572 //ok

rose_bud4201 · 04-02-2004, 11:42 AM

Hi there,

I think I know what's going on here.
When you call fork(), both processes (the original one and the newly-created child) begin execution of the rest of the code from the fork() command. And, since you're not guarding your if statements according to which process you want to execute them, you've still got 2 processes left to execute that last if statement. Try it like this:

Code:

#define PWD "/bin/pwd"
#define ID "/usr/bin/id"

int
main(void)
{
    pid_t pid_1, pid_2, pid_3;
    int status_1, status_2, status_3;
    char *en[1];
    char *in[2];

    pid_1 = fork();
    if (pid_1 == 0)  //child 1 will execute this
    { 
        fprintf(stdout,"First child with pid :: %i", getpid());
    
        en[0] = ID;
        pid_2 = fork(); //first "grandchild" will execute this
        if (pid_2 == 0) 
        {
            fprintf("Second child of child with pid :: %i, and parent pid :: %i",     getpid(),getppid());
            execv(en[0], en);
            exit(0);  //kills off grandchild
        }

        in[0] = PWD;
        pid_2 = fork();
        if (pid_2 == 0) 
        {
             fprintf("Third child of child with pid :: %i, and parent pid :: %i", getpid(), getppid());
             execv(en[0], en);
             exit(0);
          }
          wait(&status_3);  //waits for the first grandchild
          wait(&status_2);  //waits for the second grandchild
    }

    //There's no if statement guarding this block - therefore, both the parent process and the first child will execute it.
    wait(&_status1);
    fprintf(stdout,"We are in parent with pid :: %i", getpid());
    exit( EXIT_SUCCESS );
} //ends main

/*-----------------------------------------------------------------
Try: 
....
wait(&status_3);  //waits for the first grandchild
wait(&status_2);  //waits for the second grandchild
....
else if(pid_1 != 0)
    {
       wait(&status_1);  //waits for the first child
       fprintf(stdout, "We are in parent with pid:: %i", getpid());
       exit(EXIT_SUCCESS);
    } //ends if
} //ends main

--------------------------------------------------------------------/*

The best advice, when dealing with fork() and processes, is ALWAYS put EVERYTHING inside if statements. Specifically say which process you want doing things, even if you want both parent and child doing the same thing. It makes things much more readable, not to mention eliminating errors.

Post back if it's still being weird

linuxanswer · 04-02-2004, 05:05 PM

PERFECT PERFECT PERFECT : ) ) tnx & greetings

rose_bud4201 · 04-02-2004, 05:20 PM

Excellent ^_^ You're quite welcome.

Glad it turned out alright,
Laura