fork child and parent processes

Question

I am trying to create a program that uses fork   to create a new process  The sample output should look like so   This is the child process  My pid is 733 and my parent s id is 772  This is the parent process  My pid is 772 and my child s id is 773   This is how I coded my program    include  lt stdio h gt   include  lt stdlib h gt   int main         printf  This is the child process  My pid is  d and my parent s id is  d  n   getpid    fork          return 0      This results in the output   This is the child process  My pid is 22163 and my parent s id is 0  This is the child process  My pid is 22162 and my parent s id is 22163   Why is it printing the statement twice and how can I get it to properly show the parent s id after the child id displays in the first sentence   EDIT    include  lt stdio h gt   include  lt stdlib h gt   int main     int pid   fork     if  pid    0        printf  This is the child process  My pid is  d and my parent s id is  d  n   getpid    getppid       else       printf  This is the parent process  My pid is  d and my parent s id is  d  n   getpid    pid      return 0

User · Accepted Answer

Start by reading the fork man page as well as the getppid / getpid man pages.

From fork's

On success, the PID of the child process is returned in the parent's thread of execution, and a 0 is returned in the child's thread of execution. On failure, a -1 will be returned in the parent's context, no child process will be created, and errno will be set appropriately.

So this should be something down the lines of

if ((pid=fork())==0){
    printf("yada yada %u and yada yada %u",getpid(),getppid());
}
else{ /* avoids error checking*/
    printf("Dont yada yada me, im your parent with pid %u ", getpid());
}

As to your question:

This is the child process. My pid is 22163 and my parent's id is 0.

This is the child process. My pid is 22162 and my parent's id is 22163.

fork() executes before the printf. So when its done, you have two processes with the same instructions to execute. Therefore, printf will execute twice. The call to fork() will return 0 to the child process, and the pid of the child process to the parent process.

You get two running processes, each one will execute this ~~instruction~~ statement:

printf ("... My pid is %d and my parent's id is %d",getpid(),0);

and

printf ("... My pid is %d and my parent's id is %d",getpid(),22163);

~

To wrap it up, the above line is the child, specifying its pid. The second line is the parent process, specifying its id (22162) and its child's (22163).

User · Answer

We control fork   process call by if  else statement  See my code below   int main        int forkresult  parent ID      forkresult fork       if forkresult   0                printf   I am the parent my ID is    d    getpid             printf   and my child ID is    d n    forkresult           parent ID   getpid        if forkresult   0        printf   I am the  child ID is    d  getpid        else       printf   and my parent ID is    d   parent ID

User · Answer

It is printing the statement twice because it is printing it for both the parent and the child   The parent has a parent id of 0  Try something like this    pid t  pid   pid   fork     if  pid    0       printf  This is the child process  My pid is  d and my parent s id is  d  n   getpid   getppid      else      printf  This is the parent process  My pid is  d and my parent s id is  d  n   getpid    getppid

User · Answer

This is the correct way for getting the correct output     However  childs parent id maybe sometimes printed as 1 because parent process gets terminated and the root process with pid   1 controls this orphan process    pid t  pid   pid   fork     if  pid    0       printf  This is the child process  My pid is  d and my parent s id        is  d  n   getpid    getppid      else       printf  This is the parent process  My pid is  d and my parent s           id is  d  n   getpid    pid

User · Answer

It is printing twice because you are calling printf twice  once in the execution of your program and once in the fork   Try taking your fork   out of the printf call

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