I'm starting to learn some C and while studying the fork, wait functions I got to a unexpected output. At least for me.
Is there any way to create only 2 child processes from the parent?
Here my code:
#include <sys/types.h>
#include <stdio.h>
#include <unistd.h>
#include <sys/wait.h>
int main ()
{
/* Create the pipe */
int fd [2];
pipe(fd);
pid_t pid;
pid_t pidb;
pid = fork ();
pidb = fork ();
if (pid < 0)
{
printf ("Fork Failed\n");
return -1;
}
else if (pid == 0)
{
//printf("I'm the child\n");
}
else
{
//printf("I'm the parent\n");
}
printf("I'm pid %d\n",getpid());
return 0;
}
And Here is my output:
I'm pid 6763
I'm pid 6765
I'm pid 6764
I'm pid 6766
Please, ignore the pipe part, I haven't gotten that far yet. I'm just trying to create only 2 child processes so I expect 3 "I'm pid …" outputs only 1 for the parent which I will make wait and 2 child processes that will communicate through a pipe.
Let me know if you see where my error is.
Best Answer
Let us assume the parent process id is 100, the first fork creates another process 101. Now both 100 & 101 continue execution after #1, so they execute second fork. pid 100 reaches #2 creating another process 102. pid 101 reaches #2 creating another process 103. So we end up with 4 processes.
What you should do is something like this.