Question

Currently, I am doing some exercises on operating system based on UNIX. I have used the fork() system call to create a child process and the code snippet is as follows :

if(!fork())
{
   printf("I am parent process.\n");
}
 else
   printf("I am child process.\n");

And this program first executes the child process and then parent process.

But, when I replace if(!fork()) by if(fork()!=0) then the parent block and then child block executes.Here my question is – does the result should be the same in both cases or there is some reason behind this? Thanks in advance!!

Answer 1

• There is no guaranteed order of execution.
• However, if(!fork()) and if(fork()!=0) do give opposite results logically: if fork() returns zero, then !fork() is true whilst fork()!=0 is false.

Also, from the man page for fork():

On success, the PID of the child process is returned in the parent, and 0 is returned in the child. On failure, -1 is returned in the parent, no child process is created, and errno is set appropriately.

So the correct check is

pid_t pid = fork();
if(pid == -1) {
     // ERROR in PARENT
} else if(pid == 0) {
     // CHILD process
} else {
     // PARENT process, and the child has ID pid
}

EDIT: As Wyzard says, you should definitely make sure you make use of pid later as well. (Also, fixed the type to be pid_t instead of int.)