fork() outputs after the process execution is over - c

Question

Consider the below code snippet:
int main()
{
fork();
fork();
fork();
printf("Hello World\n");
}
I am getting the output:[ubuntu 12.04]
aashish#aashish-laptop:~$ ./a.out
Hello World
Hello World
Hello World
aashish#aashish-laptop:~$ Hello World <---------------------------
Hello World
Hello World
Hello World
Hello World
Why "Hello Word" is outputted after the process execution is over?

Answer 1

The short answer is that you are creating multiple processes, which run asynchronously with respect to each other. The long answer follows:
When you type ./a.out at the shell prompt, that creates a process running your program. Let's call that process 1.
Process 1 calls fork(). This creates a new child process, Process 2, and both 1 and 2 carry on execution after the first fork() call, proceeding to the second fork() call. Process 1 creates child Process 3, and Process 2 creates child process 4. All four processes carry on from after the second fork(), proceeding to the final fork() call. Process 1 creates Process 5; Process 2 creates Process 6; Process 3 creates Process 7; and Process 4 creates Process 8.
Note that these process numbers are arbitrary: there's no guarantee they'd be created in that order.
The asynchrony comes into play as soon as that first fork() gets executed. The system offers no guarantees about scheduling the parent with respect to the child. Theoretically the child could run to completion before the parent continues, the parent could finish before the child gets any resources. The most likely scenario lies somewhere in the middle: the original process shares resources with its progeny, so that all run concurrently.
The final piece of the puzzle results from the fact that the shell is waiting for Process 1 to complete. Only Process 1. The shell doesn't know (or care) that Process 1 has started other processes. So when Process 1 completes, the shell displays a prompt. As it happened, some of the descendants of Process 1 hadn't yet reached the printf() statement. By the time they got there, the shell had already displayed its prompt.
To explore this further, you might want to try changing the fork() calls to printf( "%d\n", fork() ); and/or change printf("Hello World\n") to printf("Hello from pid %d\n", getpid() )

Answer 2

The "Hello World" outputs that come after the second shell prompt come from the forked processes, not from the one that was launched by the shell (you).

Answer 3

Thats because you make 1 fork (2), then you fork again (4), then again (8) then for each fork print Hello world. That's why you got 8 outputs.

Answer 4

int main() {
pid_t c[3];
int i, n = 0;
for (i = 0; i < 3; ++i) {
switch ((c[n] = fork())) {
case 0: break;
case -1: perror("fork"); exit(EXIT_FAILURE);
default: ++n;
}
}
printf("[%d] Hello World\n", (int)getpid());
// Without waiting, some children may still be running when the
// parent exits. This makes it look like output is generated
// after the process is over, when in fact not all the processes
// are done yet.
//
// The process is not really finished until its children are
// finished. The wait call waits on a child process to finish.
for (i = 0; i < n; ++i) wait(0);
return 0;
}