Hello *NIX gurus,

I have a slight perplexing problem with multiple forks giving different results... Here is the deal.

From what I undestand, a fork() call starts executing from the next instruction that follows the fork() call. That means it inherits the PC counter register value of the parent, right?

For example

---------- -----------------

produces Hello

Now if I remove the new line character from printf, as follows:
-------------- ------------

It produces: HelloHelloHelloHello


Why does this happen? I tried fptintf to stderr (unbuffered output) and it produces just one hello, so I am thinking it has to do with when the output is flushed? But not sure how...

Could someone please shed some light on this.

Hmm.... that seems to be right. It is the buffer flushing that is responsible for that. Try replacing the printf with a write(2) and you won't see this behaviour.

It is the exit(2) syscall that causes this (by flushing the buffers).

Here is a post from linuxforums.org that describes something similar to what you have here.

Output to a tty is line buffered so the /n causes a flush. In this case, no data remains buffered and you get 4 copies of an empty buffer. Remove the /n and the data stays in the buffer. The buffer (and data) is then quadruplicated and all 4 processes flush their buffers at program exit.

Hmmmm, Thanks you guys. i read the link and your explanaitons and it makes sense to me now. What I understand now is that the data sections get copied to child processes, which means any unflushed buffers will be copied. Then on exit, every process flushes the output, hence producing a four printouts of Hello.

Thanks blowtorch for modifying the post for code clarity, i'll keep that in mind for future posts...

to flush the output buffer without using \n, after the printf use fflush(stdout) ;