Re: fd allocation [was: light weight user level semaphores]

[Edgar Toernig <froese@gmx.de>]

Linus Torvalds wrote:
> 
>         pid = fork();
>         if (!pid) {
>                 close(0);
>                 close(1);
>                 dup(pipe[0]);   /* input pipe */
>                 dup(pipe[1]);   /* output pipe */
>                 execve("child");
>                 exit(1);
>         }
> 
> The above is absolutely _standard_ behaviour. It's required to work.
> 
> And btw, it's _still_ required to work even if there happens to be a
> "malloc()" in between the close() and the dup() calls.

Right.  This is expected (and defined) behaviour.  But do you have
_any_ example where this is used for fds > 2?  I can't remember.
And IMHO that would be pretty fragile too.  Shell scripts sometimes
open temporary fds > 2 and these are passed to called programs.  I.e.

#!/bin/sh
exec 3>log
echo >&3 "script started"
ls /proc/self/fd              # gets fd3 already opened
ls /proc/self/fd 4</dev/null  # now 3 and 4 already in use...
# or look into any configure script...

So, IMHO as long as some library does not mess with fds 0, 1, and 2
it should be ok [1].  Yes, it would be against the standard but I
still have to find some code where this semantic is used for fds > 2.

Ciao, ET.


PS: I would prefer to keep the standard semantics but the reasons
for that are pretty weak ... ;-)

PPS: Even your sample code is fragile.  It breaks if I start it
with  ./a.out <&-  ;-)  (the close(0) is likely to close one end
of the pipe)

[1] Unintentionally setting the controlling tty may be a problem.