Messages in this thread | | | Date | Sat, 21 Apr 2001 06:06:25 +0200 | From | Edgar Toernig <> | Subject | Re: fd allocation [was: light weight user level semaphores] |
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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.
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