At Sunday 08 August 2010, Ralf Wildenhues wrote:
> * Stefano Lattarini wrote on Sun, Aug 08, 2010 at 07:09:46PM CEST:
> > At Sunday 08 August 2010, Ralf Wildenhues wrote:
> > > * Stefano Lattarini wrote on Sun, Aug 08, 2010 at 06:22:00PM
CEST:
> > > > Exactly! So:
> > > > foo() { foo_result=`sed ...`; }
> > > > foo; bar=$foo_result;
> > > >
> > > > instead of:
> > > > foo() { sed ...; }
> > > > bar=`foo`
> > > >
> > > > Where is the significant improvement?
> > >
> > > Two forks vs. one: the latter still forks right before exec'ing
> > > sed.
> >
> > And the first doesn't?
>
> For some shells, that is correct.
>
> Your questioning prompted me to try it out, to make sure I'm not
> telling anything wrong here. So, with bash 3.2.39 on GNU/Linux,
> the first of the scripts below calls clone once, the second twice,
> as strace -fF tells me.
Ah-ah! An optimization done under-the-hood by the shell... Now,
this makes sense.
> Cheers,
> Ralf
>
> --- script 1 ---
> #! /bin/sh
> foo () { result=`sed -n 1p $0`; }
> foo
> echo $result
>
>
> --- script 2 ---
> #! /bin/sh
> foo () { sed -n 1p $0; }
> result=`foo`
> echo $result
Thanks
Stefano
