On Thu, Apr 11, 2002 at 01:43:31PM -0400, Jeff 'japhy' Pinyan wrote:
> On Apr 11, [EMAIL PROTECTED] said:
>
> >On Thu, Apr 11, 2002 at 11:13:44AM -0400, Jeff 'japhy' Pinyan wrote:
> >> On Apr 11, Jeff 'japhy' Pinyan said:
> >>
> >> ($a^$b)=~/\0*/*$+[0]
> >
> >Yeah, but does Perl actually garantee it will evaluate the
> >left operand of arithmetic operators first? If so, I cannot
> >find it in the documentation.
>
> Well, I've never had that problem. pop() - pop() works as I expect it to,
> evaluating left-to-right.
Well, yes. But how can you be sure this will always be the case?
Perhaps in a next release, it won't. Is it documented to work this
way, or does it just happen to work?
> >Luckely, there's a 1-character length operator that is documented
> >to first evaluate the left operand, then the right: ,
> >
> > ($a^$b)=~/\0*/,$+[0]
>
> Ah, but yours cannot be dropped in as an assignment, as mine can.
>
> $VAL = ($a^$b)=~/\0*/*$+[0];
> $VAL = ($a^$b)=~/\0*/,$+[0];
Well, it can be dropped in, you just have to know how to drop it! ;-)
($a^$b)=~/\0*/,$VAL = $+[0];
Abigail
