in reply to Re: Re: Re: Resolving scalars in a scalar
in thread Resolving scalars in a scalar
I understood the context that Zaxo was putting it in (sort of), but I still don't quite follow on why "and" would be treated as an operator. Is it because Zaxo is only wrapping what I want to be an evaluated string only once in quotes? So when it evaluates it, it umm... that's the part that loses me. If Zaxo wrapped my $delta and $alpha in quotes already, why did it treat and as an operator? Does eval strip the outer nesting of quotes? I found that if I re-write Zaxon's code like this;
then I get the behavior that I want.perl -e '($alpha, $delta) = qw{A D};$test = q(qq($delta and $alpha)); +print eval $test'
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Re: Re: Re: Re: Re: Resolving scalars in a scalar
by Roger (Parson) on Oct 31, 2003 at 03:56 UTC |