I drank too much to night so I might regret this one.
Thanks for teh explanation on q, qq and qx. I understand what qw was used for, but I couldn't make the leap or connection to figure out what they meant on my own. I was sure I read a mention of it in the llama book, but I couldn't find it in the index.

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; 

perl -e '($alpha, $delta) = qw{A D};$test = q(qq($delta and $alpha)); 
+print eval $test'

[download]
then I get the behavior that I want.

In reply to Re: Re: Re: Re: Resolving scalars in a scalar by SavannahLion
in thread Resolving scalars in a scalar by Anonymous Monk

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