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 $test'
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qw is short for quoted words. It builds a list. So the code ($alpha,$delta)=qw/A D/; means: build a list with two elements A and D, and assign value A to $alpha, and value D to $delta.

In your example, $test=q{qq{$delta and $alpha}}; is equivalent to $test='"$delta and $alpha"'. Which means: give me a variable $test, and make the value equal to "$delta and $alpha" (with double quotes). When you evaluate this expression -


print eval ' "$delta and $alpha" ';
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it is equivalent to the code -


print "$delta and $alpha";
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Which will print "D and A" natually.

eval evaluates the value held inside the string given, "$delta and $alpha", in this case, which returns a scalar string.

In Zaxo's example, he was doing -

print eval '$delta and $alpha';
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which is equivalent to -

print $delta and $alpha;
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which will print the value of $alpha. Note that in the second case, there is no double quote in the equivalent code.