If either operand of scalar ".." is a constant expression, that operand is implicitly compared to the $. variable, the current line number.That's why I said that \"aString" .. __ doesn't do anything useful in scalar context. :) I should have clarified that I was referring to that expression specifically, rather than to scalar context .. in general.
In reply to Re: Re: Re: Un-obfuscate me
by chipmunk
in thread Un-obfuscate me
by Anonymous Monk
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