In S02 synopsis, section "Names and Variables", I see:

$<foo> match variable, short for $/{'foo'}

So, when you say my $<-var>;, you are actually initializing the key -var of the hash $/.

Here's what I see in recent rakudo:

>> perl6 -v This is Rakudo Perl 6, version 2011.07-2-g1b7dd12 built on parrot 3.6. +0 RELEASE_3_6_0 >> perl6 > my $<-var> = 42 42 > $/ -var 42 > $/.WHAT Hash()

In reply to Re: Perl6: S2 identifier and $ special var questions by PrakashK
in thread Perl6: S2 identifier and $ special var questions by rir

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