[perl #114432] [BUG] 'repeat' gets interpreted as a listop if there's no 'while' or 'until' in Rakudo

[Carl Mäsak]

# New Ticket Created by  "Carl Mäsak" 
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<masak> std: repeat { say "OH HAI" }
<p6eval> std f43a358: OUTPUT«===SORRY!===␤Undeclared routine:␤
'repeat' used at line 1␤Check failed␤FAILED 00:00 41m␤»
<masak> hm.
<jnthn> did you mean loop?
<masak> jnthn: no, I meant to check what happens if I do 'repeat'
without the 'until' or 'while' after the block.
<masak> std: repeat { say "OH HAI" } while True
<p6eval> std f43a358: OUTPUT«ok 00:00 41m␤»
<masak> I'm surprised it parses as a listop when the 'while' is not there.
<jnthn> Hmmmm.
<jnthn> Surely it doesn't backtrack there...
<masak> right.
<jnthn> std: repeat { my $x; say "OH HAI" }
<p6eval> std f43a358: OUTPUT«===SORRY!===␤Undeclared routine:␤
'repeat' used at line 1 [...] Check failed␤FAILED 00:00 42m␤»
<jnthn> r: repeat { say "OH HAI" } # curious
<p6eval> rakudo 34e8d4: OUTPUT«===SORRY!===␤CHECK FAILED:␤Undefined
routine '&repeat' called (line 1)␤»
<PerlJam> That could make for some interesting code.
<PerlJam> r: sub repeat { $^a.() }; repeat { say "hi" }
<p6eval> rakudo 34e8d4: OUTPUT«hi␤»
<jnthn> Wow!
<PerlJam> r:  sub repeat { $^a.() }; repeat { say "hi" }; my $i = 0;
repeat { say "foo" } while $i++ < 3;
<p6eval> rakudo 34e8d4: OUTPUT«hi␤foo␤foo␤foo␤foo␤»
<PerlJam> good for obfuscation I guess
<masak> was gonna say.
<masak> I feel 'repeat' is a badly chosen name for that sub :P
<jnthn> I think the parser is gonna need to commit after seeing repeat, fwiw
* masak submits stdbug
* masak submits rakudobug
* jnthn will wait for an STD fix and steal that