[perl #85844] "sub foo;" tries to call foo instead of giving parse error

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# New Ticket Created by  Paweł Pabian 
# Please include the string:  [perl #85844]
# in the subject line of all future correspondence about this issue. 
# <URL: http://rt.perl.org/rt3/Ticket/Display.html?id=85844 >


[16:12] <bbkr_> rakudo: sub foo; # looks bad, this should be parse error, not 
attempt to call foo.
[16:12] <p6eval> rakudo a38d45: OUTPUT«Could not find sub &foo␤  in main 
program body at line 22:/tmp/c10WrR9Dod␤»
[16:12] <moritz_> bbkr_: long known
[16:12] <jnthn> std: sub foo;
[16:12] <p6eval> std 4608239: OUTPUT«===SORRY!===␤Malformed 
block at /tmp/89UskaNhFr line 1:␤------> sub foo⏏;␤    
expecting any of:␤       new name to be defined␤ routine_def␤    trait␤Parse 
failed␤FAILED 00:01 117m␤»
[16:13] <moritz_> bbkr_: oh wait, I misparsed :-)
[16:13] <moritz_> std: sub(foo())
[16:13] <p6eval> std 4608239: OUTPUT«===SORRY!===␤Undeclared 
routines:␤        'foo' used at line 1␤   'sub' used at line 1␤Check 
failed␤FAILED 00:01 118m␤»
[16:18] <bbkr_> so it is bug or not? I don't understand the difference 
between "sub foo" and sub(foo()) in STD.
[16:18] <moritz_> the parenthesis :-)
[16:18] <moritz_> yes, bug
[16:19] --> hanekomu has joined this channel 
(~hanekomu@2001:62a:4:203:cabc:c8ff:fea1:8046).
[16:19] <bbkr_> in STD (because parenthesized syntax behaves differently) or 
in rakudo?
[16:21] <moritz_> rakudo has the bug
[16:21] * bbkr_ reports, thanks