[perl #101124] [BUG] Wrong junction autothreading evaluation order in Rakudo

[Will Coleda via RT]

On Tue Oct 11 00:23:29 2011, masak wrote:
> <tadzik> b: say ?(0|1 == 0&1); say ?(0&1 == 0|1)
> <p6eval> b 1b7dd1: OUTPUT«Bool::True␤Bool::True␤»
> <grondilu> rakudo: say ?(0|1 == 0&1), ?(0&1 == 0|1)
> <p6eval> rakudo 38907e: OUTPUT«Bool::FalseBool::True␤»
> <masak> 0|1 == 0&1 means any(0,1) == all(0,1) means all(any(0,1) == 0,
> any(0,1) == 1) means all(any(0 == 0, 1 == 0), any(0 == 1, 1 == 1))
> <masak> which is true, AFAICS.
> * masak submits rakudobug
> 
> In other words, the result of ?(0|1 == 0&1) should be Bool::True,
> because (intuitively) 0 and 1 is equal to 0 or 1. Other
> implementations agree.
> 
> <grondilu> perl6: say ?(0|1 == 0&1); say ?(0&1 == 0|1)
> <p6eval> pugs: OUTPUT«1␤1␤»
> <p6eval> ..rakudo 38907e: OUTPUT«Bool::False␤Bool::True␤»
> <p6eval> ..niecza v10-40-ga57ed3d: OUTPUT«Bool::True␤Bool::True␤»

Works again.

01:03 <[Coke]> r:  say ?(0|1 == 0&1); say ?(0&1 == 0|1)
01:03 <camelia> rakudo bfd850: OUTPUT«True␤True␤»

Closable with tests.

-- 
Will "Coke" Coleda