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(|)= doesn't parse in Rakudo #3071

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p6rt opened this issue Mar 9, 2013 · 3 comments
Closed

(|)= doesn't parse in Rakudo #3071

p6rt opened this issue Mar 9, 2013 · 3 comments

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@p6rt
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p6rt commented Mar 9, 2013

Migrated from rt.perl.org#117103 (status was 'resolved')

Searchable as RT117103$

@p6rt
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p6rt commented Mar 9, 2013

From @masak

<masak> r​: https://gist.github.com/masak/5124688
<p6eval> rakudo 5a6b52​: OUTPUT«set(abcd, bacd, acbd, abdc, badc, adbc,
dabc, adcb, dacb, acdb, cadb, cdab, cabd, cbad, bcad, cbda, bcda,
cdba, dcba, dbca, dcab, bdca, dbac, bdac)␤»
<masak> \o/
<TimToady> masak​: $set = $set (|) $d; <-- why not (|)= ?
<masak> TimToady​: didn'a work :(
<masak> r​: my $s = set; $s (|)= 5
<p6eval> rakudo 5a6b52​: OUTPUT«===SORRY!===␤Two terms in a row␤at
/tmp/VbH2WyDu5i​:1␤------> my $s = set; $s ⏏(|)= 5␤ [...]
* masak submits rakudobug
<TimToady> masak​: you should get masak to file a...
<masak> TimToady​: this time, I was faster than that guy! :P

@p6rt
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p6rt commented Jul 20, 2014

From @FROGGS

FROGGS m​: my $s = set; $s (|)= 5; say $s
camelia rakudo-moar bbdcfd​: OUTPUT«set(5)␤»

Test​: Raku/roast@2e900e5d4b

@p6rt p6rt closed this as completed Jul 20, 2014
@p6rt
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p6rt commented Jul 20, 2014

@FROGGS - Status changed from 'new' to 'resolved'

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