# New Ticket Created by "Carl Mäsak"
# Please include the string: [perl #67944]
# in the subject line of all future correspondence about this issue.
# <URL: http://rt.perl.org/rt3/Ticket/Display.html?id=67944 >
<moritz_> rakudo: sub a() { 3, 4 }; my $x; my @a = ($x = a(), 4); say @a.perl
<p6eval> rakudo 4c31fb: OUTPUT«[[3, 4, 4]]»
<moritz_> is that correct? or should it be [[3, 4], 4]?
<masak> moritz_: well, things flatten, don't they?
<moritz_> masak: there's also the question of relative precedence of , and =
<masak> moritz_: I believe S03 is very clear on that.
<b2gills> rakudo: sub a() { 3, 4 }; my $x; my @a = (($x = a()), 4);
<p6eval> rakudo 4c31fb: ( no output )
<b2gills> rakudo: sub a() { 3, 4 }; my $x; my @a = ($x = a(), 4); say $x.perl
<p6eval> rakudo 4c31fb: OUTPUT«[3, 4, 4]»
<moritz_> masak: if naively read the precedence table in S03 then it
tells me that = is tighter, and therefore $x should only get the
elements from a()
<masak> moritz_: no, there's a whole section on list assignment.
<masak> line 1785.
<masak> and it carries on explaining until S03:1861.
<pmichaud> if the left hand side of eq is a scalar, it's an item
assignment, and = binds tighter than comma
<pmichaud> at present Rakudo doesn't do that -- it treats all ='s as
list assignment
<pmichaud> sorry, it parses all ='s as list assignment
<pmichaud> so Rakudo will incorrectly see $x = a(), 4 as being $x
= (a(), 4) instead of ($x = a()), 4
<moritz_> pmichaud: ok, that answers my question
<moritz_> and that's indepent of a possible @a = to the left of the
whole expression, right?
<pmichaud> correct.
<moritz_> thank you.
<masak> so... it's a bug? :>
<moritz_> aye.
* masak submits
