# New Ticket Created by "Carl Mäsak"
# Please include the string: [perl #125505]
# in the subject line of all future correspondence about this issue.
# <URL: https://rt.perl.org/Ticket/Display.html?id=125505 >
<jaffa4> I How can I make a reference to scalar?
<masak> m: my $scalar = 42; my $ref = \$scalar; $scalar++; say $ref
<camelia> rakudo-moar 367b58: OUTPUT«\(43)»
<jaffa4> masak: is it not that capture?
<masak> jaffa4: yes, that is a capture. when someone asks about
"taking a reference" in Perl 6, I think about captures.
<jaffa4> masak : there is a problem with that reference... cannot
change variable value
<jnthn> A Capture can be indexed positionally to get the things in it
<jnthn> m: my $a = 41; my $b = \($a); $b[0]++; say $a
<camelia> rakudo-moar 367b58: OUTPUT«Cannot assign to an immutable value [...]
<jnthn> ah
<jnthn> But yeah, the usual way to do such things these days is "is
rw" parameters and so on
<Ben_Goldberg> m: my $scalar = 42; my $ref = \$scalar; { ++$_ }( |$ref
); say $scalar;
<camelia> rakudo-moar 367b58: OUTPUT«43»
<jaffa4> jnthn: so is it a bug?
<jnthn> That you can't ++ the [0]'d thing maybe is
* masak submits rakudobug
<jnthn> masak: Thanks, I'm just a bit too tired to fix that now, but
can easily do it in the morning :)
