in reply to Re^11: ref to read-only alias ... why? (notabug)
in thread ref to read-only alias ... why?
my $a = 2; $_++ for 1..$a; print "$a\n"; # 2
shows that $a is not aliased. Therefore I don't see how the original problem that needs a scalar be both aliased and read-only, is relevant here.
|
|---|
| Replies are listed 'Best First'. | |
|---|---|
|
Re^13: ref to read-only alias ... why? (notabug)
by ikegami (Patriarch) on Jan 06, 2012 at 22:16 UTC | |
by dk (Chaplain) on Jan 06, 2012 at 22:26 UTC | |
by ikegami (Patriarch) on Jan 07, 2012 at 01:09 UTC | |
by dk (Chaplain) on Jan 08, 2012 at 19:46 UTC | |
by ikegami (Patriarch) on Jan 09, 2012 at 09:59 UTC | |
| |
by LanX (Saint) on Jan 08, 2012 at 21:04 UTC |