Re^3: regexp list return 5.6 vs 5.8

Something that just crossed my mind and turned out to behave like I didn't expect:

perl -le 'sub x { my @x = qw(a b c); my @y = qw(A B C D); return (@x, 
+@y)} my $r = x(); print $r'
4
[download]

Must say it made me LOL :-)

use strict; use warnings; print "Just Another Perl Hacker\n";

Comment on Re^3: regexp list return 5.6 vs 5.8 Download Code

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Re^4: regexp list return 5.6 vs 5.8 by shmem (Chancellor) on Jan 24, 2008 at 12:48 UTC
From what we've already discussed I would have expected the outcome: the return list is constructed at compile time. at runtime, the list `(@x, @y)` is evaluated in the caller's context, so it resolves to the comma operator, returning it's last argument - `@y` then the array `@y` is evaluated in the caller's context, which happens to be scalar, hence 4 as a result. Pretty much the same as `perl -le '@x=qw(a b c);@y=qw(a b c d); $f=(@x,@y); print $f' 4` [download] ;-) --shmem _($_=" "x(1<<5)."?\n".q·/)Oo. G°\ / /\_¯/(q / ---------------------------- \__(m.====·.(_("always off the crowd"))."· ");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}	[reply] [d/l] [select]

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Re^4: regexp list return 5.6 vs 5.8
by shmem (Chancellor) on Jan 24, 2008 at 12:48 UTC

the return list is constructed at compile time.
at runtime, the list (@x, @y) is evaluated in the caller's context, so it resolves to the comma operator, returning it's last argument - @y
then the array @y is evaluated in the caller's context, which happens to be scalar, hence 4 as a result.

Pretty much the same as

perl -le '@x=qw(a b c);@y=qw(a b c d); $f=(@x,@y); print $f'
4
[download]

;-)

--shmem

_($_=" "x(1<<5)."?\n".q·/)Oo.  G°\        /
                              /\_¯/(q    /
----------------------------  \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}

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