ww has asked for the wisdom of the Perl Monks concerning the following question:
Perhaps a wiser head or three can tell me if the next statement appears to be valid:
Some of the answers to How to force one variable to be an alias to another? (in Q&A QandASection: references) have some problems. Rather than post a reply to a single answer there (and to avoid polluting Q&A if I've missed something), here's what I observed:
Starting with my OS & perl:And, my test code:Win7 Pro perl -v: This is perl 5, version 24, subversion 0 (v5.24.0) built for MSWin32- +x86-multi-thr (with 1 registered patch ....) Binary build 2400 [300560] provided by ActiveState http://www.ActiveS +tate.com Built Jul 13 2016 14:32:48
use v5.22; use warnings; # cat answer to "How to force one variable to be an alias to another? +" # indented text led by #: errors prior to commenting out the bare blo +ck my $seg = 1; say "\$seg: $seg"; =head { use feature 'refaliasing'; # Silence warnings if desired: no warnings 'experimental::refaliasing'; my ($x, $y); \$x = \$y; # alias here $x = 5; say $y; # prints nothing -- see msg above }; =cut $seg++; say "\$seg = $seg"; #2 =head { my $x = 5; my $y = 273; make_alias( $x, $y ); # $seg = 2 # Undefined subroutine &main::make_alias called at D:\_Perl_\pl_te +st\#1183761.pl line 3 +6. print $y; # prints nothing -- see msg above }; =cut $seg++; say "\$seg = $seg"; #3 { use Lexical::Alias; my $that = "not_that"; my $this = "another seg"; alias $that, $this; # $this is now an alias for $that. say $this; # prints "not_that" }; $seg++; say "\$seg = $seg"; #4 { my @x = qw(foo bar baz); my @y = qw(bat bif bim); alias @x, @y; for (@x) { say "element of \@x: " . $_; } for (@y) { say "\t element of \@y: " . $_; } }; $seg++; say "\$seg = $seg"; #5 =head { use Data::Alias; # from perldelta for 5.24: # Known Problems # Some modules have been broken by the context s +tack rework. # These modules were relying on non-guaranteed i +mplementation # details in perl. <i> (did I miss something?)< +/i> my( $this, $that ); alias $that, $this; # $this is now an alias for $that. }; =cut =head FINAL OUTPUT AFTER commenting failed bareblocks: $seg: 1 $seg = 2 $seg = 3 not_that $seg = 4 element of @x: foo element of @x: bar element of @x: baz element of @y: foo element of @y: bar element of @y: baz $seg = 5 =cut
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Re: Possible problems in answers to CatQ 586121
by choroba (Cardinal) on Mar 11, 2017 at 21:03 UTC |