Re^2: string diff and compare

So I've gone through your example now. I understand everything in it, except...

my ( $rold, $rnew ) = map scalar reverse, $old, $new;

What's up with the map scalar reverse syntax? What does the scalar do for example? I mean, I get that you're reversing two strings and putting them into two new strings, but I am really having difficulty parsing how Perl reads this to do what you've done here.

Anyone care to explain it, decompress it (for us mortals)?

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Re^3: string diff and compare by Anonymous Monk on May 04, 2012 at 15:54 UTC
From perlod ... reverse LIST In list context, returns a list value consisting of the elements of LIST in the opposite order. In scalar context, concatenates the elements of LIST and returns a string value with all characters in the opposite order. ... And, read map chain as: `map { scalar( reverse $_ ) } ( $old , $new )`.	[reply] [d/l]
Re^4: string diff and compare by Anonymous Monk on May 04, 2012 at 16:13 UTC
"perlod"? Ha! :-) sounds like "perllord" was misspelled.	[reply]
Re^3: string diff and compare by jwkrahn (Abbot) on May 04, 2012 at 21:27 UTC
`my ( $rold, $rnew ) = map scalar reverse, $old, $new;` [download] Is just a different way of saying: `my $rold = reverse $old; my $rnew = reverse $new;` [download] You need to use scalar because reverse is in list context in the map expression.	[reply] [d/l] [select]
Re^4: string diff and compare by tobyink (Canon) on May 05, 2012 at 07:17 UTC
There are other ways of enforcing scalar context which are a lot more fun. :-) `use 5.010; use strict; use warnings all => 'FATAL'; my ( $old, $new ) = qw( old new ); my ( $rold, $rnew ) = map~~reverse, $old, $new; say for "$old => $rold", "$new => $rnew";` [download] `perl -E'sub Monkey::do{say$_,for@_,do{($monkey=[caller(0)]->[3])=~s{::}{ }and$monkey}}"Monkey say"->Monkey::do'`	[reply] [d/l]