Re: Tips on how to perform this substitution?

[moritz]

The trick with the alternation is that your regex matches -s delimited by B's, thus slurping them up. The next regex search starts off after the last B, and needs to skip the following -s to find the next B.

use 5.010;
use strict;
use warnings;

$_ = '------BBBBB-----------------------------------------------------
+-------------------------------------------------------BBBBBB----BBBB
+BBBB-----------------BBBBBB------------------------------------------
+--------------BBBBBBB---------------BBBBB-----BBBBBBBBB--------------
+----------------------BBBBBBBBB------BBBBBBBBB-----------------------
+-----BBBBBBBBB--------------------------------------BBBBBBBB-------BB
+BBBB---------------------BBBBBBB--------------------------BBBBBBBB---
+-';

my $expected = 'iiiiiiMMMMMooooooooooooooooooooooooooooooooooooooooooo
+oooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooMMMM
+MMiiiiMMMMMMMMoooooooooooooooooMMMMMMiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
+iiiiiiiiiiiiiiiiiiiiiiiiMMMMMMMoooooooooooooooMMMMMiiiiiMMMMMMMMMoooo
+ooooooooooooooooooooooooooooooooMMMMMMMMMiiiiiiMMMMMMMMMooooooooooooo
+oooooooooooooooMMMMMMMMMiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiMMMMMMM
+MoooooooMMMMMMiiiiiiiiiiiiiiiiiiiiiMMMMMMMooooooooooooooooooooooooooM
+MMMMMMMiiii';

s/(^|B+)(-+)($|B+)/ $1 . ("i" x length($2)) . $3/eg;
s/-/o/g;
s/B/M/g;

say $expected;
say $_;

say $expected eq $_ ? 'yes' : 'no';

[tobyink]

It can be done in one s/// operation...

use v5.10;
use strict;
use warnings;

$_ = '--BBB----BB--B-------B--B--BBBB---B--';

s ((B+|.)) {
    state $char = 'i';
    if (index($1, 'B') == 0) {
        $char = $char eq 'i' ? 'o' : 'i';
        'M' x length($1);
    }
    else {
        $char;
    }
}eg;

say;

[Tux]

That is way too complex :)

my $p = "--BBB----BB--B-------B--B--BBBB---B--";

my @x = qw( i o );
my $x = 0;

$p =~ y/B/M/;
$p =~ s/(-+)/$x[$x++%2]x length($1)/ge;

say $p;

[Anonymous Monk]

Thank you very much for your kind tips!
Will look at them carefully!