Hello everybody.
Lets explain my problem.
Lets see the folowing perl 5 program :
#!/usr/bin/perl
my $c = "AABBCC";
my $m = "B";
$c =~ s/($m+)/XX/;
print "Chain : $1.\n";
print "Final chain : $c.\n";
Le résultat est bien celui que j'attendais :
Chain : BB.
Final chain : AAXXCC.
Lets see now Perl 6.
with a simple mach.
use v6;
my $c = "AABBCC";
my $m = "B";
$c ~~ m:c/($m+)/;
say "Chain : $0. Place : ", $/.from;
Ici aussi, le résultat est bien celui attendu :
Chain : BB. Place : 2
and the perl 5 program translated in Perl 6.
use v6;
my $c = "AABBCC";
my $m = "B";
$c ~~ s/($m+)/XX/;
say "Chain : $0.";
say "Final chain : $c.";
Oups, result is not really what I was waiting :
Chain : Any().
Final chain : AAXXCC.
The $0 var is set on a mach operation but not in a substitute.
So that its value is "Undef"
The problem can be solved using firts a m then a s
use v6;
my $c = "AABBCC";
my $m = "B";
$c ~~ m:c/($m+)/;
$c ~~ s/($m+)/XX/;
say "Chain : $0.";
say "Final chain : $c.";
The rseult is what I was waiting :
Chain : BB.
Final chain : AAXXCC.
Where is the problem ?
Is there any solution to set the $0 var in a s operation ?
Thanks
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| Christian Aperghis-Tramoni |
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print join('',map({$i=1-$i;$a=$i?10*$_."\b\b":pack"c",$a+$_+0x16}split
(//,"5110789279758710838810587992861093898779948387799310")),"...\n");
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