Re: $1[

[thanos1983]

Hello rsFalse,

Well I am having some questions based on your code.

The sample of code that you provide us it gives me a different error (sample below):

#!/usr/bin/perl

use warnings;
use strict;

$\ = $/;

$_ = reverse "(a+b)*c"; # or some longer sequence

s/\)([^()]+)\(/]$1[/g;

print scalar reverse y/][/)(/r;

__END__

$ perl test.pl
Missing right curly or square bracket at test.pl line 10, within strin
+g
syntax error at test.pl line 10, at EOF
Execution of test.pl aborted due to compilation errors.

Based on fellow Monk holli solution:

#!/usr/bin/perl
use strict;
use warnings;
use feature 'say';

my $reversed = reverse "(a+b)*c"; # or some longer sequence
$reversed =~ tr/()/[]/;
my $str = reverse $reversed;
say $str;

__END__

$ perl test.pl
[a+b]*c

Question: why you reverse the string? Am I missing something here? (Sample below):

#!/usr/bin/perl
use strict;
use warnings;
use feature 'say';

my $test = "(a+b)*c";
$test =~ tr/()/[]/;
say $test;

__END__

$ perl test.pl
[a+b]*c

An alternative way of resolving your issue (not more efficient - just for fun :D):

#!/usr/bin/perl
use strict;
use warnings;
use feature 'say';

sub replaceChar {
    my ($str, $char, $replace) = @_;
    substr($str, index($str, $char), length($char), $replace);
    return $str;
}

my $s = "(a+b)*c";
my $leftParenthesis = "(";
my $rightParenthesis = ")";
my $leftCurlyBracket = "[";
my $rightCurlyBracket = "]";

my $str = replaceChar($s, $leftParenthesis, $leftCurlyBracket);
say $str;
say replaceChar($str, $rightParenthesis, $rightCurlyBracket);

__END__

$ perl test.pl
[a+b)*c
[a+b]*c

BR / Thanos