In Perl, 'Use of uninitialized value in substitution iterator' error for $1 variable

Question

Working from an example found else where on stackoverflow.com I am attempting to replace on the Nth instance of a regex match on a string in Perl. My code is as follows:

Answer 1

I would write it like this

The while loop iterates over occurrences of the \d+ pattern in the string. When the Nth occurrence is found the last match is replaced using a call to substr using the values in built-in arrays @- (the start of the last match) and @+ (the end of the last match)

#!/usr/bin/env perl

use strict;
use warnings;

@ARGV = ( 3, 'INTEGER_PLACEHOLDER', 'method(0, 1, 6);' );

if ( @ARGV != 3 ) {
    print qq{\nUsage: replace_integer.pl occurrence replacement to_replace};
    print qq{\nE.g. `./replace_integer.pl 1 "INTEGER_PLACEHOLDER" "method(0 , 1, 6);"`};
    print qq{\nWould output: "method(INTEGER_PLACEMENT , 1, 6);"\n};
    exit;
}

my ( $occurrence, $replacement, $string ) = @ARGV;

my $n;
while ( $string =~ /\d+/g ) {

    next unless ++$n == $occurrence;

    substr $string, $-[0], $+[0]-$-[0], $replacement;
    last;
}

print "RESULT: $string\n";

output

$ replace_integer.pl 3 INTEGER_PLACEHOLDER 'method(0, 1, 6);'
RESULT: method(0, 1, INTEGER_PLACEHOLDER);

$ replace_integer.pl 2 INTEGER_PLACEHOLDER 'method(0, 1, 6);'
RESULT: method(0, INTEGER_PLACEHOLDER, 6);

$ replace_integer.pl 1 INTEGER_PLACEHOLDER 'method(0, 1, 6);'
RESULT: method(INTEGER_PLACEHOLDER, 1, 6);

Answer 2

$1 is only set after you capture a pattern, for example:

$foo =~ /([0-9]+)/;
# $1 equals whatever was matched between the parens above

Try wrapping your matching in parens to capture it to $1