Re: perl pattern match for end of string using STDIN and chomp

[AnomalousMonk]

I tend to agree with the suggestions of others that the second regex in the chomped <STDIN> section of the OPed code is simply failing to match, and the previous value of  $1 is persisting.

Try inserting the statement
    print qq{\$1 reset: '$1'} if 'foo' =~ /(foo)/;
between the first and second regex in that section and seeing what happens. If the second  'ZZZ' becomes  'foo' you will know what is happening (although not why the second regex fails to match, which I cannot understand myself).

[Marshall]

This "previous value of $1" can be problematic. One of the practices that I often use in my code is to NOT use $1 or $2, etc. I like to assign $1 right away to a variable that has more contextual meaning (like $name, $cust_id) or whatever. Why have $name = $1;?

One way to do this is illustrated below, put the match into a list context and use list slice to get $1,$2, etc. If "$1" is undef, then in this case $thing gets undef, not the previous value of $1.

print "match failed\n" unless $string =~ m/(.+)\.BBB$/; 
print "dollar $1:\n";  #prints previous $1 value
my ($thing) = ($string =~ m/(.+)\.BBB$/)[0];
print "thing =$thing\n";  #$thing is undef

[cbolcato]

it seems pretty clear that the $1 in the code is falling through since the second regex is not matching, so i will use some of the code examples to avoid that in the future. However, the big question is why the second regex does not match with a chomped <STDIN> but will match a chomped $ARGV[0], it only appears to be an issue with 5.8.0, and appears to have been fixed in atleast 5.8.3, can anyone explain this behavior, may need some ammo to get our Linux support to upgrade our standard Perl version