Steve Bertrand wrote:
I have two scenarios here, and in the first one, I am not seeing the
logic I would normally expect. I'll compact the code as to save everyone
from scrolling. I have strict and warnings enabled (as I always do). Can
someone tell me why in the first case $1 isn't initialized and in the
second case it is?
# First run(catch $1 and $2, check $2 for correctness (it is), print $1)
my $email = '[EMAIL PROTECTED]';
$email =~ /(.*)@(.*)/;
if ($2 !~ /domain\.com/) {
print "var 2 is bad\n";
}
print "$1\n";
/** prints 'uninitialized' for line 'print "$1\n";**/
# Second run(same as above, but the if will fail:
my $email = '[EMAIL PROTECTED]';
$email =~ /(.*)@(.*)/;
if ($2 !~ /domain\.com/) {
print "var 2 is bad\n";
}
print "$1\n";
/** prints
var 2 is bad
steveb
*/
Why does the $1 initialize and print only if the 'if' fails?
From perldoc perlre:
The numbered match variables ($1, $2, $3, etc.) and the related
punctuation set ($+, $&, $`, $', and $^N) are all dynamically scoped
until the end of the enclosing block or until the next successful match,
whichever comes first.
In your first example, $1 is valid until the successful match /domain\.com/,
when
it becomes undefined as there are no capturing parentheses. In the second
example
$1 retains its value since the match fails.
It's always safer to save captured strings before you use them, and I would
never
make a capture variable the target of a regex match as in your $2 !~
/domain\.com/.
Much better to write something like
use strict;
use warnings;
my $email = '[EMAIL PROTECTED]';
my ($name, $host) = $email =~ /(.*)@(.*)/;
if ($host !~ /domain\.com/) {
print "Host name is bad\n";
}
print "$name\n";
and also, in this case, the test would be better as
unless ($host eq 'domain.com') {
print "Host name is bad\n";
}
unless you really want to test whether the host name /contains/ that string?
Oh, and I think I also prefer
my ($name, $host) = split /@/, $email;
instead of the first regex.
HTH,
Rob
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