Re: doubt about $ in perldoc perlvar

[Jos I Boumans]

Sure, just try running these 2 bits of code and you'll see what that means:

### example 1 ###
$foo = 'bar';

$foo =~ /(bar)/;
$firstmatch = $1;

$foo =~ /(quux/;
$secondmatch = $1;

print "$firstmatch and $secondmatch\n";

### example 2 ###

$foo = 'bar';

{   
    $foo =~ /(bar)/;
    $firstmatch = $1;
}

{
    $foo =~ /(quux/;
    $secondmatch = $1;
}

print "$firstmatch and $secondmatch\n";


### end examples ###

you'll notice the difference when you have the vars printed out...
the { } is the block the perldoc is talking about...

another solution would be to use a statement like this:
$firstmatch = $foo =~ /(bar)/;


Hope that helps,

Jos Boumans


----- Original Message ----- 
From: <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, May 17, 2001 6:59 PM
Subject: doubt about $<digits> in perldoc perlvar


> Gurus,
>         perldoc perlvar says:
> 
>      $<digits>
>              Contains the subpattern from the corresponding set
>              of parentheses in the last pattern matched, not
>              counting patterns matched in nested blocks that have
>              been exited already.  (Mnemonic: like \digits.)
>              These variables are all read-only.
> 
> I don't understand the second part ==> "not counting patterns matched in 
> nested blocks that have been exited already." Could I get some example 
> illustrating this point?
> 
> -- Thanks,
> Atul
> 
>