I modified your code a bit so you can see what's happening: 

use strict;
my $R;
my $i=1;

sub mySub {
    printf "\n\n%d. R = \"%s\"\n",$i++,$R;
    unless ($R) {
        print "CONDITION WORKS\n";
        $R = 'some string';
    }
}

print mySub(); #prints CONDITION WORKS\nsome string
print mySub(); #prints some string
print mySub(); #again

[download]
This ends up producing this for an output: 

1. R = ""
CONDITION WORKS
some string

2. R = "some string"
some string

3. R = "some string"
some string

[download]
On the first iteration $R is undefined and there for in a "false" state. Therefore your print inside the "unless" gets executed and an assignment gets performed on $R. Where it gets trickier is the next iterations. Since you have unless($R) as the last evalutation the value of $R becomes the return value for the sub since you didn't specify one. see sub and search for "return" if you want more on this.

Peter L. Berghold -- Unix Professional
Peter -at- Berghold -dot- Net; AOL IM redcowdawg Yahoo IM: blue_cowdawg

In reply to Re: Perl: last evaluated value as a returned value for a subroutine. See inside by blue_cowdawg
in thread Perl: last evaluated value as a returned value for a subroutine. See inside by Doctrin

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