On 10/10/07, Rob Dixon <[EMAIL PROTECTED]> wrote: > > Kaushal Shriyan wrote: > > Hi, > > > > I am referring to http://www.gnulamp.com/perlscalars.html > > > > $a = $b; # Assign $b to $a > > > > Note that when Perl assigns a value with *$a = $b* it makes a copy of $b > and > > then assigns that to $a. Therefore the next time you change $b it will > not > > alter $a. > > > > I did not understand the Note:- Can some one make me understand this > > statement with examples > > All they're saying is that $a = $b doesn't tie $a and $b together in any > way, > it just copies the value of $b to $a. If you change $b after that it won't > alter $a again. > > > and also what i understand from the below statements( I have added in > the > > comments), is it correct. > > > > $a += $b; #is it $a = $a + $b; > > $a -= $b; #is it $a = $a - $b; > > $a .= $b; #is it $a = $a . $b; > > Exactly right. Well done. > > Rob >
Thanks Rob All they're saying is that $a = $b doesn't tie $a and $b together in any way, it just copies the value of $b to $a. If you change $b after that it won't alter $a again. Can you please explain me with a sample code. If I understand it correctly does the below code holds true for your explanation #!/usr/bin/perl -w $b = 2; $a = $b; print "$a\n"; $b = 4; $c = $b; print "$c\n"; Thanks again Thanks and Regards Kaushal
