Thanks for that
Jenda Krynicky wrote:
> Then how come
>
> $x = 'Hello';
> sub foo {
> $_[0] = 'Hi';
> }
> foo($x);
> print $x,"\n";
>
> prints
> Hi
> ?
>
> The parameters are passed by reference, though in little strange way.
> If for example you include an array (and I mean array, not array
> reference!) then the function gets the elements of the array ... but
> still by reference:
>
> $x = 0;
> @a = (1,2);
> sub foo {
> for (@_) {
> $_ += 10;
> }
> }
> foo( $x, @a);
> print "\$x = $x\n";
> print "\@a = @a\n";
>
> It's when you do the usual
> my $var = shift;
> or
> my ( $a, $b, $c) = @_;
> that Perl makes copies of the parameters. If you do not shift() and
> access @_ directly you CAN modify the parameters.
>
> Jenda
> =========== [EMAIL PROTECTED] == http://Jenda.Krynicky.cz ==========
> There is a reason for living. There must be. I've seen it somewhere.
> It's just that in the mess on my table ... and in my brain
> I can't find it.
> --- me
>
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