On Mon, Jul 09, 2001 at 02:39:15PM -0500, Jay Strauss wrote:
> I'm trying to check if a hash key exists and if the value is 1. I'm passing
> the reference to the hash to a subroutine like:
> my $help = 1;
> my %hash = (help=>\$help);
> somefunc(\%hash);
>
> sub somefunc {
> my ($refHash) = @_;
>
> if ((defined $refHash->{help}&& ${$refHash->{help}}) {
> # stuff
> }
> else {
> # other stuff
> }
> }
> This seems long an clumsy, I suspect there is a cleaner way to do it in perl
> but I don't know how. I played around with "exists" but that just tells me
> that the value exists. I need to do some processing if the value is "1" and
> other processing for "0";
my $help = 1;
my $hash = {help=>\$help};
#@@ change %hash to $hash, and (key=>val) to {key=>val}
somefunc($hash);
sub somefunc {
my ($refHash) = @_;
#@@ this is unchanged...
if ((defined $refHash->{help}&& ${$refHash->{help}}) {
#@@ you might try the $$refHash->{help} syntax and see if that works for you.
#@@ (defined $refHash->{help}) will always be defined w/ the reference to
#@@ the variable regardless of $help's state, it's a pointless check.
#@@ Add ${} or change to 'exists'... see below.
# stuff
}
else {
# other stuff
}
}
any "bareword" inside a {} construct is treated as a hash key... not a
problem for 'use strict' at all.
also, make sure you look at "exists" as opposed to "defined"... exists
will tell you if a given key in a hash exists, while defined tells you if
a given variable has a "defined" value. You can assign undef to a
variable, which would fail a defined check, but pass a exists check.
(looks like defined is correct for you, but you may need a ${} or $$
syntax to really check the variable and not the ref -- as the ref will
always be defined regardless of an undef state in the $help variable).
--
Ted Deppner
http://www.psyber.com/~ted/
