Re: question about data structures - what does $# mean?

On Aug 30, 5:21 pm, Ed <edwardjameshartn...@xxxxxxxxx> wrote:
Howdy all!

Here's a little program:

#!/usr/bin/perl -w
my $d = {sid=>["lll"]};
print $#{$d->{sid}}."\n";

I expect this to print 1, but it prints 0.

What's up with that?

As I read this, $d is a ref to an anonymous assoc. array, which
contains one pair of values, named sid, with an anonymous array
containing one element: "lll".

So to get the number of elements in the array I try:
$#{$d->{sid}}

But that does not work. It gives me 0.

Meanwhile, this does work:
scalar(@{$d->{sid}})

Which made me realize I don't really know what the # does when used in
$#{$d->{sid}}.

Any help would be appreciated!

Thanks,

Ed
$#array gives the last index of the array not the number of elements
in the array, so you need to add 1 to $#array to get the number of
elements.
Since you have only one element its giving 0 , if you put 3 elements
in array it will give 2
.

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