Brian Ling wrote:
> my $test1 = ${refer}->{value}->{fred};
> print Dumper($refer);
> # this doesn't change the data as i'd expect
>
> my $test2 = ${refer}->{value}->{fred}->{value};
> print Dumper($refer);
> # this actually creates a key 'fred' pointing
> # to a empty hash ref #I
>
> So my question really has two parts.
>
> Why does the above actually change the data, any
> pointers to reading material would be great. I've read
> a bit about autovification, is that what's going on.
> If so, how is this happening with an "assignment" or
> an "if".
> my $test1 = ${refer}->{value}->{fred};
Perl does not autovivify $refer->{value}{fred}.
Read perlfaq4:
"Why does passing a subroutine an undefined
element in a hash create it?"
"Normally, merely accessing a key's value for
a nonexistent key does not cause that key to
be forever there."
> my $test2 = ${refer}->{value}->{fred}->{value};
Here we are looking for a key's value. Perl
does not autovivify that key, but it does
autovivify $refer->{value}{fred} as expected.
> Given that I have a data structure similar to the
> above (returned from a module), of which I don't
> know the exact shape. How do I test for the
> existence of a key, down a branch without
> changing the data. All I could come up with is:
>
> if (${refer}->{value}->{fred} ) {
> my $test3 = ${refer}->{value}->{fred}->{value}
> }
>
> As I know if fred exists there will be a key called
> value, and testing for fred doesn't change my data.
> Is there a better way of doing this?
Read 'perlfunc' for details about the 'exists'
function. I would still test $refer->{value}{fred}{value}.
if ( exists $refer->{value}{fred}
and exists $refer->{value}{fred}{value} ) {
my $test3 = $refer->{value}{fred}{value};
}
OR:
my $test3 = $refer->{value}{fred}{value}
if exists $refer->{value}{fred}
and exists $refer->{value}{fred}{value};
HTH,
Charles K. Clarkson
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254 968-8328
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