Reference Question.

[DaLoverhino]

I'm reading about perl references in a book called "Beginning Perl" by
James Lee. I read ahead to the use of ->, and [], but I'm tripped up
on the simpler examples prior to the use of ->, [], so I don't have
much confidence in what I have read.

Anyways the book has the following example:

$ref = [ 1, 2, [ 10, 20 ] ] ;

To access 20, you can do it a number of ways:

$inside = ${$ref}[2];
$element = ${$inside}[1];

Or the following way:

Book Example> $element = ${${ref}[2]}[1];


Well, I tried the latter way, and doesn't work. So I tried this
instead:

First Try> ${${$ref}[2]}[1];

This works. But I'm wondering why?

If you have a array_ref, you access an element by doing this:

Basic Example> ${$array_ref}[0];


So if ${$array_ref}[5] is itself an array reference, if I want to
access one of it's elements, why does the First T ry work? And why
does this one fail:

${$ ${$array_ref}[0] }[5]
^^^^^^^^^^^^^^^^^ --- The reference
^^^ ^^^ -- What you do to
access elements of an array
reference.


So I put some spaces to show you I took the reference "${$array_ref}
[0]" and applied
Basic Example> to it.


thanks.

 

[xhoster]

I'm reading about perl references in a book called "Beginning Perl" by
James Lee. I read ahead to the use of ->, and [], but I'm tripped up
on the simpler examples prior to the use of ->, [], so I don't have
much confidence in what I have read.

Anyways the book has the following example:

$ref = [ 1, 2, [ 10, 20 ] ] ;

To access 20, you can do it a number of ways:

$inside = ${$ref}[2];
$element = ${$inside}[1];

Or the following way:

Book Example> $element = ${${ref}[2]}[1];

Well, I tried the latter way, and doesn't work.

Right. The inner part is the same as $ref[2] (I got a warning to that
affect, because I turned on warnings), which is trying to access an element
of the undeclared variable @ref (I got an error to that effect, because
I turned on strict).

So this is a typo, which you have corrected below.

So I tried this
instead:

First Try> ${${$ref}[2]}[1];

This works. But I'm wondering why?

Um, because that is how Perl works. I'm not sure what kind
of answer you are looking for.

If you have a array_ref, you access an element by doing this:

Basic Example> ${$array_ref}[0];

Why are you changing both the name of the variable, and the relevant
indices, in the middle of an example? What can this do other than
cause confusion?

So if ${$array_ref}[5] is itself an array reference, if I want to
access one of it's elements, why does the First T ry work? And why
does this one fail:

${$ ${$array_ref}[0] }[5]
^^^^^^^^^^^^^^^^^ --- The reference
^^^ ^^^ -- What you do to
access elements of an array

Strip off the ${...}[5] and you get

$${$array_ref}[0], which is the same as

${${$array_ref}[0]}

As you say, ${$array_ref}[0] is itself an array ref, so, lets call it
$array_ref2, yielding:

${$array_ref2}

So the extra $ is trying to dereference that inner array ref, but is trying
to do so as if it were a reference to a scalar rather than to an array.
Yielding the error "Not a SCALAR reference". If you just delete it, then
the array ref (called $array_ref2) falls into the hands of the earlier
stripped away construct, ${...}[5], which dereferences it like the array
ref it is, and all is well.

All this tedium is a very good reason to use -> instead whenever possible.

Xho

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[Bart Lateur]

$ref = [ 1, 2, [ 10, 20 ] ] ;

To access 20, you can do it a number of ways:

$inside = ${$ref}[2];
$element = ${$inside}[1];

Or the following way:

Book Example> $element = ${${ref}[2]}[1];


Well, I tried the latter way, and doesn't work.

That's because you did a poor job in copy/paste.

$element = ${${$ref}[2]}[1];

I hope the book didn't spell it the way you did.

 

[Dr.Ruud]

DaLoverhino schreef:

$ref = [ 1, 2, [ 10, 20 ] ] ;

To access 20, you can do it a number of ways:

$inside = ${$ref}[2];
$element = ${$inside}[1];

I would just get it with $ref->[2][1].