in reply to Re^3: Building a dynamic array or some other method?
in thread Building a dynamic array or some other method?

Okay.. I'm REALLY struggling to understand what's going on here - few others used similar syntax. I've tried running it through a debugger and I think that because there's SOOO much going on in individual lines, I'm missing a lot.

my $ref = \%database; $ref = $ref->{$_} //= {} for @$fields;
is puzzling. I tried to find info on //= and found only a clumping of similar nomenclature vaguely saying it transfers info into arrays. But couldn't look at intermediate values from what I think is a for loop built into the end of the end of the 2nd line. The $_ is familiar vaguely as a reference to a value within a for loop (for instance), but the $ref->{$_} part, along with the //= {} (empty set?) is throwing me as to what's happening here.

(try searching for "//=" and you get back a lot of nothing!! Yeesh... I was (somewhat) happy to have finally found the vague reference. It just didn't help me much. LOL

I'm admittedly struggling with references and dereferences, and how it's being used or addressed.

So looking at how %database gets filled isn't intuitively obvious for me. , nor the entire effort within combine and tail.

I'll study this some more. Have had to deal with my mother-in-law's recent stroke and husband ( both 90), and competing priorities in the job. It's hard to keep focused and dive slowly through this.

I did want to say thank you tybalt89!

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Re^5: Building a dynamic array or some other method?
by afoken (Chancellor) on May 11, 2024 at 10:02 UTC
    I tried to find info on //= and found only a clumping of similar nomenclature vaguely saying it transfers info into arrays.

    //= combines the // (defined-or) operator and the = (assignment) operatior into one, exactly like ||= (or + assign), += (add + assign), -= (subtract + assign), and many others.

    The // (defined-or) operator is relatively new, it was introduced in 5.10.0 and you can find a short intro in perl5100delta:

    Defined-or operator

    A new operator // (defined-or) has been implemented. The following expression:

    $a // $b

    is merely equivalent to

    defined $a ? $a : $b

    and the statement

    $c //= $d;

    can now be used instead of

    $c = $d unless defined $c;

    The // operator has the same precedence and associativity as ||. Special care has been taken to ensure that this operator Do What You Mean while not breaking old code, but some edge cases involving the empty regular expression may now parse differently. See perlop for details.


    my $ref = \%database; $ref = $ref->{$_} //= {} for @$fields;

    is puzzling.

    Yes, it is. Partly because it is optimized for minimal keystrokes, not for readability. A typical beginner's mistake.

    But it does not take much to decode. You should now know //=. $ref is initially a reference to the hash %database. (If you know C, a reference can be thought of as a pointer.) $ref->{$_} //= {} assigns a new, empty hash ({}) to $ref->{$_} unless $ref->{$_} exists and is not undef. Then, $ref is assigned whatever is stored in $ref->{$_}. And finally, the entire $ref = $ref->{$_} //= {} is executed for all elements of the $fields array reference, setting $_ to each array element. (Technically, $_ is aliased, but it does not matter here.)

    The same code, but much more verbose, looks like this:

    my $ref = \%database; foreach my $field (@$fields) { unless (defined $ref->{$field}) { $ref->{$field} = {}; } $ref = $ref->{$field}; }

    (Seven lines instead of just two. What a waste of disk space and screen estate! And all the wear on the keyboard for those unnecessary characters! A single line would have been sufficient, no whitespace needed at all: my$r=\%d;$r=$r->{$_}//={}for@$f; - SCNR)

    Alexander

    --
    Today I will gladly share my knowledge and experience, for there are no sweeter words than "I told you so". ;-)
      Alexander, THANK YOU for the pointer to the //= info, and also explaining it.

      I very much appreciate your explaining the one line of code, and breaking it out to 7 lines where I can see what's happening more fully. I'll step through it with a debugger so I can make sure I understand how this is working. (references, mostly, but it's used recursively elsewhere, and because the original code is compact and seems to work magically.)

      Your 1-line condensation (without spaces!) didn't go unnoticed. LOL

      And you had an OUTSTANDING sig there on the end of your post. (I'm gonna borrow that, if you don't mind.)

      Right up there with the coffee mug my wife gave me that says, "Of course I talk to myself. Where else am I going to get expert advice?" (I think she's riffing on my talking to myself, because she sure doesn't take any of my advice.)

      Again, THANK YOU, Alexander!