From: Pavel Smerk <smerk@...>
Date: 2009-07-23T23:25:06+09:00
Subject: Re: How to get real reference to an object?

David A. Black wrote:
> On Thu, 23 Jul 2009, Pavel Smerk wrote:
>> Assume a big hash and/or a nested structure and the need of a plenty of 
>> operations on some hash[...][...][...] which is Float. How can one avoid the 
>> repetitious evaluation of the indices? I have not been able to get a "real" 
>> reference to that variable to do _something_like_ tmp = 
>> referenceof(hash[...][...][...]) and work with the value directly through the 
>> (dereferenced) tmp variable. In Perl I would say
>>
>> $ perl -e '$x[1][2][3] = 1; $a = \$x[1][2][3]; $$a = 3; print $x[1][2][3]'
>> 3
>>
>> (where \... is a reference and $ before $a is a dereference).
> 
> h = { :a => { :b => {} } }
> tmp = h[:a][:b]
> tmp[:x] = "hi"
> tmp[:x] << " there"
> p h                 # => {:a=>{:b=>{:x=>"hi there"}}}

Yes, but I need something like:

$ ruby -e 'h = { :a => { :b => "hi" } }; tmp = h[:a][:b]; tmp << " 
there"; p h'
{:a=>{:b=>"hi there"}}

which is OK, but not for Float:

$ ruby -e 'h = { :a => { :b => 5.0 } }; tmp = h[:a][:b]; tmp += 5.0; p h'
{:a=>{:b=>5.0}}

And that's why I'm asking for a "real" reference, because tmp apparently 
is not any kind of reference to the h[:a][:b].

> You're (almost) always dealing in references in Ruby. (And the almost
> part doesn't affect you much anyway.) Every reference is exactly one
> step away from the object; there's no such thing as a reference to a
> reference. It's very different from Perl in that respect.

Mhm, an unfortunate difference, I'm afraid...

Of course, I could write tmp = h[:a] and then many times tmp[:b]. But 
something like simple tmp as in Perl would seem to me far more elegant.

>> Morover, why the return value of the assignment is not an l-value? The 
>> following is legal in Perl ($x ||= 1) *= 2 --- why it is not legal in Ruby as 
>> well?
> 
> I assume it's because (x ||= 1) evaluates to the object 1, and 1 *= 2
> doesn't make sense.

I know, my question is, why it does not evaluate to x. Then it would 
have the same value (the value of x) and moreover it could stand on the 
left side of an another assignment (as the x *= 2 does make sense).

Thanks, P.