From: David Alan Black <dblack@...>
Date: 2004-04-22T08:19:12+09:00
Subject: Re: Creating bang methods

Hi --

"Jon Hurst" <jon@jonathan.hurst.name> writes:

> On reflection, perhaps I have not been clear enough. I would like to create
> bang methods of some math functions (mostly abs), but I don't know how to
> modify the reciever in place because the value of self cannot be changed.
> It's not really important that I be able to write these bang methods, but I
> would very much like to understand how one would do it.
> 
> Class Numeric
>     def abs!
>         ??
>     end
> end

Not being able to assign to self actually isn't the issue (no method
can do that).  I think you want this:

  x = -1
  x.abs!

to result in x == 1, but that isn't how bang methods work; i.e., they
don't assign a new object to a variable, but rather they change an
object in place (and, if there are one or more variables referencing
that object, those variables don't "know" that this has happened).

  irb(main):002:0> x = [1,2]
  => [1, 2]
  irb(main):003:0> x.id
  => 537881530
  irb(main):004:0> x.reverse!
  => [2, 1]
  irb(main):005:0> x.id
  => 537881530

Nothing has happened to x in this example (no reassignment) but the
underlying array has been modified.  

The problem with abs! is that you can't modify a number in place,
because numbers are immediate values (even when referenced by
variables), and there's only one copy of each in existence.  So if you
did this:

  -1.abs!

then henceforth 1 + -1 would be 2, because you would have changed -1
*itself* to 1 (not just assigned 1 to a variable that previously
referenced -1).  The same would be true of this:

  x = -1
  x.abs!   # x is the immediate value -1

This is also why there are no ++ and -- operators in Ruby.  1++ would
mean that the actual number 1 would henceforth mean 2.  


David

-- 
David A. Black
dblack@wobblini.net