From: Tommy Morgan <tommy.morgan@...>
Date: 2008-08-14T05:23:11+09:00
Subject: Re: changing an instance variable

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Maybe you could try using a default parameter?
Like so:
def foo(x = @x)
  @z = x + @y
end

Example from irb:

irb(main):001:0> @x = 1
=> 1
irb(main):002:0> @y = 2
=> 2
irb(main):003:0> def foo(x = @x)
irb(main):004:1> @z = x + @y
irb(main):005:1> end
=> nil
irb(main):006:0> foo
=> 3
irb(main):007:0> foo(2)
=> 4
irb(main):008:0> @x
=> 1

You could then re-write your bar method as such:

def bar(new_x)
  foo(new_x)
  @z * 2
end

Also, if you've got really long complicated methods, that's usually a good
sign that you should refactor your code out a bit.

Hope that helps,

--Tommy M.

On Wed, Aug 13, 2008 at 4:05 PM, Jason Lillywhite <
jason.lillywhite@gmail.com> wrote:

> Sebastian Hungerecker wrote:
> > Jason Lillywhite wrote:
> >> What if I want to redefine @x inside method "foo" but only in there?
> >
> > Why?
> >
>
> The reason is because I have big, long algorithms inside these methods
> 'foo' and 'bar' and I don't want to re-write "@z = @x + @y" when I get
> to method 'bar'. I need to evaluate the function inside 'bar' with all
> the same variables except for @x. that one variable needs to be
> different when I run bar. Maybe I have to do this:
>
>

> def initialize(x, y)
>  @x, @y = x, y
> end
>
> def foo
>  @z = @x + @y
> end
>
> def bar(new_x)
>  @z = new_x + @y
>  @z * 2
> end
>
> ??? - Please note that my algorithms are much larger so efficiency is my
> friend in this case. Thank you!
> --
> Posted via http://www.ruby-forum.com/.
>
>

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