changing an instance variable

[Jason Lillywhite]

if I have these methods in a class:

def initialize(x, y)
@x, @y = x, y
end

def foo
@z = @x + @y
end

def bar
foo
@z * 2
end

What if I want to redefine @x inside method "foo" but only in there?

is there a simple way to do this?

 

[Sebastian Hungerecker]

What if I want to redefine @x inside method "foo" but only in there?
Why?

is there a simple way to do this?

There is no way to do that. You could change @x and then just change it back
at the end of the method, but that will still affect the value of @x for any
other method you call during the method.

HTH,
Sebastian

 

[Jason Lillywhite]

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!

 

[Tommy Morgan]

[Note: parts of this message were removed to make it a legal post.]

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.

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:
[/QUOTE]

 

[Jason Lillywhite]

exactly what I needed. Thank you for helping me in my feeble quest to do
just what you suggest: refactor my code a bit.

 

[Robert Klemme]

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

That is the most reasonable solution if I understand you correctly.
Basically you do not want to use the instance variable but an arbitrary
other value. Note that you can even call that parameter "x". It won't
interfere with "@x".

Kind regards

robert