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Message: [QUOTE="Jason Lillywhite, post: 4615484"] 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! [/QUOTE]