From: nainar <vnainar@...>
Date: 2003-12-08T16:16:52+09:00
Subject: Re: newbie question

On Mon, Dec 08, 2003 at 04:03:39AM +0900, Michael Campbell wrote:
> David A. Black wrote:
> > On Mon, 8 Dec 2003, nainar wrote:
> 
> >> I am beginning to understand this:
> >>  If you assign to the variable within the function the original is
> NOT
> >>   modified.
> >>  if you  modify the variable within the function then the original
> is modified.
> 
> That's basically it.
> 
> >>	   v=99
> >>	   def f1(n); n=n*n ; end
> >>           f1(v)
> > 
> > 
> >> it will not work.
> 
> It will, as David said, but in your case, "v" is not changed (which
> is what I think you were wondering about), where "n" is.
> 

 Yes , my last post was a bit confusing. Maybe , functions like my examples
 are  not the Ruby Way of coding.
 
 After  reading half a dozen of Perl books , I decided I was never going to
 master OO perl ( and perlXS!) and took up Python.It is much easier than perl
 but all those  'sort() is a method: len() is a function' type of rules bothered me.
 That is how I came to Ruby.It is  wonderful!.
  
 However , I am still puzzled by the rules about function parameters.
 Perl and Python  both permit you to alter the original variable, or
 work on copies of the variables inside function( as required by the
 program logic).In Ruby, outside a function an assignments and modification
 act the way you expect it to.Why should they behave differently inside a
 function?
  
 I am not a computer scientist; I presume there must  some sound logic behind
 this rule.
  
 v.nainar