From: "David A. Black" <dblack@...>
Date: 2003-12-08T03:26:35+09:00
Subject: Re: newbie question

Hi --

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.
> 
>  It means if the variable is a simple  type like a numeric it cannot be
>   modified.  If I try something like:

Variables are untyped; there's no such thing as a variable of a simple
type.  (This isn't just pickiness; it actually makes a big
difference.)

> 	   v=99
> 	   def f1(n); n=n*n ; end
>            f1(v)
>  it will not work.

Yes it will.  (Try it :-)  Any time you have a variable name on the lhs
of an assignment, the variable of that name gets assigned a new
reference.

  n = 1
  n = 2
  n = "hi"
  n = 99
  n = n * n
  n = n - 30

etc.  Every time you put 'n' on the lhs, you're starting from scratch;
previous values of 'n' have no relevance.  (On the rhs, of course, the
pre-assignment n is used to evaluate the expression.)

Literal integers are different from variables; you can't do this:

  99 = 2

since that would screw the math up quite a big :-)

>   Why can't we have something simpler/more explicit like in C or even Perl?. 

Have another look at how Ruby does it. :-)


David

-- 
David A. Black
dblack@wobblini.net