From: Sean O'Dell <sean@...>
Date: 2004-07-14T06:47:55+09:00
Subject: Re: Functional Ruby (Re: Why I don't use Ruby.)

On Tuesday 13 July 2004 14:31, Jim Weirich wrote:
> Sean O'Dell said:
> >> If you can rebind a symbol then you have states and side-effects...
> >
> > If the symbol is local to a function, what side-effect could there be?
>
> One of the benefits of functional languages is the ability to do
> algebraic-like manipulations of the source code ... kinda like provably
> correct refactorings.  For example, given X=12, within the scope of X I
> can replace all occurances of X with the value 12 (or the reverse, replace
> all occurances of 12 with X).  If X is sometimes 12 and sometimes 13, then
> the substitution is no longer valid anywhere in the scope of X.  That is
> why even local variables are immutable in a "pure" functional language.

But this is more of an implementation feature.  Tail recursion optimization 
and currying are features, not really part of the true definition of a 
functional language.  You could make C's variables totally immutable and 
pre-process C code in the same way.  It's a feature that's easy to do with 
functional languages, but I don't think truly at the heart of them.  Having 
mutable, local variables in a functional language disables such features, but 
the language is still a functional one.

> > No, functional languages are not supposed to have
> > operators.  That's why they're called "functional languages"; they work
> > like functions.
>
> Function being defined as an operation that returns the same result given
> the same input, right?  Operators in functional languages are functions in
> this sense.  The difference is a matter of syntax, not semantics.  E.g.
> 1+2 is allowed in a functional language.  The binary operator + is treated
> as a two argument function.

Yes, precisely what I meant.

	Sean O'Dell