On Wed, 2004-07-28 at 19:13, Pierre Weis wrote:
> > On Wed, 2004-07-28 at 17:26, Pierre Weis wrote:
> > 
> > > There is nothing such as your ``substitution principle'' in Caml,
> > > except if we restrict the language to truly trivial expressions (as
> > > soon as expressions may incoporate basic operators such as the integer
> > > addition your ``principle'' does not stand anymore).
> > 
> > I thought:
> > 
> > let x = expr in f x
> > 
> > and
> > 
> > f expr
> > 
> > are identical in all circumstances in Ocaml
> > for all terms 'expr', I must have missed
> > something .. do you have a counter example?
> 
> We have to precisely state the statement here:
> 
> - if you mean that ``f'' is just an ident (more precisely, a
> lowercase ident in the Caml parser parlance) bound to a unary
> function, then the two expressions are equivalent.

f is a function constant, but i didn't specify unary..

> - if f can have more than one argument, then the two expressions are
> obviously not equivalent since the first one fixes the order of
> evaluation when the second does not.

OK, thanks. You are right. Given

let x = f e in x y

and

(f e) y

might evaluate e and y in different orders.
Sorry -- you did already point this out too, i just didn't see it :)

-- 
John Skaller, mailto:skaller@users.sf.net
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