[Haskell] What makes a functional language functional?

[C.M.Brown]

Hi Jeremy,

Thanks for this very informative answer! This has certainly helped to
clear up a number of points.

Thanks,
Chris.


> Many arguments have been had about what it means for a language to be
> "functional", so that's probably not a productive line of discussion.
> (ICFP carefully doesn't stipulate language choice for the programming
> contest, for example.)
>
> Both eager and lazy evaluation can be "pure", providing referential
> transparency: all that matters of an expression is its value, and a
> subexpression may be substituted with a different one having the same
> value without changing the meaning of the surrounding context. This
> fails on languages supporting side effects.
>
> Lazy evaluation is necessary, however, in order to treat a function
> definition as a (universally applicable) equation. In Haskell, I can
> define
>
>  > three x = 3
>
> and then infer, for any expression x, that the equation
>
>    three x = 3
>
> holds. With eager evaluation, that's no longer the case: if x denotes
> a non-terminating or error-raising computation, then
>
>    three x /= 3
>
> The equation then requires a side condition:
>
>    three x = 3,  for well-defined x
>
> which complicates equational reasoning, but it doesn't break
> referential transparency.
>
> Jeremy
>
>
> On 9 Aug 2007, at 10:30, C.M.Brown wrote:
>
> > Hi,
> >
> > Is lazy evaluation necessary for a functional language to remain
> > functional?
> >
> > The reason I ask is that because it violates beta-reduction, and also
> > referential transparency (I think). In haskell, we can transform:
> >
> > g x + f x
> >
> > into:
> >
> > f x + g x
> >
> > as both f and g do not change the parameter x.
> >
> > If g always evaluates to a normal form (in both a lazy and a strict
> > world)
> >
> > g x = x
> >
> > but f is defined thus:
> >
> > f x = (\y -> if x /= 0 then x else y/x)
> >
> > And we apply f to 0 (1/0) then f becomes _|_
> >
> > therefore:
> >
> > 0 + _|_ /= _|_ + 0
> >
> > Or, does this just become:
> >
> > _|_ = _|_ ?
> >
> > Or, am I missing something totally obvious?
> >
> > Regards,
> > Chris.
> >
> >
> > _______________________________________________
> > Haskell mailing list
> > Haskell at haskell.org
> > http://www.haskell.org/mailman/listinfo/haskell
>
>
>
> Jeremy.Gibbons at comlab.ox.ac.uk
>    Oxford University Computing Laboratory,    TEL: +44 1865 283508
>    Wolfson Building, Parks Road,              FAX: +44 1865 283531
>    Oxford OX1 3QD, UK.
>    URL: http://www.comlab.ox.ac.uk/oucl/people/jeremy.gibbons.html
>
>
>