[Haskell] What makes a functional language functional?
jeremy.gibbons at comlab.ox.ac.uk
Thu Aug 9 06:49:53 EDT 2007
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
> 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
On 9 Aug 2007, at 10:30, C.M.Brown wrote:
> Is lazy evaluation necessary for a functional language to remain
> 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
> 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
> 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 _|_
> 0 + _|_ /= _|_ + 0
> Or, does this just become:
> _|_ = _|_ ?
> Or, am I missing something totally obvious?
> Haskell mailing list
> Haskell at haskell.org
Jeremy.Gibbons at comlab.ox.ac.uk
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