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.
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