On Wed, 13 Sep 1995 [EMAIL PROTECTED] wrote:
> Well, I'm glad to see I provoked some discussion!
...
> Why should foo evaluate its argument? It sounds to me like
> Lennart is right, and I should not have let Simon lead me astray!
...
> This is assuming I have understood Lennart correctly, and that
>
> foo (Age n) = (n, Age (n+1))
> foo' a = (n, Age (n+1)) where (Age n) = a
>
> are equivalent when Age is declared as a strict datatype. Unlike
> Sebastian or Simon, I believe it would be a disaster if for a newtype
> one had to distinguish these two definitions.
I don't see how these two can be equivalent, unless a special case is
made in the semantics for data types with a single constructor when the
constructor happens to be strict. Consider
data G = F !Int | D Int
f :: G -> Bool
f (D _) = True
f (F _) = False
If Lennart is right about foo, doesn't it follow that
f (D undefined) = True?
In which case, since D is strict, we have
f undefined = f (D undefined) = True
and so, by monotonicity of f, f v = True for all v and, in particular,
f (F e) = True
This can't be right, surely?
Sebastian
