At 2003-08-03 14:09, Ben Rudiak-Gould wrote:
>> ((let g = \_ _ -> [EMAIL PROTECTED] -> @x in ((g ([EMAIL PROTECTED] -> @x)) [EMAIL
>> PROTECTED] = 2})) ([EMAIL PROTECTED] ->
>> @x))[EMAIL PROTECTED] = 1}
>> ((let g = \_ _ -> [EMAIL PROTECTED] -> @x in (g 2)) ([EMAIL PROTECTED] ->
>> @x))[EMAIL PROTECTED] = 1}
>
>This reduction is incorrect. Auto-lifted parameters on the RHS of an
>application get lifted out, so
>
> g ([EMAIL PROTECTED] -> @x) => ([EMAIL PROTECTED] -> g { @x = @x } @x)
Hmm... I assume you mean specifically this:
g ([EMAIL PROTECTED] -> @x)
[EMAIL PROTECTED] -> (g { @x = @x } @x)
Supposing you have this:
g = \_ -> 3
then you have this reduction:
g ?x
g ([EMAIL PROTECTED] -> @x)
[EMAIL PROTECTED] -> (g { @x = @x } @x)
[EMAIL PROTECTED] -> ((\_ -> 3) { @x = @x } @x)
error: can't apply (\_ -> 3) { @x = @x }
Something else I haven't mentioned is that you shouldn't use (->) as such
in the type signatures. This is because (->) is an ordinary
type-constructor. For instance, if you define this:
type Func = (->)
then all these are the same type:
Func a b
(->) a b
a -> b
But in your scheme, (->) has been extended to allow certain pseudo-types
in its first position. It might be better to use a different syntax
rather than overload (->) with something that isn't a type-constructor.
--
Ashley Yakeley, Seattle WA
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