| newtype Put a = Put {
| runPut :: (a -> {-# UNPACK #-} !Buffer -> [B.ByteString])
| -> {-# UNPACK #-} !Buffer -> [B.ByteString]
| }
OK I'm beginning to get it.
Writing
data Foo a = MkFoo !a
means this: define the MkFoo constructor thus
MkFoo x = x `seq` :MkFoo x
where :MkFoo is the "real constructor". (Well that's what I think it means;
see Ian's Haskell Prime thread!)
Now you are proposing that
data Bar a = MkBar (!a -> a)
means this:
MkBar f = :MkBar (\x. x `seq` f x)
That is, even if the argument to MkBar is a lazy function, when you take a
MkBar apart you'll find a strict function.
I suppose you can combine the two notations:
data Baz a = MkBaz !(!a -> a)
means
MkBaz f = f `seq` :MkBaz (\x. x `seq` f x)
Interesting. Is that what you meant? An undesirable consequence would be that
case (MkBar bot) of MkBar f -> f `seq` 0
would return 0, because the MkBar constructor puts a lambda inside. This seems
bad. Maybe you can only put a ! inside the function type if you have a bang at
the top (like MkBaz).
Simon
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