On 7/14/06, Jenny678
Hallo
Can somebody tell me the type of the following function?
func :: ????????????? func f x = (z,y) where (z,y) = f x
Let's try to reason about it. Let's start with some approximation to the type of 'func'. It appears to take two parameters so we'll say, func :: a -> b -> c Now we look at the first parameter, f, and see that it can be applied to x to get a tuple. f :: b -> (d, e) Now we can replace the the type variable 'a' with the signature for f. func :: (b -> (d, e)) -> b -> c But we also notice that the type (d, e) which is returned by f is also returned by func. func :: (b -> (d, e)) -> b -> (d, e) So there you have it. That's how you can reason about types. (and hopefully I didn't make a clerical error...) HTH, Jason