The trick is this: add :: (b -> Int) -> (b -> Int) -> (b -> Int) is equal to add :: (b -> Int) -> (b -> Int) -> b -> Int This is because the function arrow is right-associative. Reading the function type in the latter way, you can then look at add f g b = f b + g b in a new way. f has type (b -> Int), g also has this type. The b has type b and the function result is an Int. Intuitively you can look at this function in two ways, as a function that takes two functions as arguments and then returns a function, or as a function that takes two functions and a value, and then returns a value. (You can take one more step and see it as a function that takes one argument (of type b -> Int) and then returns a function of type (b -> Int) -> b -> Int.) Similarly the function can be defined as: add f g = \b -> f b + g b which is a more direct implementation of the first type you gave. It's all equivalent though. It's actually pretty cool when you get used to it. Paul Tom Poliquin wrote:
I was reading "Arrows and Computation"
http://www.soi.city.ac.uk/~ross/papers/fop.ps.gz
(trying to lose my 'beginner' status) when I saw (on page one)
add :: (b -> Int) -> (b -> Int) -> (b -> Int) add f g b = f b + g b
It seemed like the type definition was wrong (short at least).
I tried it anyway ..
module Main where add :: (b -> Int) -> (b -> Int) -> (b -> Int) add f g b = f b + g b main = do x <- return $ add (+2) (+3) 7 print x
The program compiles and runs and produces '19' !
For fun I loaded into ghci and got what I believe is the proper type ..
*Main> :t add add :: (b -> Int) -> (b -> Int) -> b -> Int
When I try the same thing with something simpler (leaving a bit off the type definition) I get the expected error (by me) ..
module Main where dog :: Int -> Int dog a b = a + b
main = do x <- return $ dog 2 3 print x
Main.hs:5:0: The equation(s) for `dog' have two arguments, but its type `Int -> Int' has only one
What am I missing? .. Apparently something fundamental about type definitions ..
Any help appreciated.
Tom
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