This one had me puzzled too - so did a traced through the program below. Please make sure your viewing window is at least this wide: <----------------------------------------------------------------> It was more helpful to think of this as using the ((->) r) instance of Monad. It is defined in ./libraries/base/Control/Monad/Instances.hs in your GHC source as:
instance Monad ((->) r) where return = const f >>= k = \ r -> k (f r) r
And const just returns its first argument like: const 1 3 => 1 const "hello" "world" => "hello" And liftM2 is defined in ./libraries/base/Control/Monad.hs as :
liftM2 :: (Monad m) => (a1 -> a2 -> r) -> m a1 -> m a2 -> m r liftM2 f m1 m2 = do { x1 <- m1; x2 <- m2; return (f x1 x2) }
So a trace of the program goes like this:
(liftM2 (&&) (< 0.5) (> -0.5))
=> do {x1 <- (< 0.5);
x2 <- (> -0.5);
return ((&&) x1 x2)}
=> (< 0.5) >>= \x1
(> -0.5) >>= \x2
return ((&&) x1 x2)
=> \r ->(\x1 ->
(> -0.5) >>= \x2
return ((&&) x1 x2))
((< 0.5) r)
r
=> \r -> (return (> -0.5) >>= \x2
return ((&&) ((< 0.5) r) x2))
r
=> \r -> (\r' -> (\x2 ->
return ((&&) (const (< 0.5) r) x2))
((> -0.5) r')
r')
r
=> \r -> (\r' -> (return ((&&) ((< 0.5) r) ((> -0.5) r'))) r') r
=> \r -> (\r' -> (const ((&&) ((< 0.5) r) ((> -0.5) r'))) r') r
=> \r -> (\r' -> ((&&) ((< 0.5) r) ((> -0.5) r'))) r
=> \r -> (\r' -> ((r < 0.5) && (r' > -0.5))) r
hope this helps,
-deech
On Mon, Oct 19, 2009 at 10:24 AM, Jordan Cooper
Whoa... how on earth does this work? How does it interpret the sections as Reader monads?
That's a job for the reader monad.
Lambda Fu, form 53 - silent reader of truth
import Control.Monad import Control.Monad.Reader
filter (liftM2 (&&) (< 0.5) (> -0.5)) xs
Regards, apfelmus
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