Re: \x -> x < 0.5 && x > -0.5

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Jordan Cooper

19 Oct 2009 19 Oct '09

3:24 p.m.

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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Daniel Fischer

19 Oct 19 Oct

3:58 p.m.

New subject: \x -> x < 0.5 && x > -0.5

Am Montag 19 Oktober 2009 17:24:12 schrieb Jordan Cooper:

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Whoa... how on earth does this work? How does it interpret the sections as Reader monads?

liftM2 (&&) :: (Monad m) => m Bool -> m Bool -> m Bool So for (liftM2 (&&) (< 0.5) (> -0.5)) to be well typed, we must have (< 0.5) :: m Bool for some Monad m (same for (> -0.5)). Now (< 0.5) :: Fractional a => a -> Bool, hence m must be ((->) a) for some a in the Fractional class. Reader r a is just (r -> a) wrapped in a newtype, so (r -> a) 'is' the reader monad. To use it, we must import Control.Monad for liftM2 and some module which brings the instance Monad ((->) r) into scope. That could be Control.Monad.Instances or, appropriately, Control.Monad.Reader.

...

...
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

aditya siram

4:42 p.m.

New subject: \x -> x < 0.5 && x > -0.5

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 wrote:

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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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aditya siram

4:47 p.m.

New subject: \x -> x < 0.5 && x > -0.5

There was a mistake in the trace, please ignore the previous one and look at this. Again your viewing window should be this wide: <-----------------------------------------------------------------> (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 -> ((> -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 On Mon, Oct 19, 2009 at 11:42 AM, aditya siram wrote:

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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 wrote:

...
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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Re: \x -> x < 0.5 && x > -0.5

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