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Problem with do statement

Dennis Raddle

1 Jul 2011 1 Jul '11

4:08 p.m.

It's been about a year since I used Haskell and I'm rusty. I think I knew how to do this once, but I need a reminder. I've got some functions: getDeviceInfo :: Int -> IO DeviceInfo name :: DeviceInfo -> String I want to do something like func :: IO () func = do ds <- mapM getDeviceInfo [0..10] mapM_ (print . name) ds Is there a way to combine 'getDeviceInfo', 'name', and 'print' in one line? Dennis

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Alexey G

1 Jul 1 Jul

4:48 p.m.

Try this:

mapM (\x -> getDeviceInfo x >>= print . name) [0..10]

2011/7/1 Dennis Raddle

It's been about a year since I used Haskell and I'm rusty. I think I knew how to do this once, but I need a reminder.

I've got some functions:

getDeviceInfo :: Int -> IO DeviceInfo

name :: DeviceInfo -> String

I want to do something like

func :: IO () func = do ds <- mapM getDeviceInfo [0..10] mapM_ (print . name) ds

Is there a way to combine 'getDeviceInfo', 'name', and 'print' in one line?

Dennis

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Chaddaï Fouché

2 Jul 2 Jul

3:55 a.m.

On Fri, Jul 1, 2011 at 10:48 PM, Alexey G wrote:

Try this:

...
mapM (\x -> getDeviceInfo x >>= print . name) [0..10]

Though they're not very well known, Control.Monad now contains "composition" monad operators, so you can just write :

mapM_ (getDeviceInfo >=> print . name) [0..10]

which is pretty elegant (to my eyes at least). (>=>) :: (a -> m b) -> (b -> m c) -> a -> m c -- Jedaï

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Chaddaï Fouché

4 a.m.

On Sat, Jul 2, 2011 at 9:55 AM, Chaddaï Fouché wrote:

On Fri, Jul 1, 2011 at 10:48 PM, Alexey G wrote:

...
Try this:

...
mapM (\x -> getDeviceInfo x >>= print . name) [0..10]

...
mapM_ (getDeviceInfo >=> print . name) [0..10]

Or maybe, to be more consistent order-wise :

mapM_ (print . name <=< getDeviceInfo) [0..10]

(>=>) :: (a -> m b) -> (b -> m c) -> a -> m c

Oops, forgot the constraint : (>=>) :: Monad m => (a -> m b) -> (b -> m c) -> (a -> m c) -- Jedaï

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Sean Perry

1:23 a.m.

On Jul 1, 2011, at 1:08 PM, Dennis Raddle wrote:

It's been about a year since I used Haskell and I'm rusty. I think I knew how to do this once, but I need a reminder.

I've got some functions:

getDeviceInfo :: Int -> IO DeviceInfo

name :: DeviceInfo -> String

I want to do something like

func :: IO () func = do ds <- mapM getDeviceInfo [0..10] mapM_ (print . name) ds

Is there a way to combine 'getDeviceInfo', 'name', and 'print' in one line?

I end up with either: join . fmap (mapM_ (print . name)) $ mapM getDeviceInfo [1..10] or mapM getDeviceInfo [1..10] >>= mapM_ (print . name) You can move the call to name into the mapM using (fmap name . getDeviceInfo) but I left them separate to match your code better.

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Alexey G
Chaddaï Fouché
Dennis Raddle
Sean Perry