Merry monad mixup?

michael rice

28 Jan 2011 28 Jan '11

7:20 p.m.

The first and third work, but not the second. Why? Michael ============== f :: String -> IO () f s = do putStrLn s {- g :: [String] -> IO () g l = do s <- l putStrLn s -} {- h :: [Int] -> [Int] h l = do i <- l return (i+1) -} ============== serialize2.hs:29:9: Couldn't match expected type `[String]' against inferred type `IO String' In a stmt of a 'do' expression: s <- l In the expression: do { s <- l; putStrLn s } In the definition of `g': g l = do { s <- l; putStrLn s }

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David Menendez

28 Jan 28 Jan

7:36 p.m.

On Fri, Jan 28, 2011 at 2:20 PM, michael rice wrote:

...

The first and third work, but not the second. Why?

Michael

==============

f :: String -> IO () f s = do putStrLn s

{- g :: [String] -> IO () g l = do s <- l putStrLn s -}

{- h :: [Int] -> [Int] h l = do i <- l return (i+1) -}

Written without the do-notation, your example is g :: [String] -> IO () g l = l >>= \s -> putStrLn s This won't work because (>>=) has type Monad m => m a -> (a -> m b) -> m b, and your example requires m to be both [] and IO. -- Dave Menendez http://www.eyrie.org/~zednenem/

Chris Smith

7:41 p.m.

On Fri, 2011-01-28 at 11:20 -0800, michael rice wrote:

...

The first and third work, but not the second. Why?

When you use a do block, it can be the syntactic sugar for whatever monad you like; but you do have to make a choice. Your first example had a do block for the IO monad. Your third example used the [] monad. Both are fine. The second, though, wasn't clear on what monad it was using. When you used the (<-) syntax to nondeterministically choose from a list, the compiler settled upon the [] monad. But then the next line was a statement in the IO monad. That's inconsistent, hence the error. Perhaps you wanted to build a monad out of both behaviors? In this case, you should likely look into monad transformers, and in particular, the ListT monad transformer in the List package. This would allow you to write the code you did in a monad called ListT IO, except that the IO actions would need to be lifted through the use of either `lift` or `liftIO`. -- Chris Smith

Daniel Peebles

7:46 p.m.

Beware of ListT. It only works if your internal monad is commutative, which IO is not. (Maybe would work, for example) On Fri, Jan 28, 2011 at 2:41 PM, Chris Smith wrote:

...

On Fri, 2011-01-28 at 11:20 -0800, michael rice wrote:

...
The first and third work, but not the second. Why?

When you use a do block, it can be the syntactic sugar for whatever monad you like; but you do have to make a choice. Your first example had a do block for the IO monad. Your third example used the [] monad. Both are fine.

The second, though, wasn't clear on what monad it was using. When you used the (<-) syntax to nondeterministically choose from a list, the compiler settled upon the [] monad. But then the next line was a statement in the IO monad. That's inconsistent, hence the error.

Perhaps you wanted to build a monad out of both behaviors? In this case, you should likely look into monad transformers, and in particular, the ListT monad transformer in the List package. This would allow you to write the code you did in a monad called ListT IO, except that the IO actions would need to be lifted through the use of either `lift` or `liftIO`.

-- Chris Smith

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Daniel Peebles

7:47 p.m.

And by "works", I mean, ListT is is a monad only if the internal monad is commutative. On Fri, Jan 28, 2011 at 2:46 PM, Daniel Peebles wrote:

...

Beware of ListT. It only works if your internal monad is commutative, which IO is not. (Maybe would work, for example)

On Fri, Jan 28, 2011 at 2:41 PM, Chris Smith wrote:

...
On Fri, 2011-01-28 at 11:20 -0800, michael rice wrote:

...
The first and third work, but not the second. Why?

When you use a do block, it can be the syntactic sugar for whatever monad you like; but you do have to make a choice. Your first example had a do block for the IO monad. Your third example used the [] monad. Both are fine.

The second, though, wasn't clear on what monad it was using. When you used the (<-) syntax to nondeterministically choose from a list, the compiler settled upon the [] monad. But then the next line was a statement in the IO monad. That's inconsistent, hence the error.

Perhaps you wanted to build a monad out of both behaviors? In this case, you should likely look into monad transformers, and in particular, the ListT monad transformer in the List package. This would allow you to write the code you did in a monad called ListT IO, except that the IO actions would need to be lifted through the use of either `lift` or `liftIO`.

-- Chris Smith

_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

michael rice

7:51 p.m.

So, my suspicions are confirmed. Thanks, all. Michael --- On Fri, 1/28/11, Daniel Peebles wrote: From: Daniel Peebles Subject: Re: [Haskell-cafe] Merry monad mixup? To: "Chris Smith" Cc: "michael rice" , "haskell cafe" Date: Friday, January 28, 2011, 2:47 PM And by "works", I mean, ListT is is a monad only if the internal monad is commutative. On Fri, Jan 28, 2011 at 2:46 PM, Daniel Peebles wrote: Beware of ListT. It only works if your internal monad is commutative, which IO is not. (Maybe would work, for example) On Fri, Jan 28, 2011 at 2:41 PM, Chris Smith wrote: On Fri, 2011-01-28 at 11:20 -0800, michael rice wrote:

...

The first and third work, but not the second. Why?

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Chris Smith
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