From paqui.lucio at ehu.es  Fri Jan 11 10:49:26 2008
From: paqui.lucio at ehu.es (Paqui Lucio)
Date: Fri Jan 11 10:43:03 2008
Subject: [Hugs-users] wrong infered types?
Message-ID: <005501c85469$8b358960$a471e39e@SIPC64>

Hi, 
In Hugs, the infered types for the following four functions: 
f a = let m = id a in do return m 
f' a = do m <- id a 
          return m 
g a = let m =[a,a] in do return m 
g' a = do m <- [a,a] 
          return m 
respectively are 
f :: Monad a => b -> a b 
f' :: Monad a => a b -> a b 
g :: Monad a => b -> a [b] 
g' :: a -> [a] 
I didn't expect this types, do you? 
Anybody has some explanation? 
Thanks in advance, 
Paqui

--------------------------------- 
Paqui Lucio                                
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From ndmitchell at gmail.com  Fri Jan 11 10:57:21 2008
From: ndmitchell at gmail.com (Neil Mitchell)
Date: Fri Jan 11 10:50:50 2008
Subject: [Hugs-users] wrong infered types?
In-Reply-To: <005501c85469$8b358960$a471e39e@SIPC64>
References: <005501c85469$8b358960$a471e39e@SIPC64>
Message-ID: <404396ef0801110757i1b01e989n41ae9ad2b447ca66@mail.gmail.com>

Hi

In future, you are probably best off addressing these general
questions to the haskell-cafe@ mailing list.

> f a = let m = id a in do return m

We can simplify this to:

f a = let m = id a in do return m
f a = let m = a in do return m
f a = do return a
f a = return a

This shows why f has the same type as return.

>  f' a = do m <- id a
>            return m

f' a = id a >>= \m -> return m
f' a = a >>= \m -> return m

You basically return a, but have sent it through >>= which means it
must be monadic.

>  g a = let m =[a,a] in do return m

g a = return [a,a]

> g' a = do m <- [a,a]
>           return m

This is the tricky one. Here your use of m <- [a,a] has bound the
Monad in question to be the list Monad, hence you get a result in
terms of lists, not monads. do notation applies to any monad,
including list.

Thanks

Neil