]))
else do f (Roll (z ++ [throw]))
--- On *Wed, 12/22/10, Ozgur Akgun ozgurak...@gmail.com* wrote:
From: Ozgur Akgun ozgurak...@gmail.com
Subject: Re: [Haskell-cafe] Why is Haskell flagging this?
To: Ryan Ingram ryani.s...@gmail.com
Cc: haskell-cafe@haskell.org, Daniel
leim...@gmail.comhttp://mc/compose?to=leim...@gmail.com
* wrote:
From: David Leimbach
leim...@gmail.comhttp://mc/compose?to=leim...@gmail.com
Subject: Re: [Haskell-cafe] Why is Haskell flagging this?
To: michael rice nowg...@yahoo.comhttp://mc/compose?to=nowg...@yahoo.com
Cc
On Wednesday 22 December 2010 12:03:01, Ryan Ingram wrote:
Huh, that's weird, I just copy and pasted this into a new file and it
worked for me.
As a guess, you have mtl-1.*?
In mtl-2.*, State s is made a type synonym for StateT s Identity, so
there's no longer a data constructor State.
Interesting. In that case,
state f = StateT $ \s - Identity (f s)
allows state to replace State in that code.
On Wed, Dec 22, 2010 at 4:56 AM, Daniel Fischer
daniel.is.fisc...@googlemail.com wrote:
On Wednesday 22 December 2010 12:03:01, Ryan Ingram wrote:
Huh, that's weird, I just copy and
see also:
http://hackage.haskell.org/packages/archive/mtl/latest/doc/html/Control-Monad-State-Lazy.html#v:state
On 22 December 2010 20:02, Ryan Ingram ryani.s...@gmail.com wrote:
Interesting. In that case,
state f = StateT $ \s - Identity (f s)
allows state to replace State in that code.
]))
else do f (Roll (z ++ [throw]))
--- On Wed, 12/22/10, Ozgur Akgun ozgurak...@gmail.com wrote:
From: Ozgur Akgun ozgurak...@gmail.com
Subject: Re: [Haskell-cafe] Why is Haskell flagging this?
To: Ryan Ingram ryani.s...@gmail.com
Cc: haskell-cafe@haskell.org, Daniel Fischer
daniel.is.fisc
/10, David Leimbach leim...@gmail.com* wrote:
From: David Leimbach leim...@gmail.com
Subject: Re: [Haskell-cafe] Why is Haskell flagging this?
To: michael rice nowg...@yahoo.com
Cc: haskell-cafe@haskell.org, Daniel Fischer
daniel.is.fisc...@googlemail.com
Date: Friday, December 17, 2010, 7
/21/10, Ryan Ingram ryani.s...@gmail.com wrote:
From: Ryan Ingram ryani.s...@gmail.com
Subject: Re: [Haskell-cafe] Why is Haskell flagging this?
To: michael rice nowg...@yahoo.com
Cc: David Leimbach leim...@gmail.com, Daniel Fischer
daniel.is.fisc...@googlemail.com, haskell-cafe@haskell.org
Date
Ingram ryani.s...@gmail.com
Subject: Re: [Haskell-cafe] Why is Haskell flagging this?
To: michael rice nowg...@yahoo.com
Cc: David Leimbach leim...@gmail.com, Daniel Fischer
daniel.is.fisc...@googlemail.com, haskell-cafe@haskell.org
Date: Tuesday, December 21, 2010, 7:00 PM
First, let's make some
On Fri, Dec 17, 2010 at 11:04 AM, michael rice nowg...@yahoo.com wrote:
I don't understand this error message. Haskell appears not to understand
that 1 is a Num.
Prelude :t 1
1 :: (Num t) = t
Prelude :t [1,2,3,4,5]
[1,2,3,4,5] :: (Num t) = [t]
Prelude
Michael
===
f
I think it is giving you the error because you the fmap in your code is
operating on the IO monad and not the List monad. In order to get it to
work, you can remove the IO layer with = as below:
f :: [Int] - IO [Int]
f lst = do return lst
main = do let lst = f [1,2,3,4,5]
lst = return
This is a bit tricky.
The type of 'f' is '[Int] - IO [Int]', which means that the type of 'lst'
is 'IO [Int]'.
So fmap (+1) tries to add one to the [Int] underneath the 'IO' type
constructor.
You can either use two 'fmap's, the first to lift up to IO and the second to
lift into the list, or you
Hi Michael
The type of lst is IO [Int] and therefore fmap (+1) applies (+1) to
the hole lists of integers, and not to each member of the list. That is:
fmap (+1) lst =
fmap (+1) (return [1,2,3,4,5]) =
return ([1,2,3,4,5] + 1)
and you cannot say [1,2,3,4,5] + 1.
Does that make sense?
Maybe
To make that a little clearer, here is code that uses two calls to fmap to
drill through two monadic layers:
f :: [Int] - IO [Int]
f lst = do return lst
main = do let lst = f [1,2,3,4,5]
fmap (fmap (+1)) lst
So the order of operations is :
1. The first fmap converts an IO [Int] to
On Fri, Dec 17, 2010 at 9:04 AM, michael rice nowg...@yahoo.com wrote:
I don't understand this error message. Haskell appears not to understand
that 1 is a Num.
Prelude :t 1
1 :: (Num t) = t
Prelude :t [1,2,3,4,5]
[1,2,3,4,5] :: (Num t) = [t]
Prelude
Michael
===
f ::
Write out more types and it'll get more clear.
f is [Int] - IO [Int]
lst is f applied to Num a = [a], so it is of type IO [Int]
fmap is applied to lst, which means it's stepping inside the IO. That
means it's applying +1 to [1,2,3,4,5], which doesn't make much sense unless
you have a Num
On 17 December 2010 18:04, michael rice nowg...@yahoo.com wrote:
===
f :: [Int] - IO [Int]
f lst = do return lst
main = do let lst = f [1,2,3,4,5]
fmap (+1) lst
The problem is that you are applying fmap to a type IO a.
fmap (+1) (return [1,2,3])
But to achieve
On 17 Dec 2010, at 21:44, Christopher Done wrote:
On 17 December 2010 18:04, michael rice nowg...@yahoo.com wrote:
===
f :: [Int] - IO [Int]
f lst = do return lst
main = do let lst = f [1,2,3,4,5]
fmap (+1) lst
The problem is that you are applying fmap to
stumped by something simple like this, but that's
how one learns.
Thanks again,
Michael
--- On Fri, 12/17/10, Daniel Fischer daniel.is.fisc...@googlemail.com
wrote:
From: Daniel Fischer daniel.is.fisc...@googlemail.com
Subject: Re: [Haskell-cafe] Why is Haskell flagging
On Fri, Dec 17, 2010 at 09:04:20AM -0800, michael rice wrote:
I don't understand this error message. Haskell appears not to understand that
1 is a Num.
Prelude :t 1
1 :: (Num t) = t
Prelude :t [1,2,3,4,5]
[1,2,3,4,5] :: (Num t) = [t]
Prelude
Michael
===
f ::
On 17 Dec 2010, at 20:04, michael rice wrote:
I don't understand this error message. Haskell appears not to understand that
1 is a Num.
As it clearly states in the error message, it doesn't understand that [Int] is
a Num - and it's not.
No instance for Num something usually indicates that
like this, but that's how
one learns.
Thanks again,
Michael
--- On Fri, 12/17/10, Daniel Fischer daniel.is.fisc...@googlemail.com wrote:
From: Daniel Fischer daniel.is.fisc...@googlemail.com
Subject: Re: [Haskell-cafe] Why is Haskell flagging this?
To: haskell-cafe@haskell.org
...@gmail.com wrote:
From: David Leimbach leim...@gmail.com
Subject: Re: [Haskell-cafe] Why is Haskell flagging this?
To: michael rice nowg...@yahoo.com
Cc: haskell-cafe@haskell.org, Daniel Fischer
daniel.is.fisc...@googlemail.com
Date: Friday, December 17, 2010, 7:45 PM
No problem. Haskell is a different
On Friday 17 December 2010 18:04:20, michael rice wrote:
I don't understand this error message. Haskell appears not to understand
that 1 is a Num.
Prelude :t 1
1 :: (Num t) = t
Prelude :t [1,2,3,4,5]
[1,2,3,4,5] :: (Num t) = [t]
Prelude
Michael
===
f :: [Int] - IO
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