Re: [Haskell-cafe] Why is Haskell flagging this?
Haha, not exactly. You can replace sj - get let (a, sk) = runState something sj put sk with a - something Also, you don't need do notation for single statements; do return x is just return x On Wed, Dec 22, 2010 at 7:21 PM, michael rice nowg...@yahoo.com wrote: Thanks for the tip, Ozgur. It worked for me. Is this what you had in mind, Ryan? Michael == import Control.Monad.State.Lazy import Control.Monad import System.Random type GeneratorState = State StdGen data Craps a = Roll a | Win a | Lose a deriving (Show) genRandomR :: Random a = (a,a) - GeneratorState a genRandomR = state . randomR rollDie :: GeneratorState Int rollDie = genRandomR (1,6) roll2Dice :: GeneratorState Int roll2Dice = liftM2 (+) rollDie rollDie f :: Craps [Int] - GeneratorState (Craps [Int]) f (Roll []) = do g0 - get let (throw1,g1) = runState roll2Dice g0 put g1 case throw1 of 2 - return (Lose [throw1]) 3 - return (Lose [throw1]) 7 - return (Win [throw1]) 11 - return (Win [throw1]) _ - do g1 - get let (throw2,g2) = runState roll2Dice g1 put g2 if throw2 == throw1 then do return (Win [throw1,throw2]) else if throw2 == 7 then do return (Lose [throw1,throw2]) else do f (Roll [throw1,throw2]) f (Roll z@(throw1:throws)) = do g0 - get let (throw,g1) = runState roll2Dice g0 put g1 if throw == throw1 then do return (Win (z ++ [throw])) else if throw == 7 then do return (Lose (z ++ [throw])) 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...@googlemail.com Date: Wednesday, December 22, 2010, 7:37 PM 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.comhttp://mc/compose?to=ryani.s...@gmail.com wrote: Interesting. In that case, state f = StateT $ \s - Identity (f s) allows state to replace State in that code. Ozgur -Inline Attachment Follows- ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://mc/compose?to=haskell-c...@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Why is Haskell flagging this?
Huh, that's weird, I just copy and pasted this into a new file and it worked for me. I did prepend the line module RandTest where -- ryan On Tue, Dec 21, 2010 at 6:43 PM, michael rice nowg...@yahoo.com wrote: I changed your die function to rollDie in function roll2Dice (I assume that's what you meant) but get the errors listed below. Michael import Control.Monad.State import Control.Monad import System.Random type GeneratorState = State StdGen genRandom :: Random a = GeneratorState a genRandom = State random -- similar genRandomR :: Random a = (a,a) - GeneratorState a genRandomR = State . randomR rollDie :: GeneratorState Int rollDie = genRandomR (1,6) roll2Dice :: GeneratorState Int roll2Dice = liftM2 (+) rollDie rollDie === Prelude :l craps7 [1 of 1] Compiling Main ( craps7.hs, interpreted ) craps7.hs:7:12: Not in scope: data constructor `State' craps7.hs:11:13: Not in scope: data constructor `State' Failed, modules loaded: none. Prelude --- On *Tue, 12/21/10, 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: Tuesday, December 21, 2010, 7:00 PM First, let's make some useful operations in your GeneratorState monad: -- State :: (s - (a,s)) - State s a -- random :: Random a = StdGen - (a, StdGen) genRandom :: Random a = GeneratorState a genRandom = State random -- similar genRandomR :: Random a = (a,a) - GeneratorState a genRandomR = State . randomR rollDie :: GeneratorState Int rollDie = genRandomR (1,6) roll2Dice :: GeneratorState Int roll2Dice = liftM2 (+) die die These can be used to simplify a lot of the code here. -- ryan On Fri, Dec 17, 2010 at 5:55 PM, michael rice nowg...@yahoo.comhttp://mc/compose?to=nowg...@yahoo.com wrote: Paul Graham refers to all those features as orthogonality (On Lisp, pg. 63) and you're right, Haskell has it in spades, but it takes time to understand all of it and even more time to use it effectively. One almost needs a checklist. But I think I'm catching on. I programmed this craps simulation last week. It's a problem from Problems For Computer Solution, Gruenberger Jaffray, 1965, The RAND Corp. import Control.Monad.State import System.Random type GeneratorState = State StdGen data Craps a = Roll a | Win a | Lose a deriving (Show) f :: Craps [Int] - GeneratorState (Craps [Int]) f (Roll []) = do g0 - get let (d1,g1) = randomR (1,6) g0 (d2,g2) = randomR (1,6) g1 t1 = d1+d2 put g2 case t1 of 2 - return (Lose [t1]) 3 - return (Lose [t1]) 7 - return (Win [t1]) 11 - return (Win [t1]) _ - do g2 - get let (d3,g3) = randomR (1,6) g2 (d4,g4) = randomR (1,6) g3 t2 = d3+d4 put g4 if t2 == t1 then do return (Win [t1,t2]) else if t2 == 7 then do return (Lose [t1,t2]) else f (Roll [t2,t1]) f (Roll l) = do g0 - get let (d1,g1) = randomR (1,6) g0 (d2,g2) = randomR (1,6) g1 t = d1+d2 if t == (last l) then do put g2 return (Win (reverse (t:l))) else if t == 7 then do put g2 return (Lose (reverse (t:l))) else do put g2 f (Roll (t:l)) progressive (z@(x:xs),n) (Win _) = let b = x + (last xs) in (init xs,n+b) progressive (z@(x:xs),n) (Lose _) = let b = x + (last xs) in (z ++ [b],n-b) *Main let r = evalState (sequence $ replicate 6 (f (Roll []))) (mkStdGen 987) *Main r [Win [8,12,10,3,8],Win [5,9,10,11,12,11,8,9,5],Win [7],Lose [9,7],Win [5,5],Win [5,2,6,4,6,8,5]] *Main foldl progressive ([1..10],0) r ([6],49) Function f generates the roll cycle outcomes which are then folded with the progressive betting system. In the final answer, the [6] is what's left of the original betting list [1..10]. The betting list is used to determine the bet: always bet the (first + last
Re: [Haskell-cafe] Why is Haskell flagging this?
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. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Why is Haskell flagging this?
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 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. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Why is Haskell flagging this?
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. Ozgur ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Why is Haskell flagging this?
Thanks for the tip, Ozgur. It worked for me. Is this what you had in mind, Ryan? Michael == import Control.Monad.State.Lazy import Control.Monad import System.Random type GeneratorState = State StdGen data Craps a = Roll a | Win a | Lose a deriving (Show) genRandomR :: Random a = (a,a) - GeneratorState a genRandomR = state . randomR rollDie :: GeneratorState Int rollDie = genRandomR (1,6) roll2Dice :: GeneratorState Int roll2Dice = liftM2 (+) rollDie rollDie f :: Craps [Int] - GeneratorState (Craps [Int]) f (Roll []) = do g0 - get let (throw1,g1) = runState roll2Dice g0 put g1 case throw1 of 2 - return (Lose [throw1]) 3 - return (Lose [throw1]) 7 - return (Win [throw1]) 11 - return (Win [throw1]) _ - do g1 - get let (throw2,g2) = runState roll2Dice g1 put g2 if throw2 == throw1 then do return (Win [throw1,throw2]) else if throw2 == 7 then do return (Lose [throw1,throw2]) else do f (Roll [throw1,throw2]) f (Roll z@(throw1:throws)) = do g0 - get let (throw,g1) = runState roll2Dice g0 put g1 if throw == throw1 then do return (Win (z ++ [throw])) else if throw == 7 then do return (Lose (z ++ [throw])) 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...@googlemail.com Date: Wednesday, December 22, 2010, 7:37 PM 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. Ozgur -Inline Attachment Follows- ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Why is Haskell flagging this?
First, let's make some useful operations in your GeneratorState monad: -- State :: (s - (a,s)) - State s a -- random :: Random a = StdGen - (a, StdGen) genRandom :: Random a = GeneratorState a genRandom = State random -- similar genRandomR :: Random a = (a,a) - GeneratorState a genRandomR = State . randomR rollDie :: GeneratorState Int rollDie = genRandomR (1,6) roll2Dice :: GeneratorState Int roll2Dice = liftM2 (+) die die These can be used to simplify a lot of the code here. -- ryan On Fri, Dec 17, 2010 at 5:55 PM, michael rice nowg...@yahoo.com wrote: Paul Graham refers to all those features as orthogonality (On Lisp, pg. 63) and you're right, Haskell has it in spades, but it takes time to understand all of it and even more time to use it effectively. One almost needs a checklist. But I think I'm catching on. I programmed this craps simulation last week. It's a problem from Problems For Computer Solution, Gruenberger Jaffray, 1965, The RAND Corp. import Control.Monad.State import System.Random type GeneratorState = State StdGen data Craps a = Roll a | Win a | Lose a deriving (Show) f :: Craps [Int] - GeneratorState (Craps [Int]) f (Roll []) = do g0 - get let (d1,g1) = randomR (1,6) g0 (d2,g2) = randomR (1,6) g1 t1 = d1+d2 put g2 case t1 of 2 - return (Lose [t1]) 3 - return (Lose [t1]) 7 - return (Win [t1]) 11 - return (Win [t1]) _ - do g2 - get let (d3,g3) = randomR (1,6) g2 (d4,g4) = randomR (1,6) g3 t2 = d3+d4 put g4 if t2 == t1 then do return (Win [t1,t2]) else if t2 == 7 then do return (Lose [t1,t2]) else f (Roll [t2,t1]) f (Roll l) = do g0 - get let (d1,g1) = randomR (1,6) g0 (d2,g2) = randomR (1,6) g1 t = d1+d2 if t == (last l) then do put g2 return (Win (reverse (t:l))) else if t == 7 then do put g2 return (Lose (reverse (t:l))) else do put g2 f (Roll (t:l)) progressive (z@(x:xs),n) (Win _) = let b = x + (last xs) in (init xs,n+b) progressive (z@(x:xs),n) (Lose _) = let b = x + (last xs) in (z ++ [b],n-b) *Main let r = evalState (sequence $ replicate 6 (f (Roll []))) (mkStdGen 987) *Main r [Win [8,12,10,3,8],Win [5,9,10,11,12,11,8,9,5],Win [7],Lose [9,7],Win [5,5],Win [5,2,6,4,6,8,5]] *Main foldl progressive ([1..10],0) r ([6],49) Function f generates the roll cycle outcomes which are then folded with the progressive betting system. In the final answer, the [6] is what's left of the original betting list [1..10]. The betting list is used to determine the bet: always bet the (first + last) of betting list. If a win, delete the first and last. If a loss, add loss to end of betting list. The 49 is winnings, initially 0. There's no explanation in the book of what should happen if the betting list becomes empty, or a singleton, but that could be fixed by making it longer. Comments, criticism, and better ways of doing it are welcome. Michael --- On *Fri, 12/17/10, 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, 7:45 PM No problem. Haskell is a different animal than even other functional languages in my experience, and it takes time to get used to the coolness in the type system, the lazy evaluation, the point free style, functional composition and all the other interesting techniques you now have at your fingertips for writing very expressive code :-). Do that for a while then go back to algol based languages, and wonder why the heck anyone uses those on purpose :-). (yeah there's good reasons to use them, but it starts to feel confining) Dave On Fri, Dec 17, 2010 at 4:28 PM, michael rice nowg...@yahoo.comhttp://mc/compose?to=nowg...@yahoo.com wrote: Hi, all. Plenty of answers. Thank you. Putting the list in the IO monad was deliberate. Another one I was looking
Re: [Haskell-cafe] Why is Haskell flagging this?
Thanks, Ryan.
I rewrote it yesterday. Here's my updated version.
Better?
Michael
==
import Data.Functor (($))
import System.Random
data Craps a = Roll a | Win a | Lose a deriving (Show)
-- Returns an infinite list of die throws
rollDice :: IO [Int]
rollDice = randomRs (1,6) $ newStdGen
-- fmap g rollDice - an infinite list of double dice throws.
g :: [Int] - [Int]
g (x:y:rest) = (x+y) : (g rest)
h :: Craps [Int] - [Int] - [Craps [Int]]
h (Roll []) (2:ys) = (Lose [2]) : (h (Roll []) ys)
h (Roll []) (3:ys) = (Lose [3]) : (h (Roll []) ys)
h (Roll []) (7:ys) = (Win [7]) : (h (Roll []) ys)
h (Roll []) (11:ys) = (Win [11]) : (h (Roll []) ys)
h (Roll []) (y:ys) = h (Roll [y]) ys
h (Roll z@(x:xs)) (y:ys) = if y == 7
then (Lose (z ++ [y])) : (h (Roll []) ys)
else
if x == y
then (Win (z ++ [y])) : (h (Roll []) ys)
else h (Roll (z ++ [y])) ys
progressive ((x:xs),won) (Win _) = let bet = x + (last xs)
in (init xs,won+bet)
progressive (z@(x:xs),won) (Lose _) = let bet = x + (last xs)
in (z ++ [bet],won-bet)
martingale (won,lost) (Win _) = let bet = max 1 (2*lost)
in (won+bet,0)
martingale (won,lost) (Lose _) = let bet = max 1 (2*lost)
in (won,lost+bet)
-- Play
-- n : throw cycles
-- f : betting system
-- x : starting condition
playCraps n f x = let r = fmap ((take n) . (h (Roll [])) . g) rollDice
in fmap (foldl f x) r
{-
*Main playCraps 5 progressive ([1..10],0)
([5,6,7],37)
*Main playCraps 5 martingale (0,0)
(7,1)
-}
--- On Tue, 12/21/10, 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: Tuesday, December 21, 2010, 7:00 PM
First, let's make some useful operations in your GeneratorState monad:
-- State :: (s - (a,s)) - State s a
-- random :: Random a = StdGen - (a, StdGen)
genRandom :: Random a = GeneratorState a
genRandom = State random
-- similar
genRandomR :: Random a = (a,a) - GeneratorState a
genRandomR = State . randomR
rollDie :: GeneratorState Int
rollDie = genRandomR (1,6)
roll2Dice :: GeneratorState Int
roll2Dice = liftM2 (+) die die
These can be used to simplify a lot of the code here.
-- ryan
On Fri, Dec 17, 2010 at 5:55 PM, michael rice nowg...@yahoo.com wrote:
Paul Graham refers to all those features as orthogonality (On Lisp, pg. 63)
and you're right, Haskell has it in spades, but it takes time to understand all
of it and even more time to use it effectively. One almost needs a checklist.
But I think I'm catching on. I programmed this craps simulation last week. It's
a problem from Problems For Computer Solution, Gruenberger Jaffray, 1965,
The RAND Corp.
import Control.Monad.State
import System.Random
type
GeneratorState = State StdGen
data Craps a = Roll a | Win a | Lose a deriving (Show)
f :: Craps [Int] - GeneratorState (Craps [Int])
f (Roll []) = do g0 - get
let (d1,g1) = randomR (1,6) g0
(d2,g2) = randomR (1,6) g1
t1 = d1+d2
put g2
case t1 of
2 - return (Lose [t1])
3 - return (Lose [t1])
7 - return (Win [t1])
11 - return (Win [t1])
_ - do g2 - get
let (d3,g3) = randomR (1,6) g2
(d4,g4) = randomR (1,6) g3
t2 = d3+d4
put g4
if t2 == t1
then do
return (Win [t1,t2])
else
if t2 == 7
then do
return (Lose [t1,t2])
else
f (Roll [t2,t1])
f (Roll l) = do g0 - get
let (d1,g1) = randomR (1,6) g0
(d2,g2) = randomR (1,6) g1
t = d1+d2
if t == (last l)
then do
put g2
return (Win (reverse (t:l)))
else
if t == 7
then do
put g2
return (Lose (reverse (t:l)))
else do
put g2
f (Roll (t:l))
progressive (z@(x:xs),n) (Win _) = let b = x + (last
xs)
in (init
Re: [Haskell-cafe] Why is Haskell flagging this?
I changed your die function to rollDie in function roll2Dice (I assume that's what you meant) but get the errors listed below. Michael import Control.Monad.State import Control.Monad import System.Random type GeneratorState = State StdGen genRandom :: Random a = GeneratorState a genRandom = State random -- similar genRandomR :: Random a = (a,a) - GeneratorState a genRandomR = State . randomR rollDie :: GeneratorState Int rollDie = genRandomR (1,6) roll2Dice :: GeneratorState Int roll2Dice = liftM2 (+) rollDie rollDie === Prelude :l craps7 [1 of 1] Compiling Main ( craps7.hs, interpreted ) craps7.hs:7:12: Not in scope: data constructor `State' craps7.hs:11:13: Not in scope: data constructor `State' Failed, modules loaded: none. Prelude --- On Tue, 12/21/10, 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: Tuesday, December 21, 2010, 7:00 PM First, let's make some useful operations in your GeneratorState monad: -- State :: (s - (a,s)) - State s a -- random :: Random a = StdGen - (a, StdGen) genRandom :: Random a = GeneratorState a genRandom = State random -- similar genRandomR :: Random a = (a,a) - GeneratorState a genRandomR = State . randomR rollDie :: GeneratorState Int rollDie = genRandomR (1,6) roll2Dice :: GeneratorState Int roll2Dice = liftM2 (+) die die These can be used to simplify a lot of the code here. -- ryan On Fri, Dec 17, 2010 at 5:55 PM, michael rice nowg...@yahoo.com wrote: Paul Graham refers to all those features as orthogonality (On Lisp, pg. 63) and you're right, Haskell has it in spades, but it takes time to understand all of it and even more time to use it effectively. One almost needs a checklist. But I think I'm catching on. I programmed this craps simulation last week. It's a problem from Problems For Computer Solution, Gruenberger Jaffray, 1965, The RAND Corp. import Control.Monad.State import System.Random type GeneratorState = State StdGen data Craps a = Roll a | Win a | Lose a deriving (Show) f :: Craps [Int] - GeneratorState (Craps [Int]) f (Roll []) = do g0 - get let (d1,g1) = randomR (1,6) g0 (d2,g2) = randomR (1,6) g1 t1 = d1+d2 put g2 case t1 of 2 - return (Lose [t1]) 3 - return (Lose [t1]) 7 - return (Win [t1]) 11 - return (Win [t1]) _ - do g2 - get let (d3,g3) = randomR (1,6) g2 (d4,g4) = randomR (1,6) g3 t2 = d3+d4 put g4 if t2 == t1 then do return (Win [t1,t2]) else if t2 == 7 then do return (Lose [t1,t2]) else f (Roll [t2,t1]) f (Roll l) = do g0 - get let (d1,g1) = randomR (1,6) g0 (d2,g2) = randomR (1,6) g1 t = d1+d2 if t == (last l) then do put g2 return (Win (reverse (t:l))) else if t == 7 then do put g2 return (Lose (reverse (t:l))) else do put g2 f (Roll (t:l)) progressive (z@(x:xs),n) (Win _) = let b = x + (last xs) in (init xs,n+b) progressive (z@(x:xs),n) (Lose _) = let b = x + (last xs) in (z ++ [b],n-b) *Main let r = evalState (sequence $ replicate 6 (f (Roll []))) (mkStdGen 987) *Main r [Win [8,12,10,3,8],Win [5,9,10,11,12,11,8,9,5],Win [7],Lose [9,7],Win [5,5],Win [5,2,6,4,6,8,5]] *Main foldl progressive ([1..10],0) r ([6],49) Function f generates the roll cycle outcomes which are then folded with the progressive betting system. In the final answer, the [6] is what's left of the original betting list [1..10]. The betting list is used to determine the bet: always bet the (first + last) of betting list. If a win, delete the first and last. If a loss, add loss to end of betting list. The 49 is winnings, initially 0. There's no explanation in the book of what should happen if the betting list becomes empty, or a singleton, but that could be fixed by making it longer. Comments, criticism, and better ways of doing
Re: [Haskell-cafe] Why is Haskell flagging this?
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 :: [Int] - IO [Int] f lst = do return lst main = do let lst = f [1,2,3,4,5] fmap (+1) lst === Prelude :l test [1 of 1] Compiling Main ( test.hs, interpreted ) test.hs:5:17: No instance for (Num [Int]) arising from the literal `1' at test.hs:5:17 Possible fix: add an instance declaration for (Num [Int]) In the second argument of `(+)', namely `1' In the first argument of `fmap', namely `(+ 1)' In the expression: fmap (+ 1) lst Failed, modules loaded: none. Prelude ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe Excuse any inaccuracies, I'm somewhat new at Haskell myself, but what it looks like is happening is that at the point in main where you've bound lst, it will have type of IO [Int]. The signature for fmap is: fmap :: (Functor f) = (a - b) - f a - f b if you call fmap (+1) the next argument that fmap expects is something that is in just one functor, for example, this fmap (+1) [1,2,3,4,5] works fine, but, something that is IO [Int] won't. You can compose two 'fmap's to solve this: :t (fmap.fmap) (fmap.fmap) :: (Functor f, Functor f1) = (a - b) - f (f1 a) - f (f1 b) which means that 'main' looks like: main = do let lst = f [1, 2, 3, 4, 5] (fmap.fmap) (+1) lst -- Chris Wilson christopher.j.wil...@gmail.com ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Why is Haskell flagging this?
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 . fmap (+1) Or you can not wrap the list in IO to begin with, my guess is that you wrote 'f' to make the compiler happy at some point in development: main = do let lst = [1,2,3,4,5] return $ fmap (+1) lst -deech 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 :: [Int] - IO [Int] f lst = do return lst main = do let lst = f [1,2,3,4,5] fmap (+1) lst === Prelude :l test [1 of 1] Compiling Main ( test.hs, interpreted ) test.hs:5:17: No instance for (Num [Int]) arising from the literal `1' at test.hs:5:17 Possible fix: add an instance declaration for (Num [Int]) In the second argument of `(+)', namely `1' In the first argument of `fmap', namely `(+ 1)' In the expression: fmap (+ 1) lst Failed, modules loaded: none. Prelude ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Why is Haskell flagging this?
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 can use monad notation: do lst - f [1,2,3,4] return $ fmap (+1) lst Does that make sense? Take care, Antoine 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 :: [Int] - IO [Int] f lst = do return lst main = do let lst = f [1,2,3,4,5] fmap (+1) lst === Prelude :l test [1 of 1] Compiling Main ( test.hs, interpreted ) test.hs:5:17: No instance for (Num [Int]) arising from the literal `1' at test.hs:5:17 Possible fix: add an instance declaration for (Num [Int]) In the second argument of `(+)', namely `1' In the first argument of `fmap', namely `(+ 1)' In the expression: fmap (+ 1) lst Failed, modules loaded: none. Prelude ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Why is Haskell flagging this?
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 you want to say: main = do let lst = [1,2,3,4,5] print $ map (+1) lst /Mads On Fri, 2010-12-17 at 09:04 -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 :: [Int] - IO [Int] f lst = do return lst main = do let lst = f [1,2,3,4,5] fmap (+1) lst === Prelude :l test [1 of 1] Compiling Main ( test.hs, interpreted ) test.hs:5:17: No instance for (Num [Int]) arising from the literal `1' at test.hs:5:17 Possible fix: add an instance declaration for (Num [Int]) In the second argument of `(+)', namely `1' In the first argument of `fmap', namely `(+ 1)' In the expression: fmap (+ 1) lst Failed, modules loaded: none. Prelude ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Why is Haskell flagging this?
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 [Int] and hands it off to the second fmap 2. The second fmap applies the (+1) function to every element of the list. 3. The second fmap re-wraps the elements back into a [Int] 4. The first fmap re-wraps and returns the transformed [Int] into an IO [Int]. -deech On Fri, Dec 17, 2010 at 3:27 PM, aditya siram aditya.si...@gmail.comwrote: 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 . fmap (+1) Or you can not wrap the list in IO to begin with, my guess is that you wrote 'f' to make the compiler happy at some point in development: main = do let lst = [1,2,3,4,5] return $ fmap (+1) lst -deech 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 :: [Int] - IO [Int] f lst = do return lst main = do let lst = f [1,2,3,4,5] fmap (+1) lst === Prelude :l test [1 of 1] Compiling Main ( test.hs, interpreted ) test.hs:5:17: No instance for (Num [Int]) arising from the literal `1' at test.hs:5:17 Possible fix: add an instance declaration for (Num [Int]) In the second argument of `(+)', namely `1' In the first argument of `fmap', namely `(+ 1)' In the expression: fmap (+ 1) lst Failed, modules loaded: none. Prelude ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Why is Haskell flagging this?
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 :: [Int] - IO [Int] f lst = do return lst main = do let lst = f [1,2,3,4,5] fmap (+1) lst f takes [Int] and returns IO [Int] fmap is fmap :: (Functor f) = (a - b) - f a - f b That is it takes a function of a's to b's, a functor of a, and returns you a functor of b. So when you fmap (+1) to an IO [Int], it's trying to add 1 to a [Int], and [Int] is not an instance of Num, so the + does not work. Luckily you can use function composition here (fmap . fmap) (+1) $ f [1..10] [2,3,4,5,6,7,8,9,10,11] fmap . fmap is the type I think you wanted: Prelude :t fmap . fmap fmap . fmap :: (Functor f, Functor f1) = (a - b) - f (f1 a) - f (f1 b) With IO as the f Functor, and [] as the f1 Functor. === Prelude :l test [1 of 1] Compiling Main ( test.hs, interpreted ) test.hs:5:17: No instance for (Num [Int]) arising from the literal `1' at test.hs:5:17 Possible fix: add an instance declaration for (Num [Int]) In the second argument of `(+)', namely `1' In the first argument of `fmap', namely `(+ 1)' In the expression: fmap (+ 1) lst Failed, modules loaded: none. Prelude ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Why is Haskell flagging this?
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 instance for [Int]. That's what the error was saying. What you probably want is fmap (fmap (+1)) lst. Not sure why you're doing this stuff in the first place though, since the return into IO is only restricting what you can do with it. Also, the do in both cases is unnecessary (in the second case you can replace the let with a let..in) Hope this helps, Dan On Fri, Dec 17, 2010 at 12:04 PM, 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 :: [Int] - IO [Int] f lst = do return lst main = do let lst = f [1,2,3,4,5] fmap (+1) lst === Prelude :l test [1 of 1] Compiling Main ( test.hs, interpreted ) test.hs:5:17: No instance for (Num [Int]) arising from the literal `1' at test.hs:5:17 Possible fix: add an instance declaration for (Num [Int]) In the second argument of `(+)', namely `1' In the first argument of `fmap', namely `(+ 1)' In the expression: fmap (+ 1) lst Failed, modules loaded: none. Prelude ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Why is Haskell flagging this?
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 the behaviour you expect, you need another fmap: fmap (fmap (+1)) (return [1,2,3]) ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Why is Haskell flagging this?
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 a type IO a. fmap (+1) (return [1,2,3]) But to achieve the behaviour you expect, you need another fmap: fmap (fmap (+1)) (return [1,2,3]) Which can be more neatly written with Conal's semantic editor cominators as (fmap . fmap) (+1) (return [1,2,3]) Of course, I question why the list is put in the IO monad at all here... surely this would be much better return $ fmap (+1) [1,2,3] Finally, that has the wrong type for main... perhaps you meant to print it out? main :: IO () main = print $ fmap (+1) [1,2,3] Bob___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Why is Haskell flagging this?
No problem. Haskell is a different animal than even other functional languages in my experience, and it takes time to get used to the coolness in the type system, the lazy evaluation, the point free style, functional composition and all the other interesting techniques you now have at your fingertips for writing very expressive code :-). Do that for a while then go back to algol based languages, and wonder why the heck anyone uses those on purpose :-). (yeah there's good reasons to use them, but it starts to feel confining) Dave On Fri, Dec 17, 2010 at 4:28 PM, michael rice nowg...@yahoo.com wrote: Hi, all. Plenty of answers. Thank you. Putting the list in the IO monad was deliberate. Another one I was looking at was f :: String - IO String f s = do return s main = do ios - f hello fmap tail ios which worked fine So, the big error was trying to add 1 + [1,2,3,4,5]. I considered that I needed an additional fmap and thought I had tried fmap (fmap (1+)) iol but must have messed it up, because I got an error. I guess I was on the right track. I like to try various combinations to test my understanding. It's kind of embarrassing when I get 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 this? To: haskell-cafe@haskell.org Cc: michael rice nowg...@yahoo.com Date: Friday, December 17, 2010, 4:24 PM 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 [Int] f lst = do return lst main = do let lst = f [1,2,3,4,5] fmap (+1) lst The fmap is relative to IO, your code is equivalent to do let lst = (return [1,2,3,4,5]) fmap (+1) lst ~ fmap (+1) (return [1,2,3,4,5]) ~ do lst - return [1,2,3,4,5] return $ (+1) lst but there's no instance Num [Int] in scope You probably meant do let lst = f [1,2,3,4,5] fmap (map (+1)) lst === Prelude :l test [1 of 1] Compiling Main ( test.hs, interpreted ) test.hs:5:17: No instance for (Num [Int]) arising from the literal `1' at test.hs:5:17 Possible fix: add an instance declaration for (Num [Int]) In the second argument of `(+)', namely `1' In the first argument of `fmap', namely `(+ 1)' In the expression: fmap (+ 1) lst Failed, modules loaded: none. Prelude --- 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 Cc: michael rice nowg...@yahoo.com Date: Friday, December 17, 2010, 4:24 PM 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 [Int] f lst = do return lst main = do let lst = f [1,2,3,4,5] fmap (+1) lst The fmap is relative to IO, your code is equivalent to do let lst = (return [1,2,3,4,5]) fmap (+1) lst ~ fmap (+1) (return [1,2,3,4,5]) ~ do lst - return [1,2,3,4,5] return $ (+1) lst but there's no instance Num [Int] in scope You probably meant do let lst = f [1,2,3,4,5] fmap (map (+1)) lst === Prelude :l test [1 of 1] Compiling Main ( test.hs, interpreted ) test.hs:5:17: No instance for (Num [Int]) arising from the literal `1' at test.hs:5:17 Possible fix: add an instance declaration for (Num [Int]) In the second argument of `(+)', namely `1' In the first argument of `fmap', namely `(+ 1)' In the expression: fmap (+ 1) lst Failed, modules loaded: none. Prelude ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Why is Haskell flagging this?
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 :: [Int] - IO [Int] f lst = do return lst main = do let lst = f [1,2,3,4,5] fmap (+1) lst 'f' operates on '[Int]', but '(+1)' operates on 'Int'... Does the following do what you want? main = do let lst = f [1,2,3,4,5] in fmap (map (+1)) lst === Prelude :l test [1 of 1] Compiling Main ( test.hs, interpreted ) test.hs:5:17: No instance for (Num [Int]) arising from the literal `1' at test.hs:5:17 Possible fix: add an instance declaration for (Num [Int]) In the second argument of `(+)', namely `1' In the first argument of `fmap', namely `(+ 1)' In the expression: fmap (+ 1) lst Failed, modules loaded: none. Prelude ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe -- You've been warned. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Why is Haskell flagging this?
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 you're trying to use an integer literal - in this case, 1 - as this something. The problem is that your lst has the type IO [Int] (which is the same as IO ([] Int)). fmap has the type (a - b) - f a - f b, so, it tries to unify the type of (+1) with [Int] - something - which, probably, isn't what you've meant. In fact, I'm pretty sure you wanted lst to have the type [Int] (= [] Int), without IO. You can do that using - instead of let: main = do lst - fst [1,2,3,4,5] return (fmap (+1) lst) 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 [Int] f lst = do return lst main = do let lst = f [1,2,3,4,5] fmap (+1) lst === Prelude :l test [1 of 1] Compiling Main ( test.hs, interpreted ) test.hs:5:17: No instance for (Num [Int]) arising from the literal `1' at test.hs:5:17 Possible fix: add an instance declaration for (Num [Int]) In the second argument of `(+)', namely `1' In the first argument of `fmap', namely `(+ 1)' In the expression: fmap (+ 1) lst Failed, modules loaded: none. Prelude ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Why is Haskell flagging this?
Hi, all. Plenty of answers. Thank you. Putting the list in the IO monad was deliberate. Another one I was looking at was f :: String - IO String f s = do return s main = do ios - f hello fmap tail ios which worked fine So, the big error was trying to add 1 + [1,2,3,4,5]. I considered that I needed an additional fmap and thought I had tried fmap (fmap (1+)) iol but must have messed it up, because I got an error. I guess I was on the right track. I like to try various combinations to test my understanding. It's kind of embarrassing when I get 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 this? To: haskell-cafe@haskell.org Cc: michael rice nowg...@yahoo.com Date: Friday, December 17, 2010, 4:24 PM 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 [Int] f lst = do return lst main = do let lst = f [1,2,3,4,5] fmap (+1) lst The fmap is relative to IO, your code is equivalent to do let lst = (return [1,2,3,4,5]) fmap (+1) lst ~ fmap (+1) (return [1,2,3,4,5]) ~ do lst - return [1,2,3,4,5] return $ (+1) lst but there's no instance Num [Int] in scope You probably meant do let lst = f [1,2,3,4,5] fmap (map (+1)) lst === Prelude :l test [1 of 1] Compiling Main ( test.hs, interpreted ) test.hs:5:17: No instance for (Num [Int]) arising from the literal `1' at test.hs:5:17 Possible fix: add an instance declaration for (Num [Int]) In the second argument of `(+)', namely `1' In the first argument of `fmap', namely `(+ 1)' In the expression: fmap (+ 1) lst Failed, modules loaded: none. Prelude --- 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 Cc: michael rice nowg...@yahoo.com Date: Friday, December 17, 2010, 4:24 PM 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 [Int] f lst = do return lst main = do let lst = f [1,2,3,4,5] fmap (+1) lst The fmap is relative to IO, your code is equivalent to do let lst = (return [1,2,3,4,5]) fmap (+1) lst ~ fmap (+1) (return [1,2,3,4,5]) ~ do lst - return [1,2,3,4,5] return $ (+1) lst but there's no instance Num [Int] in scope You probably meant do let lst = f [1,2,3,4,5] fmap (map (+1)) lst === Prelude :l test [1 of 1] Compiling Main ( test.hs, interpreted ) test.hs:5:17: No instance for (Num [Int]) arising from the literal `1' at test.hs:5:17 Possible fix: add an instance declaration for (Num [Int]) In the second argument of `(+)', namely `1' In the first argument of `fmap', namely `(+ 1)' In the expression: fmap (+ 1) lst Failed, modules loaded: none. Prelude ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Why is Haskell flagging this?
Paul Graham refers to all those features as orthogonality (On Lisp, pg. 63) and you're right, Haskell has it in spades, but it takes time to understand all of it and even more time to use it effectively. One almost needs a checklist. But I think I'm catching on. I programmed this craps simulation last week. It's a problem from Problems For Computer Solution, Gruenberger Jaffray, 1965, The RAND Corp. import Control.Monad.State import System.Random type GeneratorState = State StdGen data Craps a = Roll a | Win a | Lose a deriving (Show) f :: Craps [Int] - GeneratorState (Craps [Int]) f (Roll []) = do g0 - get let (d1,g1) = randomR (1,6) g0 (d2,g2) = randomR (1,6) g1 t1 = d1+d2 put g2 case t1 of 2 - return (Lose [t1]) 3 - return (Lose [t1]) 7 - return (Win [t1]) 11 - return (Win [t1]) _ - do g2 - get let (d3,g3) = randomR (1,6) g2 (d4,g4) = randomR (1,6) g3 t2 = d3+d4 put g4 if t2 == t1 then do return (Win [t1,t2]) else if t2 == 7 then do return (Lose [t1,t2]) else f (Roll [t2,t1]) f (Roll l) = do g0 - get let (d1,g1) = randomR (1,6) g0 (d2,g2) = randomR (1,6) g1 t = d1+d2 if t == (last l) then do put g2 return (Win (reverse (t:l))) else if t == 7 then do put g2 return (Lose (reverse (t:l))) else do put g2 f (Roll (t:l)) progressive (z@(x:xs),n) (Win _) = let b = x + (last xs) in (init xs,n+b) progressive (z@(x:xs),n) (Lose _) = let b = x + (last xs) in (z ++ [b],n-b) *Main let r = evalState (sequence $ replicate 6 (f (Roll []))) (mkStdGen 987) *Main r [Win [8,12,10,3,8],Win [5,9,10,11,12,11,8,9,5],Win [7],Lose [9,7],Win [5,5],Win [5,2,6,4,6,8,5]] *Main foldl progressive ([1..10],0) r ([6],49) Function f generates the roll cycle outcomes which are then folded with the progressive betting system. In the final answer, the [6] is what's left of the original betting list [1..10]. The betting list is used to determine the bet: always bet the (first + last) of betting list. If a win, delete the first and last. If a loss, add loss to end of betting list. The 49 is winnings, initially 0. There's no explanation in the book of what should happen if the betting list becomes empty, or a singleton, but that could be fixed by making it longer. Comments, criticism, and better ways of doing it are welcome. Michael --- On Fri, 12/17/10, 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, 7:45 PM No problem. Haskell is a different animal than even other functional languages in my experience, and it takes time to get used to the coolness in the type system, the lazy evaluation, the point free style, functional composition and all the other interesting techniques you now have at your fingertips for writing very expressive code :-). Do that for a while then go back to algol based languages, and wonder why the heck anyone uses those on purpose :-). (yeah there's good reasons to use them, but it starts to feel confining) Dave On Fri, Dec 17, 2010 at 4:28 PM, michael rice nowg...@yahoo.com wrote: Hi, all. Plenty of answers. Thank you. Putting the list in the IO monad was deliberate. Another one I was looking at was f :: String - IO String f s = do return s main = do ios - f hello fmap tail ios which worked fine So, the big error was trying to add 1 + [1,2,3,4,5]. I considered that I needed an additional fmap and thought I had tried fmap (fmap (1+)) iol but must have messed it up, because I got an error. I guess I was on the right track. I like to try various combinations to test my understanding. It's kind of embarrassing when I get 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
Re: [Haskell-cafe] Why is Haskell flagging this?
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 [Int] f lst = do return lst main = do let lst = f [1,2,3,4,5] fmap (+1) lst The fmap is relative to IO, your code is equivalent to do let lst = (return [1,2,3,4,5]) fmap (+1) lst ~ fmap (+1) (return [1,2,3,4,5]) ~ do lst - return [1,2,3,4,5] return $ (+1) lst but there's no instance Num [Int] in scope You probably meant do let lst = f [1,2,3,4,5] fmap (map (+1)) lst === Prelude :l test [1 of 1] Compiling Main ( test.hs, interpreted ) test.hs:5:17: No instance for (Num [Int]) arising from the literal `1' at test.hs:5:17 Possible fix: add an instance declaration for (Num [Int]) In the second argument of `(+)', namely `1' In the first argument of `fmap', namely `(+ 1)' In the expression: fmap (+ 1) lst Failed, modules loaded: none. Prelude ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
