[Haskell-cafe] Small question about something easy
Hi guys I am a bit new to haskell but I am doing good till now. I have to write a function that takes 2 inputs and then reutns one composite output. Now my problem is that I have to make composition of that function meaning that I have to access in some way the output of the function before it is really computed. I will show you part of my code which is working prefectly: play :: Logo - TurtleState - (Image, TurtleState) play DoNothing (pen, (x,y), angle) = (emptyImage, (pen, (x,y), angle)) play PenDown (pen, (x,y), angle) = (emptyImage,(pen, (x,y), angle)) play PenUp (pen, (x,y), angle) = (emptyImage,(pen, (x,y), angle)) play (Forward n) (pen, (x,y), angle) |pen == True = ((line (x,y) (x+n,y+n)), (True, (x+n,y+n), angle)) |otherwise = (emptyImage,(pen, (x+n,y+n), angle)) play (Turn n) (pen, (x,y), angle) = (emptyImage, (pen, (x,y), (angle+n))) play (DoNothing :: p2) (pen, (x,y), angle) = play p2 (pen, (x,y), angle) play (p1 : DoNothing) (pen, (x,y), angle) = play p1 (pen, (x,y), angle) play (PenDown :: PenUp) (pen, (x,y), angle) = (emptyImage,(pen, (x,y), angle)) play (PenDown :: (Forward n)) (pen, (x,y), angle) = play (Forward n) (True, (x,y), angle) play (PenDown :: PenDown) (pen, (x,y), angle) = (emptyImage,(pen, (x,y), angle)) play (PenDown :: (Turn n)) (pen, (x,y), angle) = play (Turn n) (pen, (x,y), angle) play (PenUp :: PenUp) (pen, (x,y), angle) = (emptyImage,(pen, (x,y), angle)) play (PenUp :: (Forward n)) (pen, (x,y), angle) = play (Forward n) (False, (x,y), angle) play (PenUp :: (Turn n)) (pen, (x,y), angle) = play (Turn n) (pen, (x,y), angle) play (PenUp :: PenDown) (pen, (x,y), angle) = (emptyImage,(pen, (x,y), angle)) Now the problem comes here: play (p1 :: p2) state |play p1 state == (i1,state1) play p2 state1 == (i2,state2) = (i1+++i2,state2) I know that if I manage to do that function the one above with this sign :: do not need to be impelmented since this one will cater for all the cases. Can you please help me? Thanks in advance! -- View this message in context: http://www.nabble.com/Small-question-about-something-easy-tp16119618p16119618.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Small question about something easy
On 18 mar 2008, at 13.51, Luke Palmer wrote: On Tue, Mar 18, 2008 at 12:24 PM, iliali16 [EMAIL PROTECTED] wrote: Now the problem comes here: play (p1 :: p2) state |play p1 state == (i1,state1) play p2 state1 == (i2,state2) = (i1+++i2,state2) I know that if I manage to do that function the one above with this sign :: do not need to be impelmented since this one will cater for all the cases. Can you please help me? You just need a nice simple let or where clause: play (p1 :: p2) state = (i1 +++ i2, state2) where (i1,state1) = play p1 state (i2,state2) = play p2 state1 Or equivalently: play (p1 :: p2) state = let (i1, state1) = play p1 state (i2, state2) = play p2 state1 in (i1 +++ i2, state2) And there's nothing lazily recursive about these, just the information usage is a little more complex. But it could be implemented perfectly naturally in scheme, for example. For further exploration: the pattern here where the state is threaded through different computations, is captured by the module Control.Monad.State. So if play returned an object of a State monad, such as: play :: Logo - State TurtleState Image Then this case could be implemented as: play (p1 :: p2) = do i1 - play p1 i2 - play p2 return (i1 +++ i2) Pretty, ain't it? A little too pretty if you ask me. Let's make it uglier and shorter still: play (p1 :: p2) = liftM2 (+++) (play p1) (play p2) Or use Applicative directly: play (p1 :: p2) = (+++) $ play p1 * play p2 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Small question about something easy
On Tue, Mar 18, 2008 at 12:24 PM, iliali16 [EMAIL PROTECTED] wrote: Now the problem comes here: play (p1 :: p2) state |play p1 state == (i1,state1) play p2 state1 == (i2,state2) = (i1+++i2,state2) I know that if I manage to do that function the one above with this sign :: do not need to be impelmented since this one will cater for all the cases. Can you please help me? You just need a nice simple let or where clause: play (p1 :: p2) state = (i1 +++ i2, state2) where (i1,state1) = play p1 state (i2,state2) = play p2 state1 Or equivalently: play (p1 :: p2) state = let (i1, state1) = play p1 state (i2, state2) = play p2 state1 in (i1 +++ i2, state2) And there's nothing lazily recursive about these, just the information usage is a little more complex. But it could be implemented perfectly naturally in scheme, for example. For further exploration: the pattern here where the state is threaded through different computations, is captured by the module Control.Monad.State. So if play returned an object of a State monad, such as: play :: Logo - State TurtleState Image Then this case could be implemented as: play (p1 :: p2) = do i1 - play p1 i2 - play p2 return (i1 +++ i2) Pretty, ain't it? A little too pretty if you ask me. Let's make it uglier and shorter still: play (p1 :: p2) = liftM2 (+++) (play p1) (play p2) :-) Luke ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Small question about something easy
Thanks to all of you I got it I was missing the notation. Thanks again! iliali16 wrote: Hi guys I am a bit new to haskell but I am doing good till now. I have to write a function that takes 2 inputs and then reutns one composite output. Now my problem is that I have to make composition of that function meaning that I have to access in some way the output of the function before it is really computed. I will show you part of my code which is working prefectly: play :: Logo - TurtleState - (Image, TurtleState) play DoNothing (pen, (x,y), angle) = (emptyImage, (pen, (x,y), angle)) play PenDown (pen, (x,y), angle) = (emptyImage,(pen, (x,y), angle)) play PenUp (pen, (x,y), angle) = (emptyImage,(pen, (x,y), angle)) play (Forward n) (pen, (x,y), angle) |pen == True = ((line (x,y) (x+n,y+n)), (True, (x+n,y+n), angle)) |otherwise = (emptyImage,(pen, (x+n,y+n), angle)) play (Turn n) (pen, (x,y), angle) = (emptyImage, (pen, (x,y), (angle+n))) play (DoNothing :: p2) (pen, (x,y), angle) = play p2 (pen, (x,y), angle) play (p1 : DoNothing) (pen, (x,y), angle) = play p1 (pen, (x,y), angle) play (PenDown :: PenUp) (pen, (x,y), angle) = (emptyImage,(pen, (x,y), angle)) play (PenDown :: (Forward n)) (pen, (x,y), angle) = play (Forward n) (True, (x,y), angle) play (PenDown :: PenDown) (pen, (x,y), angle) = (emptyImage,(pen, (x,y), angle)) play (PenDown :: (Turn n)) (pen, (x,y), angle) = play (Turn n) (pen, (x,y), angle) play (PenUp :: PenUp) (pen, (x,y), angle) = (emptyImage,(pen, (x,y), angle)) play (PenUp :: (Forward n)) (pen, (x,y), angle) = play (Forward n) (False, (x,y), angle) play (PenUp :: (Turn n)) (pen, (x,y), angle) = play (Turn n) (pen, (x,y), angle) play (PenUp :: PenDown) (pen, (x,y), angle) = (emptyImage,(pen, (x,y), angle)) Now the problem comes here: play (p1 :: p2) state |play p1 state == (i1,state1) play p2 state1 == (i2,state2) = (i1+++i2,state2) I know that if I manage to do that function the one above with this sign :: do not need to be impelmented since this one will cater for all the cases. Can you please help me? Thanks in advance! :jumping::jumping::jumping::jumping::jumping::jumping::jumping::jumping::jumping::jumping::jumping::jumping::jumping::jumping: -- View this message in context: http://www.nabble.com/Small-question-about-something-easy-tp16119618p16121996.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
