Re: Re: [Haskell-cafe] Re: StateT and modify
stupid me, that works and is more flexible than cale's solution. thanks! On 11/8/06, Nicolas Frisby <[EMAIL PROTECTED]> wrote: Applying lift outside of modifyM is not a problem. It can seem a bit tricky with the function types around. Try >modifyM $ lift . myAdd 1 instead of >modifyM $ myAdd 1 Cale's should certainly work fine and lead to more concise code for what you're after. Just thought I'd mention this in case your needs change. Good luck, Nick On 11/8/06, Peter Steiner <[EMAIL PROTECTED]> wrote: > cale's solution worked fine for me (i forgot to cc this list in my response). > > i have troubles getting your modifyM to compile, and i do not really > understand how it might without somehow lifting the function into the > inner monad. > > > import Control.Monad.State > > > > type MyState = StateT Int IO > > > > test = evalStateT bar 0 > > > > modifyM :: (MonadState s m) => (s -> m s) -> m () > > modifyM f = do > >s <- get > >s' <- f s > >put s' > > > > bar :: MyState Int > > bar = do > >modifyM $ myAdd 1 > >get > > > > myAdd :: Int -> Int -> IO Int > > myAdd x y = do > >putStr "in myAdd\n" > >return $ x + y > > fails with: > > Couldn't match `StateT Int IO' against `IO' > Expected type: StateT Int IO > Inferred type: IO > In a 'do' expression: modifyM $ (myAdd 1) > In the definition of `bar': > bar = do > modifyM $ (myAdd 1) > get > > and applying lift is not possible outside of modifyM. > what am i doing wrong? > > regards, > peter. > > On 11/8/06, Nicolas Frisby <[EMAIL PROTECTED]> wrote: > > Regardless of what monad is transformed by StateT, I think the OP's > > issue remains. > > > > modify below is straight from Gill's source at > > http://darcs.haskell.org/packages/ > > > > modify :: (MonadState s m) => (s -> s) -> m () > > modify f = do > > s <- get > > put (f s) > > > > we could add > > > > modifyM :: (MonadState s m) => (s -> m s) -> m () > > modifyM f = do > > s <- get > > s' <- f s > >put s' > > > > which I think you could use... > > > > modifyM is just a bit more flexible than Cale's liftModify, I think. > > > > On 11/8/06, Max Vasin <[EMAIL PROTECTED]> wrote: > > > >>>>> "Peter" == Peter Steiner <[EMAIL PROTECTED]> writes: > > > > > > Peter> On 11/8/06, Bulat Ziganshin <[EMAIL PROTECTED]> wrote: > > > >> Hello Peter, > > > >> > > > >> Wednesday, November 8, 2006, 1:48:24 PM, you wrote: > > > >> > > > >> > i would like to be able to debug what's happening inside the > > > >> modifier > function. that's why i want to be able to use a > > > >> modifier that's in the > IO monad > > > >> > > > >> for debugging there is 'trace' function which don't needs IO > > > >> monad > > > > > > Peter> thanks. i am aware of trace, but the potentially messed up > > > Peter> execution order makes it very hard for me to get useful > > > Peter> information out of the resulting trace. besides, IO will > > > Peter> scale to more elaborate logging mechanisms later on... > > > > > > If all you want from IO is logging why not just use MonadWriter? > > > > > > -- > > > WBR, > > > Max Vasin. > > > > > > ___ > > > 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 > > > ___ > 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] Re: StateT and modify
On 11/8/06, Max Vasin <[EMAIL PROTECTED]> wrote: Peter> thanks. i am aware of trace, but the potentially messed up Peter> execution order makes it very hard for me to get useful Peter> information out of the resulting trace. besides, IO will Peter> scale to more elaborate logging mechanisms later on... If all you want from IO is logging why not just use MonadWriter? good question. my initial idea, being lazy, was that IO provides IORefs which might prove useful later on, but then i guess that a cleanly composed monad will behave better in the long term anyways. i have to add that this is my first large haskell project and i do many design decisions on a trial'n'error basis - naturally with a strong tendency to the error side. ;-) -- peter ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: StateT and modify
cale's solution worked fine for me (i forgot to cc this list in my response). i have troubles getting your modifyM to compile, and i do not really understand how it might without somehow lifting the function into the inner monad. import Control.Monad.State type MyState = StateT Int IO test = evalStateT bar 0 modifyM :: (MonadState s m) => (s -> m s) -> m () modifyM f = do s <- get s' <- f s put s' bar :: MyState Int bar = do modifyM $ myAdd 1 get myAdd :: Int -> Int -> IO Int myAdd x y = do putStr "in myAdd\n" return $ x + y fails with: Couldn't match `StateT Int IO' against `IO' Expected type: StateT Int IO Inferred type: IO In a 'do' expression: modifyM $ (myAdd 1) In the definition of `bar': bar = do modifyM $ (myAdd 1) get and applying lift is not possible outside of modifyM. what am i doing wrong? regards, peter. On 11/8/06, Nicolas Frisby <[EMAIL PROTECTED]> wrote: Regardless of what monad is transformed by StateT, I think the OP's issue remains. modify below is straight from Gill's source at http://darcs.haskell.org/packages/ modify :: (MonadState s m) => (s -> s) -> m () modify f = do s <- get put (f s) we could add modifyM :: (MonadState s m) => (s -> m s) -> m () modifyM f = do s <- get s' <- f s put s' which I think you could use... modifyM is just a bit more flexible than Cale's liftModify, I think. On 11/8/06, Max Vasin <[EMAIL PROTECTED]> wrote: > >>>>> "Peter" == Peter Steiner <[EMAIL PROTECTED]> writes: > > Peter> On 11/8/06, Bulat Ziganshin <[EMAIL PROTECTED]> wrote: > >> Hello Peter, > >> > >> Wednesday, November 8, 2006, 1:48:24 PM, you wrote: > >> > >> > i would like to be able to debug what's happening inside the > >> modifier > function. that's why i want to be able to use a > >> modifier that's in the > IO monad > >> > >> for debugging there is 'trace' function which don't needs IO > >> monad > > Peter> thanks. i am aware of trace, but the potentially messed up > Peter> execution order makes it very hard for me to get useful > Peter> information out of the resulting trace. besides, IO will > Peter> scale to more elaborate logging mechanisms later on... > > If all you want from IO is logging why not just use MonadWriter? > > -- > WBR, > Max Vasin. > > ___ > 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 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] StateT and modify
On 11/8/06, Bulat Ziganshin <[EMAIL PROTECTED]> wrote: Hello Peter, Wednesday, November 8, 2006, 1:48:24 PM, you wrote: > i would like to be able to debug what's happening inside the modifier > function. that's why i want to be able to use a modifier that's in the > IO monad for debugging there is 'trace' function which don't needs IO monad thanks. i am aware of trace, but the potentially messed up execution order makes it very hard for me to get useful information out of the resulting trace. besides, IO will scale to more elaborate logging mechanisms later on... -- peter ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] StateT and modify
hi haskellers, i have a basic question regarding StateT encapsulating IO and the modify function. my scenario is similar to the following simple code snippet: import Control.Monad.State type MyState = StateT Int IO test = evalStateT foo 0 foo = do modify $ (+) 1 get i would like to be able to debug what's happening inside the modifier function. that's why i want to be able to use a modifier that's in the IO monad, like in the following, obviously defunct snippet: test = evalStateT bar 0 bar = do modify $ myAdd 1 get myAdd :: Int -> Int -> IO Int myAdd x y = do putStr "in myAdd\n" return $ x + y this fails because (myAdd :: Int -> Int -> IO Int) does not match the required modify argument type (Int -> Int -> Int) for MyState. Couldn't match expected type `Int' against inferred type `IO Int' In the second argument of `($)', namely `myAdd 1' In the expression: modify $ (myAdd 1) In a 'do' expression: modify $ (myAdd 1) is it possible to 'lift' StateT modify into the inner monad (IO in my case)? regards, peter. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
