## Monad transformers: Implementation of a stack machine with MaybeT (State Stack)

Question

I'm trying to implement a Maybe-State monad transformer and use it to implement a simple stack machine. The definitions of state monad and maybe should be correct. Now I'm trying to implement pop:

```
pop :: MaybeT (State Stack) Int
```

So that if the stack is empty it returns nothing, otherwise it returns `Just <popped stack>`

.
This is what I have so far:

```
pop :: MaybeT (State Stack) Int
pop = guard True (do (r:rs) <- get
put rs
return r)
```

(Obviously `True`

is just a dummy placeholder - I'll implement the condition later, for now I want to get the other part right).

What is wrong with my code? From my understanding `guard`

takes a conditional (`True`

) and a function f. If the conditional is true it then gives `pure f`

.

In my case,

```
pure = MaybeT . return . Just
```

So shouldn't my function f just return a `State Stack Int`

?

Here is the full code, with my implementations of `MaybeT`

and `State`

:

```
import Control.Applicative (Alternative(..))
import Control.Monad (liftM, ap, guard)
import Control.Monad.Trans.Class (MonadTrans(lift))
main :: IO()
main = return ()
-- State Monad
--------------
newtype State s a = MakeState { runState :: s -> (a, s) }
instance Functor (State s) where
fmap = liftM
instance Applicative (State s) where
pure a = MakeState $ \s -> (a, s)
(<*>) = ap
instance Monad (State s) where
return a = MakeState $ \s -> (a, s)
m >>= k = MakeState $ \s -> let (x, s') = runState m s
in runState (k x) s'
get :: State s s
get = MakeState $ \s -> (s, s)
put :: s -> State s ()
put s = MakeState $ \_ -> ((), s)
modify :: (s -> s) -> State s ()
modify f = MakeState $ \s -> ((), f s)
-- MaybeT MonadTransformer
---------------------------
newtype MaybeT m a = MaybeT { runMaybeT :: m (Maybe a) }
instance Monad m => Functor (MaybeT m) where
fmap a x = MaybeT $ do e <- runMaybeT x
return $ fmap a e
instance Monad m => Applicative (MaybeT m) where
pure = MaybeT . return . Just
(<*>) a b = MaybeT $ do e <- runMaybeT a
f <- runMaybeT b
return $ e <*> f
instance Monad m => Monad (MaybeT m) where
return = pure
a >>= b = MaybeT $ do aa <- runMaybeT a
maybe (return Nothing) (runMaybeT . b) aa
instance Monad m => Alternative (MaybeT m) where
empty = MaybeT $ return Nothing
a <|> b = MaybeT $ do aa <- runMaybeT a
bb <- runMaybeT b
return $ aa <|> bb
instance MonadTrans MaybeT where
-- "herwrappen" van het argument
lift x = MaybeT $ do r <- x
return $ Just r
-- Stack Manipulation
---------------------
type Stack = [Int]
-- plaats het argument bovenop de stack
push :: Int -> State Stack ()
push x = do r <- get
put (x:r)
-- geef de grootte van de stack terug
size :: State Stack Int
size = do r <- get
return $ length r
-- neem het eerste element van de stack, als het aanwezig is
-- (hint: hoogle naar `guard`)
pop :: MaybeT (State Stack) Int
pop = guard (True) (do (r:rs) <- get
put rs
return r)
```

Show source

## Answers to Monad transformers: Implementation of a stack machine with MaybeT (State Stack) ( 3 )

First of all, you should understand if your stack is empty, your pattern

`r:rs <- get`

fails. But you write it in do-block, so the`fail`

function will be called. It is implemented for`Monad m => MaybeT m`

like this:`fail _ = MaybeT (return Nothing)`

. This means that if the pattern fails it returns`Nothing`

. That what you want.So, you can do like this:

`guard`

doesn't take two arguments, it only takes a`Bool`

argument.You also need to lift your state manipulations into

`MaybeT`

:For the sake of comparison, here is a cruder implementation which doesn't rely neither on

`guard`

nor on`fail`

:Producing

`empty`

when the stack is`[]`

amounts to the same thing that using`guard`

in the way you intend, or using`fail`

to exploit a failed pattern match (as in freestyle's answer).