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

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)

Answer 1

guard doesn't take two arguments, it only takes a Bool argument.
You also need to lift your state manipulations into MaybeT:
pop :: MaybeT (State Stack) Int
pop = do
guard True
(r:rs) <- lift get
lift $ put rs
return r

Answer 2

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

Answer 3

For the sake of comparison, here is a cruder implementation which doesn't rely neither on guard nor on fail:
pop :: MaybeT (State Stack) Int
pop = do
stk <- lift get
case stk of
[] -> empty
(r:rs) -> do
lift (put rs)
return r
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).