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I met this question today while learning haskell monad transformers.
Assume I have a type instance Monad m => Monad (CustomT m).
If there's a function f :: CustomT IO Int, and there's g :: IO (Maybe Int).
How do I access the Int of g in f?
I tried something like
f = do
mVal <- g
This didn't work because f is under CustomT IO monad while g is under MaybeT IO monad.
And then I tried
f = do
mVal <- return g
This seems to work but mVal is IO (Maybe Int) type, I eventually get nested IO like CustomT IO (IO something)
Is there a way to get that Int or Maybe Int out in f?
What knowledge is involved?
Thanks in advance.
In the general case, Jeremy's answer is what you want. But let's see if we can work with your specific case here. We have f :: CustomT IO Int and g :: IO (Maybe Int), given that there exist some instances to the effect of instance Monad m => Monad (CustomT m) and instance MonadTrans CustomT.
And what you want is to get at the Int inside of a g within the context of CustomT IO. Since we're inside of CustomT, we can basically strip that layer off trivially. Like Jeremy says, use lift to get rid of that.
lift :: (MonadTrans t, Monad m) => m a -> t m a
So now we have CustomT IO (Maybe Int). Like I said, we're inside a do-block, so using Haskell's bind (<-) syntax gets rid of the monad layer temporarily. Thus, we're dealing with Maybe Int. To get from Maybe Int to Int, the usual approach is to use maybe
maybe :: b -> (a -> b) -> Maybe a -> b
This provides a default value just in case the Maybe Int is actually Nothing. So, for instance, maybe 0 id is a function that takes a Maybe Int and yields the inner Int, or 0 if the value is Nothing. So, in the end, we have:
f = do
mVal <- maybe 0 id $ lift g
-- Other code
In the Control.Monad.Trans you have this definition for monad transformers:
class MonadTrans (t :: (* -> *) -> * -> *) where
lift :: Monad m => m a -> t m a
Which means that if CustomT has been defined properly you can do this:
f = do
mVal <- lift g
I'm working my way through some introductory Haskell materials and am currently going through Monads. I conceptually understand that the >>= operator is of type:
(Monad m) => m a -> (a -> m b) -> m b.
In that context, I'm confused as to why the following code works, i.e., why it doesn't result in a type mismatch:
main = getLine >>= \xs -> putStrLn xs
Since we know that getLine :: IO String, I'd assume that it can be 'bound' with a function of type String -> IO String. However putStrLn is of a different type: putStrLn :: String -> IO ().
So why does Haskell allow us to use >>= with these two functions?
Let's just line up the types:
(>>=) :: m a -> ( a -> m b) -> m b
getLine :: IO String
putStrLn :: (String -> IO ())
Here we have m = IO, a = String, and b = (), so we can substitute these into >>='s type signature to get a final type signature of
(>>=) :: IO String -> (String -> IO ()) -> IO ()
() is a valid type (called unit, note that it contains only one possible non-bottom value) and in the definition would be b.
a = String and b = () thus we get:
IO String -> (String -> IO ()) -> IO ()
Since we know that getLine :: IO String, I'd assume that it can be 'bound' with a function of type String -> IO String.
Why would you think that? Look again at the type signature:
(>>=) :: m a -> (a -> m b) -> m b
Thing in the left is m a, thing on the right is m b. Most particularly the bit in the middle, a -> m b, says that the function you pass to >>= takes an a and returns an m b. It doesn't say it has to return m a, it says it can be m b, where b is any random type. It doesn't have to match a.
In your example, the lambda function takes a String and returns an IO (). So a = String and b = (). And that's fine.
This is the signature of the well know >>= operator in Haskell
>>= :: Monad m => m a -> (a -> m b) -> m b
The question is why type of the function is
(a -> m b)
instead of
(a -> b)
I would say the latter one is more practical because it allows straightforward integration of existing "pure" functions in the monad being defined.
On the contrary, it seems not difficult to write a general "adapter"
adapt :: (Monad m) => (a -> b) -> (a -> m b)
but anyway I regard more probable that you already have (a -> b) instead of (a -> m b).
Note. I explain what I mean by "pratical" and "probable".
If you haven't defined any monad in a program yet, then, the functions you have are "pure" (a -> b) and you will have 0 functions of the type (a -> m b) just because you have not still defined m. If then you decide to define a monad m it comes the need of having new a -> m b functions defined.
Basically, (>>=) lets you sequence operations in such a way that latter operations can choose to behave differently based on earlier results. A more pure function like you ask for is available in the Functor typeclass and is derivable using (>>=), but if you were stuck with it alone you'd no longer be able to sequence operations at all. There's also an intermediate called Applicative which allows you to sequence operations but not change them based on the intermediate results.
As an example, let's build up a simple IO action type from Functor to Applicative to Monad.
We'll focus on a type GetC which is as follows
GetC a = Pure a | GetC (Char -> GetC a)
The first constructor will make sense in time, but the second one should make sense immediately—GetC holds a function which can respond to an incoming character. We can turn GetC into an IO action in order to provide those characters
io :: GetC a -> IO a
io (Pure a) = return a
io (GetC go) = getChar >>= (\char -> io (go char))
Which makes it clear where Pure comes from---it handles pure values in our type. Finally, we're going to make GetC abstract: we're going to disallow using Pure or GetC directly and allow our users access only to functions we define. I'll write the most important one now
getc :: GetC Char
getc = GetC Pure
The function which gets a character then immediately considers is a pure value. While I called it the most important function, it's clear that right now GetC is pretty useless. All we can possibly do is run getc followed by io... to get an effect totally equivalent to getChar!
io getc === getChar :: IO Char
But we'll build up from here.
As stated at the beginning, the Functor typeclass provides a function exactly like you're looking for called fmap.
class Functor f where
fmap :: (a -> b) -> f a -> f b
It turns out that we can instantiate GetC as a Functor so let's do that.
instance Functor GetC where
fmap f (Pure a) = Pure (f a)
fmap f (GetC go) = GetC (\char -> fmap f (go char))
If you squint, you'll notice that fmap affects the Pure constructor only. In the GetC constructor it just gets "pushed down" and deferred until later. This is a hint as to the weakness of fmap, but let's try it.
io getc :: IO Char
io (fmap ord getc) :: IO Int
io (fmap (\c -> ord + 1) getc) :: IO Int
We've gotten the ability to modify the return type of our IO interpretation of our type, but that's about it! In particular, we're still limited to getting a single character and then running back to IO to do anything interesting with it.
This is the weakness of Functor. Since, as you noted, it deals only with pure functions it gets stuck "at the end of a computation" modifying the Pure constructor only.
The next step is Applicative which extends Functor like this
class Functor f => Applicative f where
pure :: a -> f a
(<*>) :: f (a -> b) -> f a -> f b
In other words it extends the notion of injecting pure values into our context and allowing pure function application to cross over the data type. Unsurprisingly, GetC instantiates Applicative too
instance Applicative GetC where
pure = Pure
Pure f <*> Pure x = Pure (f x)
GetC gof <*> getcx = GetC (\char -> gof <*> getcx)
Pure f <*> GetC gox = GetC (\char -> fmap f (gox char))
Applicative allows us to sequence operations and that might be clear from the definition already. In fact, we can see that (<*>) pushes character application forward so that the GetC actions on either side of (<*>) get performed in order. We use Applicative like this
fmap (,) getc <*> getc :: GetC (Char, Char)
and it allows us to build incredibly interesting functions, much more complex than just Functor would. For instance, we can already form a loop and get an infinite stream of characters.
getAll :: GetC [Char]
getAll = fmap (:) getc <*> getAll
which demonstrates the nature of Applicative being able to sequence actions one after another.
The problem is that we can't stop. io getAll is an infinite loop because it just consumes characters forever. We can't tell it to stop when it sees '\n', for instance, because Applicatives sequence without noticing earlier results.
So let's go the final step an instantiate Monad
instance Monad GetC where
return = pure
Pure a >>= f = f a
GetC go >>= f = GetC (\char -> go char >>= f)
Which allows us immediately to implement a stopping getAll
getLn :: GetC String
getLn = getc >>= \c -> case c of
'\n' -> return []
s -> fmap (s:) getLn
Or, using do notation
getLn :: GetC String
getLn = do
c <- getc
case c of
'\n' -> return []
s -> fmap (s:) getLn
So what gives? Why can we now write a stopping loop?
Because (>>=) :: m a -> (a -> m b) -> m b lets the second argument, a function of the pure value, choose the next action, m b. In this case, if the incoming character is '\n' we choose to return [] and terminate the loop. If not, we choose to recurse.
So that's why you might want a Monad over a Functor. There's much more to the story, but those are the basics.
The reason is that (>>=) is more general. The function you're suggesting is called liftM and can be easily defined as
liftM :: (Monad m) => (a -> b) -> (m a -> m b)
liftM f k = k >>= return . f
This concept has its own type class called Functor with fmap :: (Functor m) => (a -> b) -> (m a -> m b). Every Monad is also a Functor with fmap = liftM, but for historical reasons this isn't (yet) captured in the type-class hierarchy.
And adapt you're suggesting can be defined as
adapt :: (Monad m) => (a -> b) -> (a -> m b)
adapt f = return . f
Notice that having adapt is equivalent to having return as return can be defined as adapt id.
So anything that has >>= can also have these two functions, but not vice versa. There are structures that are Functors but not Monads.
The intuition behind this difference is simple: A computation within a monad can depend on the results of the previous monads. The important piece is (a -> m b) which means that not just b, but also its "effect" m b can depend on a. For example, we can have
import Control.Monad
mIfThenElse :: (Monad m) => m Bool -> m a -> m a -> m a
mIfThenElse p t f = p >>= \x -> if x then t else f
but it's not possible to define this function with just Functor m constraint, using just fmap. Functors only allow us to change the value "inside", but we can't take it "out" to decide what action to take.
As others have said, your bind is the fmap function of the Functor class, a.k.a <$>.
But why is it less powerful than >>=?
it seems not difficult to write a general "adapter"
adapt :: (Monad m) => (a -> b) -> (a -> m b)
You can indeed write a function with this type:
adapt f x = return (f x)
However, this function is not able to do everything that we might want >>='s argument to do. There are useful values that adapt cannot produce.
In the list monad, return x = [x], so adapt will always return a single-element list.
In the Maybe monad, return x = Some x, so adapt will never return None.
In the IO monad, once you retrieved the result of an operation, all you can do is compute a new value from it, you can't run a subsequent operation!
etc. So in short, fmap is able to do fewer things than >>=. That doesn't mean it's useless -- it wouldn't have a name if it was :) But it is less powerful.
The whole 'point' of the monad really (that puts it above functor or applicative) is that you can determine the monad you 'return' based on the values/results of the left hand side.
For example, >>= on a Maybe type allows us to decide to return Just x or Nothing. You'll note that using functors or applicative, it is impossible to "choose" to return Just x or Nothing based on the "sequenced" Maybe.
Try implementing something like:
halve :: Int -> Maybe Int
halve n | even n = Just (n `div` 2)
| otherwise = Nothing
return 24 >>= halve >>= halve >>= halve
with only <$> (fmap1) or <*> (ap).
Actually the "straightforward integration of pure code" that you mention is a significant aspect of the functor design pattern, and is very useful. However, it is in many ways unrelated to the motivation behind >>= --- they are meant for different applications and things.
I had the same question for a while and was thinking why bother with a -> m b once mapping a -> b to m a -> m b looks more natural. This is simialr to asking "why we need a monad given the functor".
The little answer that I give to myself is that a -> m b accounts for side-effects or other complexities that you would not capture with function a -> b.
Even better wording form here (highly recommend):
monadic value M a can itself be seen as a computation. Monadic functions represent computations that are, in some way, non-standard, i.e. not naturally supported by the programming language. This can mean side effects in a pure functional language or asynchronous execution in an impure functional language. An ordinary function type cannot encode such computations and they are, instead, encoded using a datatype that has the monadic structure.
I'd put emphasis on ordinary function type cannot encode such computations, where ordinary is a -> b.
I think that J. Abrahamson's answer points to the right reason:
Basically, (>>=) lets you sequence operations in such a way that latter operations can choose to behave differently based on earlier results. A more pure function like you ask for is available in the Functor typeclass and is derivable using (>>=), but if you were stuck with it alone you'd no longer be able to sequence operations at all.
And let me show a simple counterexample against >>= :: Monad m => m a -> (a -> b) -> m b.
It is clear that we want to have values bound to a context. And perhaps we will need to sequentially chain functions over such "context-ed values". (This is just one use case for Monads).
Take Maybe simply as a case of "context-ed value".
Then define a "fake" monad class:
class Mokad m where
returk :: t -> m t
(>>==) :: m t1 -> (t1 -> t2) -> m t2
Now let's try to have Maybe be an instance of Mokad
instance Mokad Maybe where
returk x = Just x
Nothing >>== f = Nothing
Just x >>== f = Just (f x) -- ????? always Just ?????
The first problem appears: >>== is always returning Just _.
Now let's try to chain functions over Maybe using >>==
(we sequentially extract the values of three Maybes just to add them)
chainK :: Maybe Int -> Maybe Int -> Maybe Int -> Maybe Int
chainK ma mb mc = md
where
md = ma >>== \a -> mb >>== \b -> mc >>== \c -> returk $ a+b+c
But, this code doesn't compile: md type is Maybe (Maybe (Maybe Int)) because every time >>== is used, it encapsulates the previous result into a Maybe box.
And on the contrary >>= works fine:
chainOK :: Maybe Int -> Maybe Int -> Maybe Int -> Maybe Int
chainOK ma mb mc = md
where
md = ma >>= \a -> mb >>= \b -> mc >>= \c -> return (a+b+c)
Why does this function have the type:
deleteAllMp4sExcluding :: [Char] -> IO (IO ())
instead of deleteAllMp4sExcluding :: [Char] -> IO ()
Also, how could I rewrite this so that it would have a simpler definition?
Here is the function definition:
import System.FilePath.Glob
import qualified Data.String.Utils as S
deleteAllMp4sExcluding videoFileName =
let dirGlob = globDir [compile "*"] "."
f = filter (\s -> S.endswith ".mp4" s && (/=) videoFileName s) . head . fst
lst = f <$> dirGlob
in mapM_ removeFile <$> lst
<$> when applied to IOs has type (a -> b) -> IO a -> IO b. So since mapM_ removeFile has type [FilePath] -> IO (), b in this case is IO (), so the result type becomes IO (IO ()).
To avoid nesting like this, you should not use <$> when the function you're trying to apply produces an IO value. Rather you should use >>= or, if you don't want to change the order of the operands, =<<.
Riffing on sepp2k's answer, this is an excellent example to show the difference between Functor and Monad.
The standard Haskell definition of Monad goes something like this (simplified):
class Monad m where
return :: a -> m a
(>>=) :: m a -> (a -> m b) -> m b
However, this is not the only way the class could have been defined. An alternative runs like this:
class Functor m => Monad m where
return :: a -> m a
join :: m (m a) -> m a
Given that, you can define >>= in terms of fmap and join:
(>>=) :: Monad m => m a -> (a -> m b) -> m b
ma >>= f = join (f <$> ma)
We'll look at this in a simplified sketch of the problem you're running into. What you're doing can be schematized like this:
ma :: IO a
f :: a -> IO b
f <$> ma :: IO (IO b)
Now you're stuck because you need an IO b, and the Functor class has no operation that will get you there from IO (IO b). The only way to get where you want is to dip into Monad, and the join operation is precisely what solves it:
join (f <$> ma) :: IO b
But by the join/<$> definition of >>=, this is the same as:
ma >>= f :: IO a
Note that the Control.Monad library comes with a version of join (written in terms of return and (>>=)); you could put that in your function to get the result you want. But the better thing to do is to recognize that what you're trying to do is fundamentally monadic, and thus that <$> is not the right tool for the job. You're feeding the result of one action to another; that intrinsically requires you to use Monad.
Let's say that we have two monadic functions:
f :: a -> m b
g :: b -> m c
h :: a -> m c
The bind function is defined as
(>>=) :: m a -> (a -> m b) -> m b
My question is why can not we do something like below. Declare a function which would take a monadic value and returns another monadic value?
f :: a -> m b
g :: m b -> m c
h :: a -> m c
The bind function is defined as
(>>=) :: m a -> (ma -> m b) -> m b
What is in the haskell that restricts a function from taking a monadic value as it's argument?
EDIT: I think I did not make my question clear. The point is, when you are composing functions using bind operator, why is that the second argument for bind operator is a function which takes non-monadic value (b)? Why can't it take a monadic value (mb) and give back mc . Is it that, when you are dealing with monads and the function you would compose will always have the following type.
f :: a -> m b
g :: b -> m c
h :: a -> m c
and h = f 'compose' g
I am trying to learn monads and this is something I am not able to understand.
A key ability of Monad is to "look inside" the m a type and see an a; but a key restriction of Monad is that it must be possible for monads to be "inescapable," i.e., the Monad typeclass operations should not be sufficient to write a function of type Monad m => m a -> a. (>>=) :: Monad m => m a -> (a -> m b) -> m b gives you exactly this ability.
But there's more than one way to achieve that. The Monad class could be defined like this:
class Functor f where
fmap :: (a -> b) -> f a -> f b
class Functor f => Monad m where
return :: a -> m a
join :: m (m a) -> m a
You ask why could we not have a Monad m => m a -> (m a -> m b) -> m b function. Well, given f :: a -> b, fmap f :: ma -> mb is basically that. But fmap by itself doesn't give you the ability to "look inside" a Monad m => m a yet not be able to escape from it. However join and fmap together give you that ability. (>>=) can be written generically with fmap and join:
(>>=) :: Monad m => m a -> (a -> m b) -> m b
ma >>= f = join (fmap f ma)
In fact this is a common trick for defining a Monad instance when you're having trouble coming up with a definition for (>>=)—write the join function for your would-be monad, then use the generic definition of (>>=).
Well, that answers the "does it have to be the way it is" part of the question with a "no." But, why is it the way it is?
I can't speak for the designers of Haskell, but I like to think of it this way: in Haskell monadic programming, the basic building blocks are actions like these:
getLine :: IO String
putStrLn :: String -> IO ()
More generally, these basic building blocks have types that look like Monad m => m a, Monad m => a -> m b, Monad m => a -> b -> m c, ..., Monad m => a -> b -> ... -> m z. People informally call these actions. Monad m => m a is a no-argument action, Monad m => a -> m b is a one-argument action, and so on.
Well, (>>=) :: Monad m => m a -> (a -> m b) -> m b is basically the simplest function that "connects" two actions. getLine >>= putStrLn is the action that first executes getLine, and then executes putStrLn passing it the result that was obtained from executing getLine. If you had fmap and join and not >>= you'd have to write this:
join (fmap putStrLn getLine)
Even more generally, (>>=) embodies a notion much like a "pipeline" of actions, and as such is the more useful operator for using monads as a kind of programming language.
Final thing: make sure you are aware of the Control.Monad module. While return and (>>=) are the basic functions for monads, there's endless other more high-level functions that you can define using those two, and that module gathers a few dozen of the more common ones. Your code should not be forced into a straitjacket by (>>=); it's a crucial building block that's useful both on its own and as a component for larger building blocks.
why can not we do something like below. Declare a function which would take a monadic value and returns another monadic value?
f :: a -> m b
g :: m b -> m c
h :: a -> m c
Am I to understand that you wish to write the following?
compose :: (a -> m b) -> (m b -> m c) -> (a -> m c)
compose f g = h where
h = ???
It turns out that this is just regular function composition, but with the arguments in the opposite order
(.) :: (y -> z) -> (x -> y) -> (x -> z)
(g . f) = \x -> g (f x)
Let's choose to specialize (.) with the types x = a, y = m b, and z = m c
(.) :: (m b -> m c) -> (a -> m b) -> (a -> m c)
Now flip the order of the inputs, and you get the desired compose function
compose :: (a -> m b) -> (m b -> m c) -> (a -> m c)
compose = flip (.)
Notice that we haven't even mentioned monads anywhere here. This works perfectly well for any type constructor m, whether it is a monad or not.
Now let's consider your other question. Suppose we want to write the following:
composeM :: (a -> m b) -> (b -> m c) -> (a -> m c)
Stop. Hoogle time. Hoogling for that type signature, we find there is an exact match! It is >=> from Control.Monad, but notice that for this function, m must be a monad.
Now the question is why. What makes this composition different from the other one such that this one requires m to be a Monad, while the other does not? Well, the answer to that question lies at the heart of understanding what the Monad abstraction is all about, so I'll leave a more detailed answer to the various internet resources that speak about the subject. Suffice it to say that there is no way to write composeM without knowing something about m. Go ahead, try it. You just can't write it without some additional knowledge about what m is, and the additional knowledge necessary to write this function just happens to be that m has the structure of a Monad.
Let me paraphrase your question a little bit:
why can't don't we use functions of type g :: m a -> m b with Monads?
The answer is, we do already, with Functors. There's nothing especially "monadic" about fmap f :: Functor m => m a -> m b where f :: a -> b. Monads are Functors; we get such functions just by using good old fmap:
class Functor f a where
fmap :: (a -> b) -> f a -> f b
If you have two functions f :: m a -> m b and a monadic value x :: m a, you can simply apply f x. You don't need any special monadic operator for that, just function application. But a function such as f can never "see" a value of type a.
Monadic composition of functions is much stronger concept and functions of type a -> m b are the core of monadic computations. If you have a monadic value x :: m a, you cannot "get into it" to retrieve some value of type a. But, if you have a function f :: a -> m b that operates on values of type a, you can compose the value with the function using >>= to get x >>= f :: m b. The point is, f "sees" a value of type a and can work with it (but it cannot return it, it can only return another monadic value). This is the benefit of >>= and each monad is required to provide its proper implementation.
To compare the two concepts:
If you have g :: m a -> m b, you can compose it with return to get g . return :: a -> m b (and then work with >>=), but
not vice versa. In general there is no way of creating a function of type m a -> m b from a function of type a -> m b.
So composing functions of types like a -> m b is a strictly stronger concept than composing functions of types like m a -> m b.
For example: The list monad represents computations that can give a variable number of answers, including 0 answers (you can view it as non-deterministic computations). The key elements of computing within list monad are functions of type a -> [b]. They take some input and produce a variable number of answers. Composition of these functions takes the results from the first one, applies the second function to each of the results, and merges it into a single list of all possible answers.
Functions of type [a] -> [b] would be different: They'd represent computations that take multiple inputs and produce multiple answers. They can be combined too, but we get something less strong than the original concept.
Perhaps even more distinctive example is the IO monad. If you call getChar :: IO Char and used only functions of type IO a -> IO b, you'd never be able to work with the character that was read. But >>= allows you to combine such a value with a function of type a -> IO b that can "see" the character and do something with it.
As others have pointed out, there is nothing that restricts a function to take a monadic value as argument. The bind function itself takes one, but not the function that is given to bind.
I think you can make this understandable to yourself with the "Monad is a Container" metaphor. A good example for this is Maybe. While we know how to unwrap a value from the Maybe conatiner, we do not know it for every monad, and in some monads (like IO) it is entirely impossible.
The idea is now that the Monad does this behind the scenes in a way you don't have to know about. For example, you indeed need to work with a value that was returned in the IO monad, but you cannot unwrap it, hence the function that does this needs to be in the IO monad itself.
I like to think of a monad as a recipe for constructing a program with a specific context. The power that a monad provides is the ability to, at any stage within your constructed program, branch depending upon the previous value. The usual >>= function was chosen as being the most generally useful interface to this branching ability.
As an example, the Maybe monad provides a program that may fail at some stage (the context is the failure state). Consider this psuedo-Haskell example:
-- take a computation that produces an Int. If the current Int is even, add 1.
incrIfEven :: Monad m => m Int -> m Int
incrIfEven anInt =
let ourInt = currentStateOf anInt
in if even ourInt then return (ourInt+1) else return ourInt
In order to branch based on the current result of a computation, we need to be able to access that current result. The above psuedo-code would work if we had access to currentStateOf :: m a -> a, but that isn't generally possible with monads. Instead we write our decision to branch as a function of type a -> m b. Since the a isn't in a monad in this function, we can treat it like a regular value, which is much easier to work with.
incrIfEvenReal :: Monad m => m Int -> m Int
incrIfEvenReal anInt = anInt >>= branch
where branch ourInt = if even ourInt then return (ourInt+1) else return ourInt
So the type of >>= is really for ease of programming, but there are a few alternatives that are sometimes more useful. Notably the function Control.Monad.join, which when combined with fmap gives exactly the same power as >>= (either can be defined in terms of the other).
The reason (>>=)'s second argument does not take a monad as input is because there is no need to bind such a function at all. Just apply it:
m :: m a
f :: a -> m b
g :: m b -> m c
h :: c -> m b
(g (m >>= f)) >>= h
You don't need (>>=) for g at all.
The function can take a monadic value if it wants. But it is not forced to do so.
Consider the following contrived definitions, using the list monad and functions from Data.Char:
m :: [[Int]]
m = [[71,72,73], [107,106,105,104]]
f :: [Int] -> [Char]
f mx = do
g <- [toUpper, id, toLower]
x <- mx
return (g $ chr x)
You can certainly run m >>= f; the result will have type [Char].
(It's important here that m :: [[Int]] and not m :: [Int]. >>= always "strips off" one monadic layer from its first argument. If you don't want that to happen, do f m instead of m >>= f.)
As others have mentioned, nothing restricts such functions from being written.
There is, in fact, a large family of functions of type :: m a -> (m a -> m b) -> m b:
f :: Monad m => Int -> m a -> (m a -> m b) -> m b
f n m mf = replicateM_ n m >>= mf m
where
f 0 m mf = mf m
f 1 m mf = m >> mf m
f 2 m mf = m >> m >> mf m
... etc. ...
(Note the base case: when n is 0, it's simply normal functional application.)
But what does this function do? It performs a monadic action multiple times, finally throwing away all the results, and returning the application of mf to m.
Useful sometimes, but hardly generally useful, especially compared to >>=.
A quick Hoogle search doesn't turn up any results; perhaps a telling result.