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)
Related
I'm new to functional programming (coming from javascript), and I'm having a hard time telling the difference between the two, which is also messing with my understand of functors vs. monads.
Functor:
class Functor f where
fmap :: (a -> b) -> f a -> f b
Monad (simplified):
class Monad m where
(>>=) :: m a -> (a -> m b) -> m b
fmap takes a function and a functor, and returns a functor.
>>= takes a function and a monad, and returns a monad.
The difference between the two is in the function parameter:
fmap - (a -> b)
>>= - (a -> m b)
>>= takes a function parameter that returns a monad. I know that this is significant, but I'm having difficulty seeing how this one slight thing makes monads much more powerful than functors. Can someone explain?
Well, (<$>) is an alias for fmap, and (=<<) is the same as (>>=) with the arguments swapped:
(<$>) :: (x -> y) -> b x -> b y
(=<<) :: (x -> b y) -> b x -> b y
The difference is now fairly clear: with the bind function, we apply a function that returns a b y rather than a y. So what difference does that make?
Consider this small example:
foo <$> Just 3
Notice that (<$>) will apply foo to 3, and put the result back into a Just. In other words, the result of this computation cannot be Nothing. On the contrary:
bar =<< Just 3
This computation can return Nothing. (For example, bar x = Nothing will do it.)
We can do a similar thing with the list monad:
foo <$> [Red, Yellow, Blue] -- Result is guaranteed to be a 3-element list.
bar =<< [Red, Yellow, Blue] -- Result can be ANY size.
In short, with (<$>) (i.e., fmap), the "structure" of the result is always identical to the input. But with (=<<) (i.e., (>>=)), the structure of the result can change. This allows conditional execution, reacting to input, and a whole bunch of other things.
Short answer is that if you can turn m (m a) into m a in a way which makes sense then it's a Monad. This is possible for all Monads but not necessarily for Functors.
I think the most confusing thing is that all of the common examples of Functors (e.g. List, Maybe, IO) are also Monads. We need an example of something that is a Functor but not a Monad.
I'll use an example from a hypothetical calendar program. The following code defines an Event Functor which stores some data that goes with the event and the time that it occurs.
import Data.Time.LocalTime
data Event a = MkEvent LocalTime a
instance Functor Event where
fmap f (MkEvent time a) = MkEvent time (f a)
The Event object stores the time that the event occurs and some extra data that can be changed using fmap. Now lets try and make it a Monad:
instance Monad Event where
(>>=) (MkEvent timeA a) f = let (MkEvent timeB b) = f a in
MkEvent <notSureWhatToPutHere> b
We find that we can't because you will end up with two LocalTime objects. timeA from the given Event and timeB from the Event given by the result of f a. Our Event type is defined as having only one LocalTime (time) that it occurs at and so making it a Monad isn't possible without turning two LocalTimes into one. (There may be some case where doing so might make sense and so you could turn this into a monad if you really wanted to).
Assume for a moment that IO were just a Functor, and not a Monad. How could we sequence two actions? Say, like getChar :: IO Char and putChar :: Char -> IO ().
We could try mapping over getChar (an action that, when executed, reads a Char from stdin) using putChar.
fmap putChar getChar :: IO (IO ())
Now we have a program that, when executed, reads a Char from stdin and produces a program that, when executed, writes the Char to stdout. But what we actually want is a program that, when executed, reads a Char from stdin and writes the Char to stdout. So we need a "flattening" (in the IO case, "sequencing") function with type:
join :: IO (IO ()) -> IO ()
Functor by itself does not provide this function. But it is a function of Monad, where it has the more general type:
join :: Monad m => m (m a) -> m a
What does all of this have to do with >>=? As it happens, monadic bind is just a combination of fmap and join:
:t \m f -> join (fmap f m)
(Monad m) => m a1 -> (a1 -> m a) -> m a
Another way of seeing the difference is that fmap never changes the overall structure of the mapped value, but join (and therefore >>= as well) can do that.
In terms of IO actions, fmap will never cause addicional reads/writes or other effects. But join sequences the read/writes of the inner action after those of the outer action.
The typical monad bind function has the following signature:
m a -> (a -> m b) -> m b
As I understand it (and I might well be wrong,) the function (a -> m b) is just a mapping function from one structure a to another b. Assuming that is correct begs the question why bind's signature is not simply:
m a -> (a -> b) -> m b
Given that unit is part of a monad's definition; why give the function (a -> m b) the responsibility to call unit on whatever value b it produced – wouldn't it be more sensible to make it part of bind?
A function like m a -> (a -> b) -> m b would be equivalent to fmap :: (a -> b) -> f a -> f b. All fmap can do is change the values inside the action, it can't perform new actions. With m a -> (a -> m b) -> m b, you can "run" the m a, feed that value into (a -> m b), then return a new effect of m b. Without this, you would only ever be able to have one effect in your program, you couldn't have two print statements sequentially, you couldn't connect to a network and then download a URL, and you couldn't respond to user input, you would only be able to transform the value returned from each primitive operation. It's this operation that allows monads to be more powerful than either functors or applicatives.
Another detail here is that you aren't necessarily just wrapping a value with unit, that m b could represent an action, not just returning something. For example, where is the call to return in the action putStrLn :: String -> m ()? This function's signature is compatible with the second argument to >>=, with a ~ String and b ~ (), but there is not call to return anywhere in its body. The point of >>= is to sequence two actions together, not just to wrap values in a context.
Because m a -> (a -> b) -> m b is just fmap which a Monas has, being functor.
What a Monad add to a functor is the ability to join (or squash) a nested Monad to a simple one.
Example a list of list to simple list , or [[1,2], [3]] to [1,2,3].
If you replace b with m b in the fmap signature you end up with
m a -> (a -> m b) -> m (m b)
With a normal functor, you are stuck with your double layer of container (m (m b)). With a Monad,
using the join function, you can squash the m (m b) to m b. So bind is in fact join.fmap.
In fact, join and fmap can be written using only bind (and return), so in practice, it's easier to only define one function bind, instead of two join and fmap, even though it often simpler to write the laters.
So basically, bind is a mix of fmap and join.
As I understand it (and I might well be wrong,) the function (a -> m b) is just a mapping function from one structure a to another b
You're quite right about this – if you change the word "mapping" to morphism. For functions a -> m b are morphisms of the monad's Kleisli category. In that light, the characteristic feature of monads is that you can compose Kleislis in the same way you can compose functions:
type Kleisli m a b = a -> m b -- `Control.Arrow` has this as a `newtype` with `Category` instance.
-- compare (.) :: (b->c) -> (a->b) -> a->c
(<=<) :: Kleisli m b c -> Kleisli m a b -> Kleisli m a c
(f<=<g) x = f =<< g x
Also, you can use ordinary functions as Kleislis:
(return.) :: (a->b) -> Kleisli m a b
However, Kleislis are strictly more powerful than functions. E.g. for m ≡ IO, they are basically functions which can have side-effects, which as you know ordinary Haskell functions can't. So you can't turn a Kleisli back into a function – and if >>= accepted an a->b rather than a Kleisli m a b, but all you had was a Kleisli, there would be no way to use it.
A function of type a -> m b has potentially many more capabilities than one of type a -> b followed by return (or as you call it, "unit"). In fact no "effectful" operation can be expressed in the latter form.
Another take on this: any useful monad will have a number of operations specific to it, beyond just the ones that come from the monadic interface. For example, the IO monad has getLine :: IO String. Consider this very simple program:
main :: IO ()
main = do name <- prompt "What's your name?"
putStrLn ("Hello " ++ name ++ "!")
prompt :: String -> IO String
prompt str = do putStrLn str
getLine
Note that the type of prompt fits the a -> m b mold, but it doesn't use return (a.k.a. unit) anywhere. This is because it uses getLine :: IO String, an opaque operation provided by the IO monad and which cannot be defined in terms of return and >>=.
Think of it this way: ultimately, Monad is never something you use on its own; it's an interface for plugging together things that are extrinsic to it, like getLine and putStrLn.
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What is a monad?
I am learning to program in the functional language of Haskell and I came across Monads when studying parsers. I had never heard of them before and so I did some extra studying to find out what they are.
Everywhere I look in order to learn this topic just confuses me. I can't really find a simple definition of what a Monad is and how to use them. "A monad is a way to structure computations in terms of values and sequences of computations using those values" - eh???
Can someone please provide a simple definition of what a Monad is in Haskell, the laws associated with them and give an example?
Note: I know how to use the do syntax as I have had a look at I/O actions and functions with side-effects.
Intuition
A rough intuition would be that a Monad is a particular kind of container (Functor), for which you have two operations available. A wrapping operation return that takes a single element into a container. An operation join that merges a container of containers into a single container.
return :: Monad m => a -> m a
join :: Monad m => m (m a) -> m a
So for the Monad Maybe you have:
return :: a -> Maybe a
return x = Just x
join :: Maybe (Maybe a) -> Maybe a
join (Just (Just x) = Just x
join (Just Nothing) = Nothing
join Nothing = Nothing
Likewise for the Monad [ ] these operations are defined to be:
return :: a -> [a]
return x = [x]
join :: [[a]] -> [a]
join xs = concat xs
The standard mathematical definition of Monad is based on these return and join operators. However in Haskell the definition of the class Monad substitutes a bind operator for join.
Monads in Haskell
In functional programming languages these special containers are typically used to denote effectful computations. The type Maybe a would represent a computation that may or may not succeed, and the type [a] a computation that is non-deterministic. Particularly we're interested in functions with effects, i.e.those with types a->m b for some Monad m. And we need to be able to compose them. This can be done using either a monadic composition or bind operator.
(>=>) :: Monad m => (a -> m b) -> (b -> m c) -> a -> m c
(>>=) :: Monad m => m a -> (a -> m b) -> m b
In Haskell the latter is the standard one. Note that its type is very similar to the type of the application operator (but with flipped arguments):
(>>=) :: Monad m => m a -> (a -> m b) -> m b
flip ($) :: a -> (a -> b) -> b
It takes an effectful function f :: a -> m b and a computation mx :: m a returning values of type a, and performs the application mx >>= f. So how do we do this with Monads? Containers (Functors) can be mapped, and in this case the result is a computation within a computation which can then be flattened:
fmap f mx :: m (m b)
join (fmap f mx) :: m b
So we have:
(mx >>= f) = join (fmap f mx) :: m b
To see this working in practise consider a simple example with lists (non-deterministic functions). Suppose you have a list of possible results mx = [1,2,3] and a non-deterministic function f x = [x-1, x*2]. To calculate mx >>= f you begin by mapping mx with f and then you merge the results::
fmap f mx = [[0,2],[1,4],[2,6]]
join [[0,2],[1,4],[2,6]] = [0,2,1,4,2,6]
Since in Haskell the bind operator (>>=) is more important than join, for efficiency reasons in the latter is defined from the former and not the other way around.
join mx = mx >>= id
Also the bind operator, being defined with join and fmap, can also be used to define a mapping operation. For this reason Monads are not required to be instances of the class Functor. The equivalent operation to fmap is called liftM in the Monad library.
liftM f mx = mx >>= \x-> return (f x)
So the actual definitions for the Monads Maybe becomes:
return :: a -> Maybe a
return x = Just x
(>>=) :: Maybe a -> (a -> Maybe b) -> Maybe b
Nothing >>= f = Nothing
Just x >>= f = f x
And for the Monad [ ]:
return :: a -> [a]
return x = [x]
(>>=) :: [a] -> (a -> [b]) -> [b]
xs >>= f = concat (map f xs)
= concatMap f xs -- same as above but more efficient
When designing your own Monads you may find it easier to, instead of trying to directly define (>>=), split the problem in parts and figure out what how to map and join your structures. Having map and join can also be useful to verify that your Monad is well defined, in the sense that it satisfy the required laws.
Monad Laws
Your Monad should be a Functor, so the mapping operation should satisfy:
fmap id = id
fmap g . fmap f = fmap (g . f)
The laws for return and join are:
join . return = id
join . fmap return = id
join . join = join . fmap join
The first two laws specify that merging undoes wrapping. If you wrap a container in another one, join gives you back the original. If you map the contents of a container with a wrapping operation, join again gives you back what you initially had. The last law is the associativity of join. If you have three layers of containers you get the same result by merging from the inside or the outside.
Again you can work with bind instead of join and fmap. You get fewer but (arguably) more complicated laws:
return a >>= f = f a
m >>= return = m
(m >>= f) >>= g = m >>= (\x -> f x >>= g)
A monad in Haskell is something that has two operations defined:
(>>=) :: Monad m => m a -> (a -> m b) -> m b -- also called bind
return :: Monad m => a -> m a
These two operations need to satisfy certain laws that really might just confuse you at this point, if you don't have a knack for mathy ideas. Conceptually, you use bind to operate on values on a monadic level and return to create monadic values from "trivial" ones. For instance,
getLine :: IO String,
so you cannot modify and putStrLn this String -- because it's not a String but an IO String!
Well, we have an IO Monad handy, so not to worry. All we have to do is use bind to do what we want. Let's see what bind looks like in the IO Monad:
(>>=) :: IO a -> (a -> IO b) -> IO b
And if we place getLine at the left hand side of bind, we can make it more specific yet.
(>>=) :: IO String -> (String -> IO b) -> IO b
Okay, so getLine >>= putStrLn . (++ ". No problem after all!") would print the entered line with the extra content added. The right hand side is a function that takes a String and produces an IO () - that wasn't hard at all! We just go by the types.
There are Monads defined for a lot of different types, for instance Maybe and [a], and they behave conceptually in the same way.
Just 2 >>= return . (+2) would yield Just 4, as you might expect. Note that we had to use return here, because otherwise the function on the right hand side would not match the return type m b, but just b, which would be a type error. It worked in the case of putStrLn because it already produces an IO something, which was exactly what our type needed to match. (Spoiler: Expressions of shape foo >>= return . bar are silly, because every Monad is a Functor. Can you figure out what that means?)
I personally think that this is as far as intuition will get you on the topic of monads, and if you want to dive deeper, you really do need to dive into the theory. I liked getting a hang of just using them first. You can look up the source for various Monad instances, for instance the List ([]) Monad or Maybe Monad on Hoogle and get a bit smarter on the exact implementations. Once you feel comfortable with that, have a go at the actual monad laws and try to gain a more theoretical understanding for them!
The Typeclassopedia has a section about Monad (but do read the preceding sections about Functor and Applicative first).
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.
I've written code with the following pattern several times recently, and was wondering if there was a shorter way to write it.
foo :: IO String
foo = do
x <- getLine
putStrLn x >> return x
To make things a little cleaner, I wrote this function (though I'm not sure it's an appropriate name):
constM :: (Monad m) => (a -> m b) -> a -> m a
constM f a = f a >> return a
I can then make foo like this:
foo = getLine >>= constM putStrLn
Does a function/idiom like this already exist? And if not, what's a better name for my constM?
Well, let's consider the ways that something like this could be simplified. A non-monadic version would I guess look something like const' f a = const a (f a), which is clearly equivalent to flip const with a more specific type. With the monadic version, however, the result of f a can do arbitrary things to the non-parametric structure of the functor (i.e., what are often called "side effects"), including things that depend on the value of a. What this tells us is that, despite pretending like we're discarding the result of f a, we're actually doing nothing of the sort. Returning a unchanged as the parametric part of the functor is far less essential, and we could replace return with something else and still have a conceptually similar function.
So the first thing we can conclude is that it can be seen as a special case of a function like the following:
doBoth :: (Monad m) => (a -> m b) -> (a -> m c) -> a -> m c
doBoth f g a = f a >> g a
From here, there are two different ways to look for an underlying structure of some sort.
One perspective is to recognize the pattern of splitting a single argument among multiple functions, then recombining the results. This is the concept embodied by the Applicative/Monad instances for functions, like so:
doBoth :: (Monad m) => (a -> m b) -> (a -> m c) -> a -> m c
doBoth f g = (>>) <$> f <*> g
...or, if you prefer:
doBoth :: (Monad m) => (a -> m b) -> (a -> m c) -> a -> m c
doBoth = liftA2 (>>)
Of course, liftA2 is equivalent to liftM2 so you might wonder if lifting an operation on monads into another monad has something to do with monad transformers; in general the relationship there is awkward, but in this case it works easily, giving something like this:
doBoth :: (Monad m) => ReaderT a m b -> ReaderT a m c -> ReaderT a m c
doBoth = (>>)
...modulo appropriate wrapping and such, of course. To specialize back to your original version, the original use of return now needs to be something with type ReaderT a m a, which shouldn't be too hard to recognize as the ask function for reader monads.
The other perspective is to recognize that functions with types like (Monad m) => a -> m b can be composed directly, much like pure functions. The function (<=<) :: Monad m => (b -> m c) -> (a -> m b) -> (a -> m c) gives a direct equivalent to function composition (.) :: (b -> c) -> (a -> b) -> (a -> c), or you can instead make use of Control.Category and the newtype wrapper Kleisli to work with the same thing in a generic way.
We still need to split the argument, however, so what we really need here is a "branching" composition, which Category alone doesn't have; by using Control.Arrow as well we get (&&&), letting us rewrite the function as follows:
doBoth :: (Monad m) => Kleisli m a b -> Kleisli m a c -> Kleisli m a (b, c)
doBoth f g = f &&& g
Since we don't care about the result of the first Kleisli arrow, only its side effects, we can discard that half of the tuple in the obvious way:
doBoth :: (Monad m) => Kleisli m a b -> Kleisli m a c -> Kleisli m a c
doBoth f g = f &&& g >>> arr snd
Which gets us back to the generic form. Specializing to your original, the return now becomes simply id:
constKleisli :: (Monad m) => Kleisli m a b -> Kleisli m a a
constKleisli f = f &&& id >>> arr snd
Since regular functions are also Arrows, the definition above works there as well if you generalize the type signature. However, it may be enlightening to expand the definition that results for pure functions and simplify as follows:
\f x -> (f &&& id >>> arr snd) x
\f x -> (snd . (\y -> (f y, id y))) x
\f x -> (\y -> snd (f y, y)) x
\f x -> (\y -> y) x
\f x -> x.
So we're back to flip const, as expected!
In short, your function is some variation on either (>>) or flip const, but in a way that relies on the differences--the former using both a ReaderT environment and the (>>) of the underlying monad, the latter using the implicit side-effects of the specific Arrow and the expectation that Arrow side effects happen in a particular order. Because of these details, there's not likely to be any generalization or simplification available. In some sense, the definition you're using is exactly as simple as it needs to be, which is why the alternate definitions I gave are longer and/or involve some amount of wrapping and unwrapping.
A function like this would be a natural addition to a "monad utility library" of some sort. While Control.Monad provides some combinators along those lines, it's far from exhaustive, and I could neither find nor recall any variation on this function in the standard libraries. I would not be at all surprised to find it in one or more utility libraries on hackage, however.
Having mostly dispensed with the question of existence, I can't really offer much guidance on naming beyond what you can take from the discussion above about related concepts.
As a final aside, note also that your function has no control flow choices based on the result of a monadic expression, since executing the expressions no matter what is the main goal. Having a computational structure independent of the parametric content (i.e., the stuff of type a in Monad m => m a) is usually a sign that you don't actually need a full Monad, and could get by with the more general notion of Applicative.
Hmm, I don't think constM is appropriate here.
map :: (a -> b) -> [a] -> [b]
mapM :: (Monad m) => (a -> m b) -> [a] -> m b
const :: b -> a -> b
So perhaps:
constM :: (Monad m) => b -> m a -> m b
constM b m = m >> return b
The function you are M-ing seems to be:
f :: (a -> b) -> a -> a
Which has no choice but to ignore its first argument. So this function does not have much to say purely.
I see it as a way to, hmm, observe a value with a side effect. observe, effect, sideEffect may be decent names. observe is my favorite, but maybe just because it is catchy, not because it is clear.
I don't really have a clue this exactly allready exists, but you see this a lot in parser-generators only with different names (for example to get the thing inside brackets) - there it's normaly some kind of operator (>>. and .>> in fparsec for example) for this. To really give a name I would maybe call it ignore?
There's interact:
http://hackage.haskell.org/packages/archive/haskell98/latest/doc/html/Prelude.html#v:interact
It's not quite what you're asking for, but it's similar.