Related
So, I have this datatype (it's from here: https://wiki.haskell.org/IO_Semantics):
data IO a = Done a
| PutChar Char (IO a)
| GetChar (Char -> IO a)
and I thought of writing a StateMonad instance for it. I have already written Monad and Applicative instances for it.
instance MonadState (IO s) where
get = GetChar (\c -> Done c)
put = PutChar c (Done ())
state f = Done (f s)
I don't think I fully understand what state (it was named 'modify' before) is supposed to do here.
state :: (s -> (a, s)) -> m a
I also have messed up with declarations. I don't really understand what is wrong, let alone how to fix it. Would appreciate your help.
Expecting one more argument to ‘MonadState (IO s)’
Expected a constraint,
but ‘MonadState (IO s)’ has kind ‘(* -> *) -> Constraint’
In the instance declaration for ‘MonadState (IO s)’
As I mentioned in the comments, your type doesn't really hold any state, so a StateMonad instance would be nonsensical for it.
However, since this is just an exercise (also based on the comments), I guess it's ok to implement the instance technically, even if it doesn't do what you'd expect it to do.
First, the compiler error you're getting tells you that the MonadState class actually takes two arguments - the type of the state and the type of the monad, where monad has to have kind * -> *, that is, have a type parameter, like Maybe, or list, or Identity.
In your case, the monad in question (not really a monad, but ok) is IO, and the type of your "state" is Char, since that's what you're getting and putting. So the declaration has to look like this:
instance MonadState Char IO where
Second, the state method doesn't have the signature (s -> s) -> m s as you claim, but rather (s -> (a, s)) -> m a. See its definition. What it's supposed to do is create a computation in monad m out of a function that takes a state and returns "result" plus new (updated) state.
Note also that this is the most general operation on the State monad, and both get and put can be expressed in terms of state:
get = state $ \s -> (s, s)
put s = state $ \_ -> ((), s)
This means that you do not have to implement get and put yourself. You only need to implement the state function, and get/put will come from default implementations.
Incidentally, this works the other way around as well: if you define get and put, the definition of state will come from the default:
state f = do
s <- get
let (a, s') = f s
put s'
return a
And now, let's see how this can actually be implemented.
The semantics of the state's parameter is this: it's a function that takes some state as input, then performs some computation based on that state, and this computation has some result a, and it also may modify the state in some way; so the function returns both the result and the new, modified state.
In your case, the way to "get" the state from your "monad" is via GetChar, and the way in which is "returns" the Char is by calling a function that you pass to it (such function is usually referred to as "continuation").
The way to "put" the state back into your "monad" is via PutChar, which takes the Char you want to "put" as a parameter, plus some IO a that represents the computation "result".
So, the way to implement state would be to (1) first "get" the Char, then (2) apply the function to it, then (3) "put" the resulting new Char, and then (3) return the "result" of the function. Putting it all together:
state f = GetChar $ \c -> let (a, c') = f c in PutChar c' (Done a)
As further exercise, I encourage you to see how get and put would unfold, starting from the definitions I gave above, and performing step-by-step substitution. Here, I'll give you a couple first steps:
get = state $ \s -> (s, s)
-- Substituting definition of `state`
= GetChar $ \c -> let (a, c') = (\s -> (s, s)) c in PutChar c' (Done a)
-- Substituting (\s -> (s, s)) c == (c, c)
= GetChar $ \c -> let (a, c') = (c, c) in PutChar c' (Done a)
= <and so on...>
MonadState takes two arguments, in this order:
the type of the state
the monad
In this case the monad is IO:
instance MonadState _ IO where
...
And you need to figure out what goes in place of the underscore
Is it possible to print the result of a state monad in Haskell?
I'm trying to understand state monads and in a book I have been following supplies the code below for creating a state monad, but I am struggling with this topic as I am unable to view the process visually i.e. see the end result.
newtype State s a = State { runState :: s -> (a,s)}
instance Monad (State s) where
return x = State $ \s -> (x,s)
(State h) >>= f = State $ \s -> let (a, newState) = h s
(State g) = f a
in g newState
It is generally not possible to print functions in a meaningful way. If the domain of the function is small, you can import Data.Universe.Instances.Show from the universe-reverse-instances package to get a Show instance that prints a lookup table that is semantically equivalent to the function. With that module imported, you could simply add deriving Show to your newtype declaration to be able to print State actions over small state spaces.
The code you've supplied defines the kind of thing State s a is. And it also says that State s is a monad - that is, the kind of thing State s is conforms to the Monad typeclass/interface. This means you can bind one State s computation to another (as long as the type s is the same in each).
So your situation is analogous to that of someone who has defined the kind of thing that a Map is, and has also written code that says a Map conforms to such and such interfaces, but who doesn't have any maps, and hasn't yet run any computation with them. There's nothing to print then.
I take it you want to see the result of evaluating or executing your state actions, but you have not defined any actual state actions yet, nor have you called runState (or evalState or execState) on them. Don't forget you also need to supply an initial state to run the computation.
So maybe start by letting s and a be some particular types. E.g. let s be Int and let a be Int. Now you could go write some fns, e.g. f :: Int -> (Int, Int), and g :: Int -> (Int, Int). Maybe one function decrements the state, returning the new state and value, and another function increments the state, returning the new state and value. Then you could make a State Int Int out of f by wrapping it in the State constructor. And you could use >>= to chain as many state actions together as you like. Finally, you can use runState on this, to get the resulting value and resulting state, as long as you also supply an initial state (e.g. 0).
If it's just the result you want, and if you're just debugging:
import Debug.Trace
import Control.Monad.Trans.State
action :: State [Int] ()
action = do
put [0]
modify (1:)
modify (2:)
get >>= traceShowM
modify (3:)
modify (4:)
get >>= traceShowM
This is a type declaration of a bind method:
(>>=) :: (Monad m) => m a -> (a -> m b) -> m b
I read this as follows: apply a function that returns a wrapped value, to a wrapped value.
This method was included to Prelude as part of Monad typeclass. That means there are a lot of cases where it's needed.
OK, but I don't understand why it's a typical solution of a typical case at all.
If you already created a function which returns a wrapped value, why that function doesn't already take a wrapped value?
In other words, what are typical cases where there are many functions which take a normal value, but return a wrapped value? (instead of taking a wrapped value and return a wrapped value)
The 'unwrapping' of values is exactly what you want to keep hidden when dealing with monads, since it is this that causes a lot of boilerplate.
For example, if you have a sequence of operations which return Maybe values that you want to combine, you have to manually propagate Nothing if you receive one:
nested :: a -> Maybe b
nested x = case f x of
Nothing -> Nothing
Just r ->
case g r of
Nothing -> Nothing
Just r' ->
case h r' of
Nothing -> Nothing
r'' -> i r''
This is what bind does for you:
Nothing >>= _ = Nothing
Just a >>= f = f a
so you can just write:
nested x = f x >>= g >>= h >>= i
Some monads don't allow you to manually unpack the values at all - the most common example is IO. The only way to get the value from an IO is to map or >>= and both of these require you to propagate IO in the output.
Everyone focuses on IO monad and inability to "unwrap".
But a Monad is not always a container, so you can't unwrap.
Reader r a == r->a such that (Reader r) is a Monad
to my mind is the simplest best example of a Monad that is not a container.
You can easily write a function that can produce m b given a: a->(r->b). But you can't easily "unwrap" the value from m a, because a is not wrapped in it. Monad is a type-level concept.
Also, notice that if you have m a->m b, you don't have a Monad. What Monad gives you, is a way to build a function m a->m b from a->m b (compare: Functor gives you a way to build a function m a->m b from a->b; ApplicativeFunctor gives you a way to build a function m a->m b from m (a->b))
If you already created a function which returns a wrapped value, why that function doesn't already take a wrapped value?
Because that function would have to unwrap its argument in order to do something with it.
But for many choices of m, you can only unwrap a value if you will eventually rewrap your own result. This idea of "unwrap, do something, then rewrap" is embodied in the (>>=) function which unwraps for you, let's you do something, and forces you to rewrap by the type a -> m b.
To understand why you cannot unwrap without eventually rewrapping, we can look at some examples:
If m a = Maybe a, unwrapping for Just x would be easy: just return x. But how can we unwrap Nothing? We cannot. But if we know that we will eventually rewrap, we can skip the "do something" step and return Nothing for the overall operation.
If m a = [a], unwrapping for [x] would be easy: just return x. But for unwrapping [], we need the same trick as for Maybe a. And what about unwrapping [x, y, z]? If we know that we will eventually rewrap, we can execute the "do something" three times, for x, y and z and concat the results into a single list.
If m a = IO a, no unwrapping is easy because we only know the result sometimes in the future, when we actually run the IO action. But if we know that we will eventually rewrap, we can store the "do something" inside the IO action and perform it later, when we execute the IO action.
I hope these examples make it clear that for many interesting choices of m, we can only implement unwrapping if we know that we are going to rewrap. The type of (>>=) allows precisely this assumption, so it is cleverly chosen to make things work.
While (>>=) can sometimes be useful when used directly, its main purpose is to implement the <- bind syntax in do notation. It has the type m a -> (a -> m b) -> m b mainly because, when used in a do notation block, the right hand side of the <- is of type m a, the left hand side "binds" an a to the given identifier and, when combined with remainder of the do block, is of type a -> m b, the resulting monadic action is of type m b, and this is the only type it possibly could have to make this work.
For example:
echo = do
input <- getLine
putStrLn input
The right hand side of the <- is of type IO String
The left hands side of the <- with the remainder of the do block are of type String -> IO (). Compare with the desugared version using >>=:
echo = getLine >>= (\input -> putStrLn input)
The left hand side of the >>= is of type IO String. The right hand side is of type String -> IO (). Now, by applying an eta reduction to the lambda we can instead get:
echo = getLine >>= putStrLn
which shows why >>= is sometimes used directly rather than as the "engine" that powers do notation along with >>.
I'd also like to provide what I think is an important correction to the concept of "unwrapping" a monadic value, which is that it doesn't happen. The Monad class does not provide a generic function of type Monad m => m a -> a. Some particular instances do but this is not a feature of monads in general. Monads, generally speaking, cannot be "unwrapped".
Remember that m >>= k = join (fmap k m) is a law that must be true for any monad. Any particular implementation of >>= must satisfy this law and so must be equivalent to this general implementation.
What this means is that what really happens is that the monadic "computation" a -> m b is "lifted" to become an m a -> m (m b) using fmap and then applied the m a, giving an m (m b); and then join :: m (m a) -> m a is used to squish the two ms together to yield a m b. So the a never gets "out" of the monad. The monad is never "unwrapped". This is an incorrect way to think about monads and I would strongly recommend that you not get in the habit.
I will focus on your point
If you already created a function which returns a wrapped value, why
that function doesn't already take a wrapped value?
and the IO monad. Suppose you had
getLine :: IO String
putStrLn :: IO String -> IO () -- "already takes a wrapped value"
how one could write a program which reads a line and print it twice? An attempt would be
let line = getLine
in putStrLn line >> putStrLn line
but equational reasoning dictates that this is equivalent to
putStrLn getLine >> putStrLn getLine
which reads two lines instead.
What we lack is a way to "unwrap" the getLine once, and use it twice. The same issue would apply to reading a line, printing "hello", and then printing a line:
let line = getLine in putStrLn "hello" >> putStrLn line
-- equivalent to
putStrLn "hello" >> putStrLn getLine
So, we also lack a way to specify "when to unwrap" the getLine. The bind >>= operator provides a way to do this.
A more advanced theoretical note
If you swap the arguments around the (>>=) bind operator becomes (=<<)
(=<<) :: (a -> m b) -> (m a -> m b)
which turns any function f taking an unwrapped value into a function g taking a wrapped
value. Such g is known as the Kleisli extension of f. The bind operator guarantees
such an extension always exists, and provides a convenient way to use it.
Because we like to be able to apply functions like a -> b to our m as. Lifting such a function to m a -> m b is trivial (liftM, liftA, >>= return ., fmap) but the opposite is not necessarily possible.
You want some typical examples? How about putStrLn :: String -> IO ()? It would make no sense for this function to have the type IO String -> IO () because the origin of the string doesn't matter.
Anyway: You might have the wrong idea because of your "wrapped value" metaphor; I use it myself quite often, but it has its limitations. There isn't necessarily a pure way to get an a out of an m a - for example, if you have a getLine :: IO String, there's not a great deal of interesting things you can do with it - you can put it in a list, chain it in a row and other neat things, but you can't get any useful information out of it because you can't look inside an IO action. What you can do is use >>= which gives you a way to use the result of the action.
Similar things apply to monads where the "wrapping" metaphor applies too; For example the point Maybe monad is to avoid manually wrapping and unwrapping values with and from Just all the time.
My two most common examples:
1) I have a series of functions that generate a list of lists, but I finally need a flat list:
f :: a -> [a]
fAppliedThrice :: [a] -> [a]
fAppliedThrice aList = concat (map f (concat (map f (concat (map f a)))))
fAppliedThrice' :: [a] -> [a]
fAppliedThrice' aList = aList >>= f >>= f >>= f
A practical example of using this was when my functions fetched attributes of a foreign key relationship. I could just chain them together to finally obtain a flat list of attributes. Eg: Product hasMany Review hasMany Tag type relationship, and I finally want a list of all the tag names for a product. (I added some template-haskell and got a very good generic attribute fetcher for my purposes).
2) Say you have a series of filter-like functions to apply to some data. And they return Maybe values.
case (val >>= filter >>= filter2 >>= filter3) of
Nothing -> putStrLn "Bad data"
Just x -> putStrLn "Good data"
Since the following do block:
do
x <- foo
y <- bar
return x + y
is desugared to the following form:
foo >>= (\x -> bar >>= (\y -> return x + y))
aren't \x -> ... and y -> ... actually continuations here?
I was wondering if there is a way to capture the continuations in the definition of bind, but I can't get the types right. I.e:
data Pause a = Pause a | Stop
instance Monad Pause where
return x = Stop
m >>= k = Pause k -- this doesn't work of course
Now I tried muddling around with the types:
data Pause a = Pause a (a -> Pause ???) | Stop
------- k ------
But this doesn't work either. Is there no way to capture these implicit continuations?
Btw, I know about the Cont monad, I'm just experimenting and trying out stuff.
OK I'm not really sure, but let me put a few thoughts. I'm not quite sure what it
ought to mean to capture the continuation. You could, for example, capture the whole
do block in a structure:
{-# LANGUAGE ExistentialQuantification #-}
import Control.Monad
data MonadExp b = Return b | forall a. Bind (MonadExp a) (a -> MonadExp b)
instance Monad MonadExp where
return x = Return x
f >>= g = Bind f g
For example:
block :: MonadExp Int
block = do
x <- return 1
y <- return 2
return $ x + y
instance Show (MonadExp a) where
show (Return _) = "Return _"
show (Bind _ _) = "Bind _ _"
print block
>> Bind _ _
And then evaluate the whole thing:
finish :: MonadExp a -> a
finish (Return x) = x
finish (Bind f g) = finish $ g (finish f)
print $ finish block
>> 3
Or step through it and see the parts
step :: MonadExp a -> MonadExp a
step (Return _) = error "At the end"
step (Bind f g) = g $ finish f
print $ step block
>> Bind _ _
print $ step $ step block
>> Return _
Well, now that I think about it more, that's probably not what you're asking. But
maybe it'll help you think.
Well, I don't know if your lambdas are continuations in the strictest sense of the term, but they also look similar to this concept in my eye.
But note that if they are continuations, then the desugared monadic code is already written in continuation passing style (CPS). The usual notion of a control operator that "captures" a continuation is based on direct-style programs; the "captured" continuation is only implicit in the direct-style program, but the CPS transformation makes it explicit.
Since the desugared monadic code is already in CPS or something like it, well, maybe a way to frame your question is whether monadic code can express some of the control flow tricks that CPS code can. Typically, those tricks boil down to the idea that while under the CPS regime it is conventional for a function to finish by invoking its continuation, a function can choose to replace its continuation with another of its choosing. This replacement continuation can be constructed with a reference to the original continuation, so that it can in turn "restore" that original one if it chooses. So for example, coroutines are implemented as a mutual "replace/restore" cycle.
And looked at in this light, I think your answer is mostly no; CPS requires that in foo >>= bar, foo must be able to choose whether bar will be invoked at all, and foo must be abble to supply a substitute for bar, but (>>=) by itself does not offer a mechanism for foo to do this, and more importantly, (>>=) is in control of the execution flow, not foo. Some specific monads implement parts or all of it (for example, the Maybe monad allows foo to forego execution of bar by producing a Nothing result), but others don't.
Closest I could get is to forego (>>=) and use this instead:
-- | Execute action #foo# with its "continuation" #bar#.
callCC :: Monad m => ((a -> m b) -> m b) -> (a -> m b) -> m b
foo `callCC` bar = foo bar
Here foo can choose whether bar will be used at all. But notice that this callCC is really just ($)!
I am learning Haskell and trying to understand Monads. I have two questions:
From what I understand, Monad is just another typeclass that declares ways to interact with data inside "containers", including Maybe, List, and IO. It seems clever and clean to implement these 3 things with one concept, but really, the point is so there can be clean error handling in a chain of functions, containers, and side effects. Is this a correct interpretation?
How exactly is the problem of side-effects solved? With this concept of containers, the language essentially says anything inside the containers is non-deterministic (such as i/o). Because lists and IOs are both containers, lists are equivalence-classed with IO, even though values inside lists seem pretty deterministic to me. So what is deterministic and what has side-effects? I can't wrap my head around the idea that a basic value is deterministic, until you stick it in a container (which is no special than the same value with some other values next to it, e.g. Nothing) and it can now be random.
Can someone explain how, intuitively, Haskell gets away with changing state with inputs and output? I'm not seeing the magic here.
The point is so there can be clean error handling in a chain of functions, containers, and side effects. Is this a correct interpretation?
Not really. You've mentioned a lot of concepts that people cite when trying to explain monads, including side effects, error handling and non-determinism, but it sounds like you've gotten the incorrect sense that all of these concepts apply to all monads. But there's one concept you mentioned that does: chaining.
There are two different flavors of this, so I'll explain it two different ways: one without side effects, and one with side effects.
No Side Effects:
Take the following example:
addM :: (Monad m, Num a) => m a -> m a -> m a
addM ma mb = do
a <- ma
b <- mb
return (a + b)
This function adds two numbers, with the twist that they are wrapped in some monad. Which monad? Doesn't matter! In all cases, that special do syntax de-sugars to the following:
addM ma mb =
ma >>= \a ->
mb >>= \b ->
return (a + b)
... or, with operator precedence made explicit:
ma >>= (\a -> mb >>= (\b -> return (a + b)))
Now you can really see that this is a chain of little functions, all composed together, and its behavior will depend on how >>= and return are defined for each monad. If you're familiar with polymorphism in object-oriented languages, this is essentially the same thing: one common interface with multiple implementations. It's slightly more mind-bending than your average OOP interface, since the interface represents a computation policy rather than, say, an animal or a shape or something.
Okay, let's see some examples of how addM behaves across different monads. The Identity monad is a decent place to start, since its definition is trivial:
instance Monad Identity where
return a = Identity a -- create an Identity value
(Identity a) >>= f = f a -- apply f to a
So what happens when we say:
addM (Identity 1) (Identity 2)
Expanding this, step by step:
(Identity 1) >>= (\a -> (Identity 2) >>= (\b -> return (a + b)))
(\a -> (Identity 2) >>= (\b -> return (a + b)) 1
(Identity 2) >>= (\b -> return (1 + b))
(\b -> return (1 + b)) 2
return (1 + 2)
Identity 3
Great. Now, since you mentioned clean error handling, let's look at the Maybe monad. Its definition is only slightly trickier than Identity:
instance Monad Maybe where
return a = Just a -- same as Identity monad!
(Just a) >>= f = f a -- same as Identity monad again!
Nothing >>= _ = Nothing -- the only real difference from Identity
So you can imagine that if we say addM (Just 1) (Just 2) we'll get Just 3. But for grins, let's expand addM Nothing (Just 1) instead:
Nothing >>= (\a -> (Just 1) >>= (\b -> return (a + b)))
Nothing
Or the other way around, addM (Just 1) Nothing:
(Just 1) >>= (\a -> Nothing >>= (\b -> return (a + b)))
(\a -> Nothing >>= (\b -> return (a + b)) 1
Nothing >>= (\b -> return (1 + b))
Nothing
So the Maybe monad's definition of >>= was tweaked to account for failure. When a function is applied to a Maybe value using >>=, you get what you'd expect.
Okay, so you mentioned non-determinism. Yes, the list monad can be thought of as modeling non-determinism in a sense... It's a little weird, but think of the list as representing alternative possible values: [1, 2, 3] is not a collection, it's a single non-deterministic number that could be either one, two or three. That sounds dumb, but it starts to make some sense when you think about how >>= is defined for lists: it applies the given function to each possible value. So addM [1, 2] [3, 4] is actually going to compute all possible sums of those two non-deterministic values: [4, 5, 5, 6].
Okay, now to address your second question...
Side Effects:
Let's say you apply addM to two values in the IO monad, like:
addM (return 1 :: IO Int) (return 2 :: IO Int)
You don't get anything special, just 3 in the IO monad. addM does not read or write any mutable state, so it's kind of no fun. Same goes for the State or ST monads. No fun. So let's use a different function:
fireTheMissiles :: IO Int -- returns the number of casualties
Clearly the world will be different each time missiles are fired. Clearly. Now let's say you're trying to write some totally innocuous, side effect free, non-missile-firing code. Perhaps you're trying once again to add two numbers, but this time without any monads flying around:
add :: Num a => a -> a -> a
add a b = a + b
and all of a sudden your hand slips, and you accidentally typo:
add a b = a + b + fireTheMissiles
An honest mistake, really. The keys were so close together. Fortunately, because fireTheMissiles was of type IO Int rather than simply Int, the compiler is able to avert disaster.
Okay, totally contrived example, but the point is that in the case of IO, ST and friends, the type system keeps effects isolated to some specific context. It doesn't magically eliminate side effects, making code referentially transparent that shouldn't be, but it does make it clear at compile time what scope the effects are limited to.
So getting back to the original point: what does this have to do with chaining or composition of functions? Well, in this case, it's just a handy way of expressing a sequence of effects:
fireTheMissilesTwice :: IO ()
fireTheMissilesTwice = do
a <- fireTheMissiles
print a
b <- fireTheMissiles
print b
Summary:
A monad represents some policy for chaining computations. Identity's policy is pure function composition, Maybe's policy is function composition with failure propogation, IO's policy is impure function composition and so on.
Let me start by pointing at the excellent "You could have invented monads" article. It illustrates how the Monad structure can naturally manifest while you are writing programs. But the tutorial doesn't mention IO, so I will have a stab here at extending the approach.
Let us start with what you probably have already seen - the container monad. Let's say we have:
f, g :: Int -> [Int]
One way of looking at this is that it gives us a number of possible outputs for every possible input. What if we want all possible outputs for the composition of both functions? Giving all possibilities we could get by applying the functions one after the other?
Well, there's a function for that:
fg x = concatMap g $ f x
If we put this more general, we get
fg x = f x >>= g
xs >>= f = concatMap f xs
return x = [x]
Why would we want to wrap it like this? Well, writing our programs primarily using >>= and return gives us some nice properties - for example, we can be sure that it's relatively hard to "forget" solutions. We'd explicitly have to reintroduce it, say by adding another function skip. And also we now have a monad and can use all combinators from the monad library!
Now, let us jump to your trickier example. Let's say the two functions are "side-effecting". That's not non-deterministic, it just means that in theory the whole world is both their input (as it can influence them) as well as their output (as the function can influence it). So we get something like:
f, g :: Int -> RealWorld# -> (Int, RealWorld#)
If we now want f to get the world that g left behind, we'd write:
fg x rw = let (y, rw') = f x rw
(r, rw'') = g y rw'
in (r, rw'')
Or generalized:
fg x = f x >>= g
x >>= f = \rw -> let (y, rw') = x rw
(r, rw'') = f y rw'
in (r, rw'')
return x = \rw -> (x, rw)
Now if the user can only use >>=, return and a few pre-defined IO values we get a nice property again: The user will never actually see the RealWorld# getting passed around! And that is a very good thing, as you aren't really interested in the details of where getLine gets its data from. And again we get all the nice high-level functions from the monad libraries.
So the important things to take away:
The monad captures common patterns in your code, like "always pass all elements of container A to container B" or "pass this real-world-tag through". Often, once you realize that there is a monad in your program, complicated things become simply applications of the right monad combinator.
The monad allows you to completely hide the implementation from the user. It is an excellent encapsulation mechanism, be it for your own internal state or for how IO manages to squeeze non-purity into a pure program in a relatively safe way.
Appendix
In case someone is still scratching his head over RealWorld# as much as I did when I started: There's obviously more magic going on after all the monad abstraction has been removed. Then the compiler will make use of the fact that there can only ever be one "real world". That's good news and bad news:
It follows that the compiler must guarantuee execution ordering between functions (which is what we were after!)
But it also means that actually passing the real world isn't necessary as there is only one we could possibly mean: The one that is current when the function gets executed!
Bottom line is that once execution order is fixed, RealWorld# simply gets optimized out. Therefore programs using the IO monad actually have zero runtime overhead. Also note that using RealWorld# is obviously only one possible way to put IO - but it happens to be the one GHC uses internally. The good thing about monads is that, again, the user really doesn't need to know.
You could see a given monad m as a set/family (or realm, domain, etc.) of actions (think of a C statement). The monad m defines the kind of (side-)effects that its actions may have:
with [] you can define actions which can fork their executions in different "independent parallel worlds";
with Either Foo you can define actions which can fail with errors of type Foo;
with IO you can define actions which can have side-effects on the "outside world" (access files, network, launch processes, do a HTTP GET ...);
you can have a monad whose effect is "randomness" (see package MonadRandom);
you can define a monad whose actions can make a move in a game (say chess, Go…) and receive move from an opponent but are not able to write to your filesystem or anything else.
Summary
If m is a monad, m a is an action which produces a result/output of type a.
The >> and >>= operators are used to create more complex actions out of simpler ones:
a >> b is a macro-action which does action a and then action b;
a >> a does action a and then action a again;
with >>= the second action can depend on the output of the first one.
The exact meaning of what an action is and what doing an action and then another one is depends on the monad: each monad defines an imperative sublanguage with some features/effects.
Simple sequencing (>>)
Let's say with have a given monad M and some actions incrementCounter, decrementCounter, readCounter:
instance M Monad where ...
-- Modify the counter and do not produce any result:
incrementCounter :: M ()
decrementCounter :: M ()
-- Get the current value of the counter
readCounter :: M Integer
Now we would like to do something interesting with those actions. The first thing we would like to do with those actions is to sequence them. As in say C, we would like to be able to do:
// This is C:
counter++;
counter++;
We define an "sequencing operator" >>. Using this operator we can write:
incrementCounter >> incrementCounter
What is the type of "incrementCounter >> incrementCounter"?
It is an action made of two smaller actions like in C you can write composed-statements from atomic statements :
// This is a macro statement made of several statements
{
counter++;
counter++;
}
// and we can use it anywhere we may use a statement:
if (condition) {
counter++;
counter++;
}
it can have the same kind of effects as its subactions;
it does not produce any output/result.
So we would like incrementCounter >> incrementCounter to be of type M (): an (macro-)action with the same kind of possible effects but without any output.
More generally, given two actions:
action1 :: M a
action2 :: M b
we define a a >> b as the macro-action which is obtained by doing (whatever that means in our domain of action) a then b and produces as output the result of the execution of the second action. The type of >> is:
(>>) :: M a -> M b -> M b
or more generally:
(>>) :: (Monad m) => m a -> m b -> m b
We can define bigger sequence of actions from simpler ones:
action1 >> action2 >> action3 >> action4
Input and outputs (>>=)
We would like to be able to increment by something else that 1 at a time:
incrementBy 5
We want to provide some input in our actions, in order to do this we define a function incrementBy taking an Int and producing an action:
incrementBy :: Int -> M ()
Now we can write things like:
incrementCounter >> readCounter >> incrementBy 5
But we have no way to feed the output of readCounter into incrementBy. In order to do this, a slightly more powerful version of our sequencing operator is needed. The >>= operator can feed the output of a given action as input to the next action. We can write:
readCounter >>= incrementBy
It is an action which executes the readCounter action, feeds its output in the incrementBy function and then execute the resulting action.
The type of >>= is:
(>>=) :: Monad m => m a -> (a -> m b) -> m b
A (partial) example
Let's say I have a Prompt monad which can only display informations (text) to the user and ask informations to the user:
-- We don't have access to the internal structure of the Prompt monad
module Prompt (Prompt(), echo, prompt) where
-- Opaque
data Prompt a = ...
instance Monad Prompt where ...
-- Display a line to the CLI:
echo :: String -> Prompt ()
-- Ask a question to the user:
prompt :: String -> Prompt String
Let's try to define a promptBoolean message actions which asks for a question and produces a boolean value.
We use the prompt (message ++ "[y/n]") action and feed its output to a function f:
f "y" should be an action which does nothing but produce True as output;
f "n" should be an action which does nothing but produce False as output;
anything else should restart the action (do the action again);
promptBoolean would look like this:
-- Incomplete version, some bits are missing:
promptBoolean :: String -> M Boolean
promptBoolean message = prompt (message ++ "[y/n]") >>= f
where f result = if result == "y"
then ???? -- We need here an action which does nothing but produce `True` as output
else if result=="n"
then ???? -- We need here an action which does nothing but produce `False` as output
else echo "Input not recognised, try again." >> promptBoolean
Producing a value without effect (return)
In order to fill the missing bits in our promptBoolean function, we need a way to represent dummy actions without any side effect but which only outputs a given value:
-- "return 5" is an action which does nothing but outputs 5
return :: (Monad m) => a -> m a
and we can now write out promptBoolean function:
promptBoolean :: String -> Prompt Boolean
promptBoolean message :: prompt (message ++ "[y/n]") >>= f
where f result = if result=="y"
then return True
else if result=="n"
then return False
else echo "Input not recognised, try again." >> promptBoolean message
By composing those two simple actions (promptBoolean, echo) we can define any kind of dialogue between the user and your program (the actions of the program are deterministic as our monad does not have a "randomness effect").
promptInt :: String -> M Int
promptInt = ... -- similar
-- Classic "guess a number game/dialogue"
guess :: Int -> m()
guess n = promptInt "Guess:" m -> f
where f m = if m == n
then echo "Found"
else (if m > n
then echo "Too big"
then echo "Too small") >> guess n
The operations of a monad
A Monad is a set of actions which can be composed with the return and >>= operators:
>>= for action composition;
return for producing a value without any (side-)effect.
These two operators are the minimal operators needed to define a Monad.
In Haskell, the >> operator is needed as well but it can in fact be derived from >>=:
(>>): Monad m => m a -> m b -> m b
a >> b = a >>= f
where f x = b
In Haskell, an extra fail operator is need as well but this is really a hack (and it might be removed from Monad in the future).
This is the Haskell definition of a Monad:
class Monad m where
return :: m a
(>>=) :: m a -> (a -> m b) -> m b
(>>) :: m a -> m b -> m b -- can be derived from (>>=)
fail :: String -> m a -- mostly a hack
Actions are first-class
One great thing about monads is that actions are first-class. You can take them in a variable, you can define function which take actions as input and produce some other actions as output. For example, we can define a while operator:
-- while x y : does action y while action x output True
while :: (Monad m) => m Boolean -> m a -> m ()
while x y = x >>= f
where f True = y >> while x y
f False = return ()
Summary
A Monad is a set of actions in some domain. The monad/domain define the kind of "effects" which are possible. The >> and >>= operators represent sequencing of actions and monadic expression may be used to represent any kind of "imperative (sub)program" in your (functional) Haskell program.
The great things are that:
you can design your own Monad which supports the features and effects that you want
see Prompt for an example of a "dialogue only subprogram",
see Rand for an example of "sampling only subprogram";
you can write your own control structures (while, throw, catch or more exotic ones) as functions taking actions and composing them in some way to produce a bigger macro-actions.
MonadRandom
A good way of understanding monads, is the MonadRandom package. The Rand monad is made of actions whose output can be random (the effect is randomness). An action in this monad is some kind of random variable (or more exactly a sampling process):
-- Sample an Int from some distribution
action :: Rand Int
Using Rand to do some sampling/random algorithms is quite interesting because you have random variables as first class values:
-- Estimate mean by sampling nsamples times the random variable x
sampleMean :: Real a => Int -> m a -> m a
sampleMean n x = ...
In this setting, the sequence function from Prelude,
sequence :: Monad m => [m a] -> m [a]
becomes
sequence :: [Rand a] -> Rand [a]
It creates a random variable obtained by sampling independently from a list of random variables.
There are three main observations concerning the IO monad:
1) You can't get values out of it. Other types like Maybe might allow to extract values, but neither the monad class interface itself nor the IO data type allow it.
2) "Inside" IO is not only the real value but also that "RealWorld" thing. This dummy value is used to enforce the chaining of actions by the type system: If you have two independent calculations, the use of >>= makes the second calculation dependent on the first.
3) Assume a non-deterministic thing like random :: () -> Int, which isn't allowed in Haskell. If you change the signature to random :: Blubb -> (Blubb, Int), it is allowed, if you make sure that nobody ever can use a Blubb twice: Because in that case all inputs are "different", it is no problem that the outputs are different as well.
Now we can use the fact 1): Nobody can get something out of IO, so we can use the RealWord dummy hidden in IO to serve as a Blubb. There is only one IOin the whole application (the one we get from main), and it takes care of proper sequentiation, as we have seen in 2). Problem solved.
One thing that often helps me to understand the nature of something is to examine it in the most trivial way possible. That way, I'm not getting distracted by potentially unrelated concepts. With that in mind, I think it may be helpful to understand the nature of the Identity Monad, as it's the most trivial implementation of a Monad possible (I think).
What is interesting about the Identity Monad? I think it is that it allows me to express the idea of evaluating expressions in a context defined by other expressions. And to me, that is the essence of every Monad I've encountered (so far).
If you already had a lot of exposure to 'mainstream' programming languages before learning Haskell (like I did), then this doesn't seem very interesting at all. After all, in a mainstream programming language, statements are executed in sequence, one after the other (excepting control-flow constructs, of course). And naturally, we can assume that every statement is evaluated in the context of all previously executed statements and that those previously executed statements may alter the environment and the behavior of the currently executing statement.
All of that is pretty much a foreign concept in a functional, lazy language like Haskell. The order in which computations are evaluated in Haskell is well-defined, but sometimes hard to predict, and even harder to control. And for many kinds of problems, that's just fine. But other sorts of problems (e.g. IO) are hard to solve without some convenient way to establish an implicit order and context between the computations in your program.
As far as side-effects go, specifically, often they can be transformed (via a Monad) in to simple state-passing, which is perfectly legal in a pure functional language. Some Monads don't seem to be of that nature, however. Monads such as the IO Monad or the ST monad literally perform side-effecting actions. There are many ways to think about this, but one way that I think about it is that just because my computations must exist in a world without side-effects, the Monad may not. As such, the Monad is free to establish a context for my computation to execute that is based on side-effects defined by other computations.
Finally, I must disclaim that I am definitely not a Haskell expert. As such, please understand that everything I've said is pretty much my own thoughts on this subject and I may very well disown them later when I understand Monads more fully.
the point is so there can be clean error handling in a chain of functions, containers, and side effects
More or less.
how exactly is the problem of side-effects solved?
A value in the I/O monad, i.e. one of type IO a, should be interpreted as a program. p >> q on IO values can then be interpreted as the operator that combines two programs into one that first executes p, then q. The other monad operators have similar interpretations. By assigning a program to the name main, you declare to the compiler that that is the program that has to be executed by its output object code.
As for the list monad, it's not really related to the I/O monad except in a very abstract mathematical sense. The IO monad gives deterministic computation with side effects, while the list monad gives non-deterministic (but not random!) backtracking search, somewhat similar to Prolog's modus operandi.
With this concept of containers, the language essentially says anything inside the containers is non-deterministic
No. Haskell is deterministic. If you ask for integer addition 2+2 you will always get 4.
"Nondeterministic" is only a metaphor, a way of thinking. Everything is deterministic under the hood. If you have this code:
do x <- [4,5]
y <- [0,1]
return (x+y)
it is roughly equivalent to Python code
l = []
for x in [4,5]:
for y in [0,1]:
l.append(x+y)
You see nondeterminism here? No, it's deterministic construction of a list. Run it twice, you'll get the same numbers in the same order.
You can describe it this way: Choose arbitrary x from [4,5]. Choose arbitrary y from [0,1]. Return x+y. Collect all possible results.
That way seems to involve nondeterminism, but it's only a nested loop (list comprehension). There is no "real" nondeterminism here, it's simulated by checking all possibilities. Nondeterminism is an illusion. The code only appears to be nondeterministic.
This code using State monad:
do put 0
x <- get
put (x+2)
y <- get
return (y+3)
gives 5 and seems to involve changing state. As with lists it's an illusion. There are no "variables" that change (as in imperative languages). Everything is nonmutable under the hood.
You can describe the code this way: put 0 to a variable. Read the value of a variable to x. Put (x+2) to the variable. Read the variable to y, and return y+3.
That way seems to involve state, but it's only composing functions passing additional parameter. There is no "real" mutability here, it's simulated by composition. Mutability is an illusion. The code only appears to be using it.
Haskell does it this way: you've got functions
a -> s -> (b,s)
This function takes and old value of state and returns new value. It does not involve mutability or change variables. It's a function in mathematical sense.
For example the function "put" takes new value of state, ignores current state and returns new state:
put x _ = ((), x)
Just like you can compose two normal functions
a -> b
b -> c
into
a -> c
using (.) operator you can compose "state" transformers
a -> s -> (b,s)
b -> s -> (c,s)
into a single function
a -> s -> (c,s)
Try writing the composition operator yourself. This is what really happens, there are no "side effects" only passing arguments to functions.
From what I understand, Monad is just another typeclass that declares ways to interact with data [...]
...providing an interface common to all those types which have an instance. This can then be used to provide generic definitions which work across all monadic types.
It seems clever and clean to implement these 3 things with one concept [...]
...the only three things that are implemented are the instances for those three types (list, Maybe and IO) - the types themselves are defined independently elsewhere.
[...] but really, the point is so there can be clean error handling in a chain of functions, containers, and side effects.
Not just error handling e.g. consider ST - without the monadic interface, you would have to pass the encapsulated-state directly and correctly...a tiresome task.
How exactly is the problem of side-effects solved?
Short answer: Haskell solves manages them by using types to indicate their presence.
Can someone explain how, intuitively, Haskell gets away with changing state with inputs and output?
"Intuitively"...like what's available over here? Let's try a simple direct comparison instead:
From How to Declare an Imperative by Philip Wadler:
(* page 26 *)
type 'a io = unit -> 'a
infix >>=
val >>= : 'a io * ('a -> 'b io) -> 'b io
fun m >>= k = fn () => let
val x = m ()
val y = k x ()
in
y
end
val return : 'a -> 'a io
fun return x = fn () => x
val putc : char -> unit io
fun putc c = fn () => putcML c
val getc : char io
val getc = fn () => getcML ()
fun getcML () =
valOf(TextIO.input1(TextIO.stdIn))
(* page 25 *)
fun putcML c =
TextIO.output1(TextIO.stdOut,c)
Based on these two answers of mine, this is my Haskell translation:
type IO a = OI -> a
(>>=) :: IO a -> (a -> IO b) -> IO b
m >>= k = \ u -> let !(u1, u2) = part u in
let !x = m u1 in
let !y = k x u2 in
y
return :: a -> IO a
return x = \ u -> let !_ = part u in x
putc :: Char -> IO ()
putc c = \ u -> putcOI c u
getc :: IO Char
getc = \ u -> getcOI u
-- primitives
data OI
partOI :: OI -> (OI, OI)
putcOI :: Char -> OI -> ()
getcOI :: OI -> Char
Now remember that short answer about side-effects?
Haskell manages them by using types to indicate their presence.
Data.Char.chr :: Int -> Char -- no side effects
getChar :: IO Char -- side effects at
{- :: OI -> Char -} -- work: beware!