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I'm learning Haskell through learnyouahaskell.com and wanted to test some of the concepts before finishing the input/output module. I haven't been able to google or hoogle my way out of this question, though, even though it seems quite simple.
When I try to run the following code
getName = do
name <- getLine
return name
the output of getName becomes an element of type IO String instead of String, even though name is definitely a String
By reading the documentation and other StackVverflow's questions I couldn't figure out why is this happening when I declare getName as a function (when I use the bind <- operation directly on main there's no problem whatsoever).
The return function is not conceptually the same as what return does in languages like C++, Java and Python. return :: Monad m => a -> m a takes an a (here a String), and produces an m a (here IO a).
The do notation is syntacticual sugar. If we desugar the statement, you wrote:
getName = getLine >>= (\name -> return name)
or cleaner:
getName = getLine >>= return
The bind function (>>=) :: Monad m => m a -> (a -> m b) -> m b thus has as first operand an m a, and as second a function a -> m b, and produces an m b. Since getLine :: IO String is an IO String, that thus means that m is the same as IO, and a is the same as String. The return :: Monad m => a -> m a, makes it clear that here b is the same as a.
Then what is IO here. A metaphor that is frequently used is the one of a recipe. In this metaphor an IO a is a set of instructions that when you follow these, you will get an a. But that does not mean that that recipe is an a.
(>>=) here basically says that, on the left hand I have a recipe to make a, on the right hand I have a function that converts that a into a recipe to make b, so we can construct a recipe to make b with these two.
People often ask how to unwrap an a out of an IO a, but conceptually it makes not much sense. You can not "unwrap" the cake out of a recipe to make a cake. You can follow the instructions to make a cake. Following instructions is something the main will eventually do. We thus can construct a (long) recipe the main will do. But we can not unwrap the values.
Strictly speaking there is a function unsafePerformIO :: IO a -> a that can do that. But it is strongly adviced not to use that. Functions in Haskell are supposed to be pure that means that for the same input, we always retrieve the same output. getLine itself is a pure, since it always produces the same recipe (IO String).
type InterpreterMonad = ErrorT String ((StateT (Stack EnvEval)) IO)
argsToContext :: [DefArg] -> [CallArg] -> InterpreterMonad ()
argsToContext xs ys = argsToContext' xs ys Map.empty where
argsToContext' ((ArgForDefinition t name):xs) ((ArgForCall e):ys) m = get >>= \contextStack -> (argsToContext' xs ys (Map.insert name e m ))
argsToContext' [] [] m = get >>= \contextStack -> (put (push contextStack m)) >>= \_ -> return ()
data DefArg =
ArgForDefinition Type VarName
data CallArg =
ArgForCall Evaluable
Hi,
I've got a problem with understanding above piece of Haskell code
Especially, I cannot understand how does it return () and why it is placed here.
Please explain.
return in Haskell does not mean what it means in imperative languages like C or Java. In fact its really the wrong name, but we are stuck with it for historical reasons. Better names would be "pure" or "wrap", meaning that it takes a pure value and wraps it up into the monadic context.
The () type is known as "unit". Its not actually empty because it has one value, also called () (hence the name). However it is sort-of empty because you use it when you don't want to convey any information: there is only one value, so it needs zero bits to represent it. (There is a "Void" type available for when you really don't want it to have any values).
So return () means that this monadic action wraps a unit up in the monadic context. In effect its a no-op: do nothing and return no information.
In this case its being used in the case where the arguments are two empty lists. argsToContext' has the job of pairing up the DefArg list with the CallArg list. The first definition takes the head of each list, does its thing with them, and then calls itself with the tails of each list. When the tails are both empty it calls the second version which puts the resulting context on top of the context stack. If the two lists are of different length then it will throw an exception because neither pattern matches. If its your code then you ought to put in a defensive case, partly to help you debug, partly to show that you actually thought about the case, and partly to stop the compiler nagging you about it.
The 'context' in this case is the interpreter's variables, which are held in the Map. Hence the only effect of argsToContext' is to add these pairs to the context and then return a 'Unit' value. The InterpreterMonad is the monadic type, and the return value has the type InterpreterMonad (), meaning that no information is returned, it just has side effects within the monadic context.
In fact I don't think you need the return () here because put already has type m () for whatever monad you are in. So just delete the >>= \_ -> return () and I think it will work fine.
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"
Here's a simple, barebones example of how the code that I'm trying to do would look in C++.
while (state == true) {
a = function1();
b = function2();
state = function3();
}
In the program I'm working on, I have some functions that I need to loop through until bool state equals false (or until one of the variables, let's say variable b, equals 0).
How would this code be done in Haskell? I've searched through here, Google, and even Bing and haven't been able to find any clear, straight forward explanations on how to do repetitive actions with functions.
Any help would be appreciated.
Taking Daniels comment into account, it could look something like this:
f = loop init_a init_b true
where
loop a b True = loop a' b' (fun3 a' b')
where
a' = fun1 ....
b' = fun2 .....
loop a b False = (a,b)
Well, here's a suggestion of how to map the concepts here:
A C++ loop is some form of list operation in Haskell.
One iteration of the loop = handling one element of the list.
Looping until a certain condition becomes true = base case of a function that recurses on a list.
But there is something that is critically different between imperative loops and functional list functions: loops describe how to iterate; higher-order list functions describe the structure of the computation. So for example, map f [a0, a1, ..., an] can be described by this diagram:
[a0, a1, ..., an]
| | |
f f f
| | |
v v v
[f a0, f a1, ..., f an]
Note that this describes how the result is related to the arguments f and [a0, a1, ..., an], not how the iteration is performed step by step.
Likewise, foldr f z [a0, a1, ..., an] corresponds to this:
f a0 (f a1 (... (f an z)))
filter doesn't quite lend itself to diagramming, but it's easy to state many rules that it satisfies:
length (filter pred xs) <= length xs
For every element x of filter pred xs, pred x is True.
If x is an element of filter pred xs, then x is an element of xs
If x is not an element of xs, then x is not an element of filter pred xs
If x appears before x' in filter pred xs, then x appears before x' in xs
If x appears before x' in xs, and both x and x' appear in filter pred xs, then x appears before x' in filter pred xs
In a classic imperative program, all three of these cases are written as loops, and the difference between them comes down to what the loop body does. Functional programming, on the contrary, insists that this sort of structural pattern does not belong in "loop bodies" (the functions f and pred in these examples); rather, these patterns are best abstracted out into higher-order functions like map, foldr and filter. Thus, every time you see one of these list functions you instantly know some important facts about how the arguments and the result are related, without having to read any code; whereas in a typical imperative program, you must read the bodies of loops to figure this stuff out.
So the real answer to your question is that it's impossible to offer an idiomatic translation of an imperative loop into functional terms without knowing what the loop body is doing—what are the preconditions supposed to be before the loop runs, and what the postconditions are supposed to be when the loop finishes. Because that loop body that you only described vaguely is going to determine what the structure of the computation is, and different such structures will call for different higher-order functions in Haskell.
First of all, let's think about a few things.
Does function1 have side effects?
Does function2 have side effects?
Does function3 have side effects?
The answer to all of these is a resoundingly obvious YES, because they take no inputs, and presumably there are circumstances which cause you to go around the while loop more than once (rather than def function3(): return false). Now let's remodel these functions with explicit state.
s = initialState
sentinel = true
while(sentinel):
a,b,s,sentinel = function1(a,b,s,sentinel)
a,b,s,sentinel = function2(a,b,s,sentinel)
a,b,s,sentinel = function3(a,b,s,sentinel)
return a,b,s
Well that's rather ugly. We know absolutely nothing about what inputs each function draws from, nor do we know anything about how these functions might affect the variables a, b, and sentinel, nor "any other state" which I have simply modeled as s.
So let's make a few assumptions. Firstly, I am going to assume that these functions do not directly depend on nor affect in any way the values of a, b, and sentinel. They might, however, change the "other state". So here's what we get:
s = initState
sentinel = true
while (sentinel):
a,s2 = function1(s)
b,s3 = function2(s2)
sentinel,s4 = function(s3)
s = s4
return a,b,s
Notice I've used temporary variables s2, s3, and s4 to indicate the changes that the "other state" goes through. Haskell time. We need a control function to behave like a while loop.
myWhile :: s -- an initial state
-> (s -> (Bool, a, s)) -- given a state, produces a sentinel, a current result, and the next state
-> (a, s) -- the result, plus resultant state
myWhile s f = case f s of
(False, a, s') -> (a, s')
(True, _, s') -> myWhile s' f
Now how would one use such a function? Well, given we have the functions:
function1 :: MyState -> (AType, MyState)
function2 :: MyState -> (BType, MyState)
function3 :: MyState -> (Bool, MyState)
We would construct the desired code as follows:
thatCodeBlockWeAreTryingToSimulate :: MyState -> ((AType, BType), MyState)
thatCodeBlockWeAreTryingToSimulate initState = myWhile initState f
where f :: MyState -> (Bool, (AType, BType), MyState)
f s = let (a, s2) = function1 s
(b, s3) = function2 s2
(sentinel, s4) = function3 s3
in (sentinel, (a, b), s4)
Notice how similar this is to the non-ugly python-like code given above.
You can verify that the code I have presented is well-typed by adding function1 = undefined etc for the three functions, as well as the following at the top of the file:
{-# LANGUAGE EmptyDataDecls #-}
data MyState
data AType
data BType
So the takeaway message is this: in Haskell, you must explicitly model the changes in state. You can use the "State Monad" to make things a little prettier, but you should first understand the idea of passing state around.
Lets take a look at your C++ loop:
while (state == true) {
a = function1();
b = function2();
state = function3();
}
Haskell is a pure functional language, so it won't fight us as much (and the resulting code will be more useful, both in itself and as an exercise to learn Haskell) if we try to do this without side effects, and without using monads to make it look like we're using side effects either.
Lets start with this structure
while (state == true) {
<<do stuff that updates state>>
}
In Haskell we're obviously not going to be checking a variable against true as the loop condition, because it can't change its value[1] and we'd either evaluate the loop body forever or never. So instead, we'll want to be evaluating a function that returns a boolean value on some argument:
while (check something == True) {
<<do stuff that updates state>>
}
Well, now we don't have a state variable, so that "do stuff that updates state" is looking pretty pointless. And we don't have a something to pass to check. Lets think about this a bit more. We want the something to be checked to depend on what the "do stuff" bit is doing. We don't have side effects, so that means something has to be (or be derived from) returned from the "do stuff". "do stuff" also needs to take something that varies as an argument, or it'll just keep returning the same thing forever, which is also pointless. We also need to return a value out all this, otherwise we're just burning CPU cycles (again, with no side effects there's no point running a function if we don't use its output in some way, and there's even less point running a function repeatedly if we never use its output).
So how about something like this:
while check func state =
let next_state = func state in
if check next_state
then while check func next_state
else next_state
Lets try it in GHCi:
*Main> while (<20) (+1) 0
20
This is the result of applying (+1) repeatedly while the result is less than 20, starting from 0.
*Main> while ((<20) . length) (++ "zob") ""
"zobzobzobzobzobzobzob"
This is the result of concatenating "zob" repeatedly while the result's length is less than 20, starting from the empty string.
So you can see I've defined a function that is (sort of a bit) analogous to a while loop from imperative languages. We didn't even need dedicated loop syntax for it! (which is the real reason Haskell has no such syntax; if you need this kind of thing you can express it as a function). It's not the only way to do so, and experienced Haskell programmers would probably use other standard library functions to do this kind of job, rather than writing while.
But I think it's useful to see how you can express this kind of thing in Haskell. It does show that you can't translate things like imperative loops directly into Haskell; I didn't end up translating your loop in terms of my while because it ends up pretty pointless; you never use the result of function1 or function2, they're called with no arguments so they'd always return the same thing in every iteration, and function3 likewise always returns the same thing, and can only return true or false to either cause while to keep looping or stop, with no information resulting.
Presumably in the C++ program they're all using side effects to actually get some work done. If they operate on in-memory things then you need to translate a bigger chunk of your program at once to Haskell for the translation of this loop to make any sense. If those functions are doing IO then you'll need to do this in the IO monad in Haskell, for which my while function doesn't work, but you can do something similar.
[1] As an aside, it's worth trying to understand that "you can't change variables" in Haskell isn't just an arbitrary restriction, nor is it just an acceptable trade off for the benefits of purity, it is a concept that doesn't make sense the way Haskell wants you to think about Haskell code. You're writing down expressions that result from evaluating functions on certain arguments: in f x = x + 1 you're saying that f x is x + 1. If you really think of it that way rather than thinking "f takes x, then adds one to it, then returns the result" then the concept of "having side effects" doesn't even apply; how could something existing and being equal to something else somehow change a variable, or have some other side effect?
You should write a solution to your problem in a more functional approach.
However, some code in haskell works a lot like imperative looping, take for example state monads, terminal recursivity, until, foldr, etc.
A simple example is the factorial. In C, you would write a loop where in haskell you can for example write fact n = foldr (*) 1 [2..n].
If you've two functions f :: a -> b and g :: b -> c where a, b, and c are types like String or [Int] then you can compose them simply by writing f . b.
If you wish them to loop over a list or vector you could write map (f . g) or V.map (f . g), assuming you've done Import qualified Data.Vector as V.
Example : I wish to print a list of markdown headings like ## <number>. <heading> ## but I need roman numerals numbered from 1 and my list headings has type type [(String,Double)] where the Double is irrelevant.
Import Data.List
Import Text.Numeral.Roman
let fun = zipWith (\a b -> a ++ ". " ++ b ++ "##\n") (map toRoman [1..]) . map fst
fun [("Foo",3.5),("Bar",7.1)]
What the hell does this do?
toRoman turns a number into a string containing the roman numeral. map toRoman does this to every element of a loop. map toRoman [1..] does it to every element of the lazy infinite list [1,2,3,4,..], yielding a lazy infinite list of roman numeral strings
fst :: (a,b) -> a simply extracts the first element of a tuple. map fst throws away our silly Meow information along the entire list.
\a b -> "##" ++ show a ++ ". " ++ b ++ "##" is a lambda expression that takes two strings and concatenates them together within the desired formatting strings.
zipWith :: (a -> b -> c) -> [a] -> [b] -> [c] takes a two argument function like our lambda expression and feeds it pairs of elements from it's own second and third arguments.
You'll observe that zip, zipWith, etc. only read as much of the lazy infinite list of Roman numerals as needed for the list of headings, meaning I've number my headings without maintaining any counter variable.
Finally, I have declared fun without naming it's argument because the compiler can figure it out from the fact that map fst requires one argument. You'll notice that put a . before my second map too. I could've written (map fst h) or $ map fst h instead if I'd written fun h = ..., but leaving the argument off fun meant I needed to compose it with zipWith after applying zipWith to two arguments of the three arguments zipWith wants.
I'd hope the compiler combines the zipWith and maps into one single loop via inlining.
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!