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I have to implement a small programm in Haskell that increments/decrements a result by what in the console line is. For example if we have -a in the console the results must be 0, if -b the result must be incremented with 6 and so on. I have to do this with pattern matching.
I haven't used Haskell until now and I find it pretty hard to understand. I have this to start with:
import System.Environment
main = getArgs >>= print . (foldr apply 0) . reverse
apply :: String -> Integer -> Integer
I don't understand what in the main is. What does it make and the reverse from end, what does it do? As I've read on the internet the getArgs function gives me the values from the console line. But how can I use them? Are there are equivalent functions like for/while in Haskell?
Also, if you have some examples or maybe could help me, I will be very thankful.
Thanks!
This is not beginner-friendly code. Several shortcuts are taken there to keep the code very compact (and in pointfree form). The code
main = getArgs >>= print . (foldr apply 0) . reverse
can be expanded as follows
main = do
args <- getArgs
let reversedArgs = reverse args
result = foldr apply 0 reversedArgs
print result
The result of this can be seen as follows. If the command line arguments are, say, args = ["A","B","C"], then we get reversedArgs = ["C","B","A"] and finally
result = apply "C" (apply "B" (apply "A" 0))
since foldr applies the function apply in such way.
Honestly, I'm unsure about why the code uses reverse and foldr for your task. I would have considered foldl (or, to improve performance, foldl') instead.
I expect the exercise is not to touch the given code, but to expand on it to perform your function. It defines a complicated-looking main function and declares the type of a more straight forward apply, which is called but not defined.
import System.Environment -- contains the function getArgs
-- main gets arguments, does something to them using apply, and prints
main = getArgs >>= print . (foldr apply 0) . reverse
-- apply must have this type, but what it does must be elsewhere
apply :: String -> Integer -> Integer
If we concentrate on apply, we see that it receives a string and an integer, and returns an integer. This is the function we have to write, and it can't decide control flow, so we can just get to it while hoping the argument handling works out.
If we do want to figure out what main is up to, we can make a few observations. The only integer in main is 0, so the first call must get that as its second argument; later ones will be chained with whatever is returned, as that's how foldr operates. r stands for from the right, but the arguments are reversed, so this still processes arguments from the left.
So I could go ahead and just write a few apply bindings to make the program compile:
apply "succ" n = succ n
apply "double" n = n + n
apply "div3" n = n `div` 3
This added a few usable operations. It doesn't handle all possible strings.
$ runhaskell pmb.hs succ succ double double succ div3
3
$ runhaskell pmb.hs hello?
pmb.hs: pmb.hs:(5,1)-(7,26): Non-exhaustive patterns in function apply
The exercise should be about how you handle the choice of operation based on the string argument. There are several options, including distinct patterns as above, pattern guards, case and if expressions.
It can be useful to examine the used functions to see how they might fit together. Here's a look at a few of the used functions in ghci:
Prelude> import System.Environment
Prelude System.Environment> :t getArgs
getArgs :: IO [String]
Prelude System.Environment> :t (>>=)
(>>=) :: Monad m => m a -> (a -> m b) -> m b
Prelude System.Environment> :t print
print :: Show a => a -> IO ()
Prelude System.Environment> :t (.)
(.) :: (b -> c) -> (a -> b) -> a -> c
Prelude System.Environment> :t foldr
foldr :: Foldable t => (a -> b -> b) -> b -> t a -> b
Prelude System.Environment> :t reverse
reverse :: [a] -> [a]
This shows that all the strings come out of getArgs, it and print operate in the IO monad, which must be the m in >>=, and . transfers results from the right function into arguments for the left function. The type signature alone doesn't tell us what order foldr handles things, though, or what reverse does (though it can't create new values, only reorder including repetition).
As a last exercise, I'll rewrite the main function in a form that doesn't switch directions as many times:
main = print . foldl (flip apply) 0 =<< getArgs
This reads from right to left in a data flow sense and handles arguments from left to right because foldl performs left-associative folding. flip is just there to match the argument order for apply.
As suggested in the comment, hoogle is a great tool.
To find out what exactly you get from getArgs you can search for it on hoogle:
https://hackage.haskell.org/package/base-4.11.1.0/docs/System-Environment.html#v:getArgs
As you can see, it's of type IO [String].
Since I don't know how familiar you are with the IO abstractions yet, we'll just say that the right part of >>= gets those as argument.
The arguments for a call like ./a.out -a -b --asdf Hi will then be a list of strings:
["-a", "-b", "--asdf", "Hi"].
The fold + reverse in the main will then do some magic, and your apply function will be called with each string in the list and the previous return value (0 for the first invocation).
In Haskell, String is the same as [Char] with a bit of compiler sugar, so you can match on strings like you would on regular lists in your definition of apply.
What is the intuitive meaning of join for a Monad?
The monads-as-containers analogies make sense to me, and inside these analogies join makes sense. A value is double-wrapped and we unwrap one layer. But as we all know, a monad is not a container.
How might one write sensible, understandable code using join in normal circumstances, say when in IO?
An action :: IO (IO a) is a way of producing a way of producing an a. join action, then, is a way of producing an a by running the outermost producer of action, taking the producer it produced and then running that as well, to finally get to that juicy a.
join collapses consecutive layers of the type constructor.
A valid join must satisfy the property that, for any number of consecutive applications of the type constructor, it shouldn't matter the order in which we collapse the layers.
For example
ghci> let lolol = [[['a'],['b','c']],[['d'],['e']]]
ghci> lolol :: [[[Char]]]
ghci> lolol :: [] ([] ([] Char)) -- the type can also be expressed like this
ghci> join (fmap join lolol) -- collapse inner layers first
"abcde"
ghci> join (join lolol) -- collapse outer layers first
"abcde"
(We used fmap to "get inside" the outer monadic layer so that we could collapse the inner layers first.)
A small non container example where join is useful: for the function monad (->) a, join is equivalent to \f x -> f x x, a function of type (a -> a -> b) -> a -> b that applies two times the same argument to another function.
For the List monad, join is simply concat, and concatMap is join . fmap.
So join implicitly appears in any list expression which uses concat
or concatMap.
Suppose you were asked to find all of the numbers which are divisors of any
number in an input list. If you have a divisors function:
divisors :: Int -> [Int]
divisors n = [ d | d <- [1..n], mod n d == 0 ]
you might solve the problem like this:
foo xs = concat $ (map divisors xs)
Here we are thinking of solving the problem by first mapping the
divisors function over all of the input elements and then concatenating
all of the resulting lists. You might even think that this is a very
"functional" way of solving the problem.
Another approch would be to write a list comprehension:
bar xs = [ d | x <- xs, d <- divisors x ]
or using do-notation:
bar xs = do x <- xs
d <- divisors
return d
Here it might be said we're thinking a little more
imperatively - first draw a number from the list xs; then draw
a divisors from the divisors of the number and yield it.
It turns out, though, that foo and bar are exactly the same function.
Morever, these two approaches are exactly the same in any monad.
That is, for any monad, and appropriate monadic functions f and g:
do x <- f
y <- g x is the same as: (join . fmap g) f
return y
For instance, in the IO monad if we set f = getLine and g = readFile,
we have:
do x <- getLine
y <- readFile x is the same as: (join . fmap readFile) getLine
return y
The do-block is a more imperative way of expressing the action: first read a
line of input; then treat returned string as a file name, read the contents
of the file and finally return the result.
The equivalent join expression seems a little unnatural in the IO-monad.
However it shouldn't be as we are using it in exactly the same way as we
used concatMap in the first example.
Given an action that produces another action, run the action and then run the action that it produces.
If you imagine some kind of Parser x monad that parses an x, then Parser (Parser x) is a parser that does some parsing, and then returns another parser. So join would flatten this into a Parser x that just runs both actions and returns the final x.
Why would you even have a Parser (Parser x) in the first place? Basically, because fmap. If you have a parser, you can fmap a function that changes the result over it. But if you fmap a function that itself returns a parser, you end up with a Parser (Parser x), where you probably want to just run both actions. join implements "just run both actions".
I like the parsing example because a parser typically has a runParser function. And it's clear that a Parser Int is not an integer. It's something that can parse an integer, after you give it some input to parse from. I think a lot of people end up thinking of an IO Int as being just a normal integer but with this annoying IO bit that you can't get rid of. It isn't. It's an unexecuted I/O operation. There's no integer "inside" it; the integer doesn't exist until you actually perform the I/O.
I find these things easier to interpret by writing out the types and refactoring them a bit to reveal what the functions do.
Reader monad
The Reader type is defined thus, and its join function has the type shown:
newtype Reader r a = Reader { runReader :: r -> a }
join :: Reader r (Reader r a) -> Reader r a
Since this is a newtype, this means that the type Reader r a is isomorphic to r -> a. So we can refactor the type definition to give us this type that, albeit it's not the same, it's really "the same" with scare quotes:
In the (->) r monad, which is isomorphic to Reader r, join is the function:
join :: (r -> r -> a) -> r -> a
So the Reader join is the function that takes a two-place function (r -> r -> a) and applies to the same value at both its argument positions.
Writer monad
Since the Writer type has this definition:
newtype Writer w a = Writer { runWriter :: (a, w) }
...then when we remove the newtype, its join function has a type isomorphic to:
join :: Monoid w => ((a, w), w) -> (a, w)
The Monoid constraint needs to be there because the Monad instance for Writer requires it, and it lets us guess right away what the function does:
join ((a, w0), w1) = (a, w0 <> w1)
State monad
Similarly, since State has this definition:
newtype State s a = State { runState :: s -> (a, s) }
...then its join is like this:
join :: (s -> (s -> (a, s), s)) -> s -> (a, s)
...and you can also venture just writing it directly:
join f s0 = (a, s2)
where
(g, s1) = f s0
(a, s2) = g s1
{- Here's the "map" to the variable names in the function:
f g s2 s1 s0 s2
join :: (s -> (s -> (a, s ), s )) -> s -> (a, s )
-}
If you stare at this type a bit, you might think that it bears some resemblance to both the Reader and Writer's types for their join operations. And you'd be right! The Reader, Writer and State monads are all instances of a more general pattern called update monads.
List monad
join :: [[a]] -> [a]
As other people have pointed out, this is the type of the concat function.
Parsing monads
Here comes a really neat thing to realize. Very often, "fancy" monads turn out to be combinations or variants of "basic" ones like Reader, Writer, State or lists. So often what I do when confronted with a novel monad is ask: which of the basic monads does it resemble, and how?
Take for example parsing monads, which have been brought up in other answers here. A simplistic parser monad (with no support for important things like error reporting) looks like this:
newtype Parser a = Parser { runParser :: String -> [(a, String)] }
A Parser is a function that takes a string as input, and returns a list of candidate parses, where each candidate parse is a pair of:
A parse result of type a;
The leftovers (the suffix of the input string that was not consumed in that parse).
But notice that this type looks very much like the state monad:
newtype Parser a = Parser { runParser :: String -> [(a, String)] }
newtype State s a = State { runState :: s -> (a, s) }
And this is no accident! Parser monads are nondeterministic state monads, where the state is the unconsumed portion of the input string, and parse steps generate alternatives that may be later rejected in light of further input. List monads are often called "nondeterminism" monads, so it's no surprise that a parser resembles a mix of the state and list monads.
And this intuition can be systematized by using monad transfomers. The state monad transformer is defined like this:
newtype StateT s m a = StateT { runStateT :: s -> m (a, s) }
Which means that the Parser type from above can be written like this as well:
type Parser a = StateT String [] a
...and its Monad instance follows mechanically from those of StateT and [].
The IO monad
Imagine we could enumerate all of the possible primitive IO actions, somewhat like this:
{-# LANGUAGE GADTs #-}
data Command a where
-- An action that writes a char to stdout
putChar :: Char -> Command ()
-- An action that reads a char from stdin
getChar :: Command Char
-- ...
Then we could think of the IO type as this (which I've adapted from the highly-recommended Operational monad tutorial):
data IO a where
-- An `IO` action that just returns a constant value.
Return :: a -> IO a
-- An action that binds the result of a `Command` to
-- a function that computes the next step after it.
Bind :: Command x -> (x -> IO a) -> IO a
instance Monad IO where ...
Then join action would then look like this:
join :: IO (IO a) -> IO a
-- If the action is just `Return`, then its payload already
-- is what we need to return.
join (Return ioa) = ioa
-- If the action is a `Bind`, then its "next step" function
-- `f` produces `IO (IO a)`, so we can just recursively stick
-- a `join` to its result end.
join (Bind cmd f) = Bind cmd (join . f)
So all that the join does here is "chase down" the IO action until it sees a result that fits the pattern Return (ma :: IO a), and strip out the outer Return.
So what did I do here? Just like for parser monads, I just defined (or rather copied) a toy model of the IO type that has the virtue of being transparent. Then I work out the behavior of join from the toy model.
In my humble opinion the answers to the famous question "What is a monad?", especially the most voted ones, try to explain what is a monad without clearly explaining why monads are really necessary. Can they be explained as the solution to a problem?
Why do we need monads?
We want to program only using functions. ("functional programming (FP)" after all).
Then, we have a first big problem. This is a program:
f(x) = 2 * x
g(x,y) = x / y
How can we say what is to be executed first? How can we form an ordered sequence of functions (i.e. a program) using no more than functions?
Solution: compose functions. If you want first g and then f, just write f(g(x,y)). This way, "the program" is a function as well: main = f(g(x,y)). OK, but ...
More problems: some functions might fail (i.e. g(2,0), divide by 0). We have no "exceptions" in FP (an exception is not a function). How do we solve it?
Solution: Let's allow functions to return two kind of things: instead of having g : Real,Real -> Real (function from two reals into a real), let's allow g : Real,Real -> Real | Nothing (function from two reals into (real or nothing)).
But functions should (to be simpler) return only one thing.
Solution: let's create a new type of data to be returned, a "boxing type" that encloses maybe a real or be simply nothing. Hence, we can have g : Real,Real -> Maybe Real. OK, but ...
What happens now to f(g(x,y))? f is not ready to consume a Maybe Real. And, we don't want to change every function we could connect with g to consume a Maybe Real.
Solution: let's have a special function to "connect"/"compose"/"link" functions. That way, we can, behind the scenes, adapt the output of one function to feed the following one.
In our case: g >>= f (connect/compose g to f). We want >>= to get g's output, inspect it and, in case it is Nothing just don't call f and return Nothing; or on the contrary, extract the boxed Real and feed f with it. (This algorithm is just the implementation of >>= for the Maybe type). Also note that >>= must be written only once per "boxing type" (different box, different adapting algorithm).
Many other problems arise which can be solved using this same pattern: 1. Use a "box" to codify/store different meanings/values, and have functions like g that return those "boxed values". 2. Have a composer/linker g >>= f to help connecting g's output to f's input, so we don't have to change any f at all.
Remarkable problems that can be solved using this technique are:
having a global state that every function in the sequence of functions ("the program") can share: solution StateMonad.
We don't like "impure functions": functions that yield different output for same input. Therefore, let's mark those functions, making them to return a tagged/boxed value: IO monad.
Total happiness!
The answer is, of course, "We don't". As with all abstractions, it isn't necessary.
Haskell does not need a monad abstraction. It isn't necessary for performing IO in a pure language. The IO type takes care of that just fine by itself. The existing monadic desugaring of do blocks could be replaced with desugaring to bindIO, returnIO, and failIO as defined in the GHC.Base module. (It's not a documented module on hackage, so I'll have to point at its source for documentation.) So no, there's no need for the monad abstraction.
So if it's not needed, why does it exist? Because it was found that many patterns of computation form monadic structures. Abstraction of a structure allows for writing code that works across all instances of that structure. To put it more concisely - code reuse.
In functional languages, the most powerful tool found for code reuse has been composition of functions. The good old (.) :: (b -> c) -> (a -> b) -> (a -> c) operator is exceedingly powerful. It makes it easy to write tiny functions and glue them together with minimal syntactic or semantic overhead.
But there are cases when the types don't work out quite right. What do you do when you have foo :: (b -> Maybe c) and bar :: (a -> Maybe b)? foo . bar doesn't typecheck, because b and Maybe b aren't the same type.
But... it's almost right. You just want a bit of leeway. You want to be able to treat Maybe b as if it were basically b. It's a poor idea to just flat-out treat them as the same type, though. That's more or less the same thing as null pointers, which Tony Hoare famously called the billion-dollar mistake. So if you can't treat them as the same type, maybe you can find a way to extend the composition mechanism (.) provides.
In that case, it's important to really examine the theory underlying (.). Fortunately, someone has already done this for us. It turns out that the combination of (.) and id form a mathematical construct known as a category. But there are other ways to form categories. A Kleisli category, for instance, allows the objects being composed to be augmented a bit. A Kleisli category for Maybe would consist of (.) :: (b -> Maybe c) -> (a -> Maybe b) -> (a -> Maybe c) and id :: a -> Maybe a. That is, the objects in the category augment the (->) with a Maybe, so (a -> b) becomes (a -> Maybe b).
And suddenly, we've extended the power of composition to things that the traditional (.) operation doesn't work on. This is a source of new abstraction power. Kleisli categories work with more types than just Maybe. They work with every type that can assemble a proper category, obeying the category laws.
Left identity: id . f = f
Right identity: f . id = f
Associativity: f . (g . h) = (f . g) . h
As long as you can prove that your type obeys those three laws, you can turn it into a Kleisli category. And what's the big deal about that? Well, it turns out that monads are exactly the same thing as Kleisli categories. Monad's return is the same as Kleisli id. Monad's (>>=) isn't identical to Kleisli (.), but it turns out to be very easy to write each in terms of the other. And the category laws are the same as the monad laws, when you translate them across the difference between (>>=) and (.).
So why go through all this bother? Why have a Monad abstraction in the language? As I alluded to above, it enables code reuse. It even enables code reuse along two different dimensions.
The first dimension of code reuse comes directly from the presence of the abstraction. You can write code that works across all instances of the abstraction. There's the entire monad-loops package consisting of loops that work with any instance of Monad.
The second dimension is indirect, but it follows from the existence of composition. When composition is easy, it's natural to write code in small, reusable chunks. This is the same way having the (.) operator for functions encourages writing small, reusable functions.
So why does the abstraction exist? Because it's proven to be a tool that enables more composition in code, resulting in creating reusable code and encouraging the creation of more reusable code. Code reuse is one of the holy grails of programming. The monad abstraction exists because it moves us a little bit towards that holy grail.
Benjamin Pierce said in TAPL
A type system can be regarded as calculating a kind of static
approximation to the run-time behaviours of the terms in a program.
That's why a language equipped with a powerful type system is strictly more expressive, than a poorly typed language. You can think about monads in the same way.
As #Carl and sigfpe point, you can equip a datatype with all operations you want without resorting to monads, typeclasses or whatever other abstract stuff. However monads allow you not only to write reusable code, but also to abstract away all redundant detailes.
As an example, let's say we want to filter a list. The simplest way is to use the filter function: filter (> 3) [1..10], which equals [4,5,6,7,8,9,10].
A slightly more complicated version of filter, that also passes an accumulator from left to right, is
swap (x, y) = (y, x)
(.*) = (.) . (.)
filterAccum :: (a -> b -> (Bool, a)) -> a -> [b] -> [b]
filterAccum f a xs = [x | (x, True) <- zip xs $ snd $ mapAccumL (swap .* f) a xs]
To get all i, such that i <= 10, sum [1..i] > 4, sum [1..i] < 25, we can write
filterAccum (\a x -> let a' = a + x in (a' > 4 && a' < 25, a')) 0 [1..10]
which equals [3,4,5,6].
Or we can redefine the nub function, that removes duplicate elements from a list, in terms of filterAccum:
nub' = filterAccum (\a x -> (x `notElem` a, x:a)) []
nub' [1,2,4,5,4,3,1,8,9,4] equals [1,2,4,5,3,8,9]. A list is passed as an accumulator here. The code works, because it's possible to leave the list monad, so the whole computation stays pure (notElem doesn't use >>= actually, but it could). However it's not possible to safely leave the IO monad (i.e. you cannot execute an IO action and return a pure value — the value always will be wrapped in the IO monad). Another example is mutable arrays: after you have leaved the ST monad, where a mutable array live, you cannot update the array in constant time anymore. So we need a monadic filtering from the Control.Monad module:
filterM :: (Monad m) => (a -> m Bool) -> [a] -> m [a]
filterM _ [] = return []
filterM p (x:xs) = do
flg <- p x
ys <- filterM p xs
return (if flg then x:ys else ys)
filterM executes a monadic action for all elements from a list, yielding elements, for which the monadic action returns True.
A filtering example with an array:
nub' xs = runST $ do
arr <- newArray (1, 9) True :: ST s (STUArray s Int Bool)
let p i = readArray arr i <* writeArray arr i False
filterM p xs
main = print $ nub' [1,2,4,5,4,3,1,8,9,4]
prints [1,2,4,5,3,8,9] as expected.
And a version with the IO monad, which asks what elements to return:
main = filterM p [1,2,4,5] >>= print where
p i = putStrLn ("return " ++ show i ++ "?") *> readLn
E.g.
return 1? -- output
True -- input
return 2?
False
return 4?
False
return 5?
True
[1,5] -- output
And as a final illustration, filterAccum can be defined in terms of filterM:
filterAccum f a xs = evalState (filterM (state . flip f) xs) a
with the StateT monad, that is used under the hood, being just an ordinary datatype.
This example illustrates, that monads not only allow you to abstract computational context and write clean reusable code (due to the composability of monads, as #Carl explains), but also to treat user-defined datatypes and built-in primitives uniformly.
I don't think IO should be seen as a particularly outstanding monad, but it's certainly one of the more astounding ones for beginners, so I'll use it for my explanation.
Naïvely building an IO system for Haskell
The simplest conceivable IO system for a purely-functional language (and in fact the one Haskell started out with) is this:
main₀ :: String -> String
main₀ _ = "Hello World"
With lazyness, that simple signature is enough to actually build interactive terminal programs – very limited, though. Most frustrating is that we can only output text. What if we added some more exciting output possibilities?
data Output = TxtOutput String
| Beep Frequency
main₁ :: String -> [Output]
main₁ _ = [ TxtOutput "Hello World"
-- , Beep 440 -- for debugging
]
cute, but of course a much more realistic “alterative output” would be writing to a file. But then you'd also want some way to read from files. Any chance?
Well, when we take our main₁ program and simply pipe a file to the process (using operating system facilities), we have essentially implemented file-reading. If we could trigger that file-reading from within the Haskell language...
readFile :: Filepath -> (String -> [Output]) -> [Output]
This would use an “interactive program” String->[Output], feed it a string obtained from a file, and yield a non-interactive program that simply executes the given one.
There's one problem here: we don't really have a notion of when the file is read. The [Output] list sure gives a nice order to the outputs, but we don't get an order for when the inputs will be done.
Solution: make input-events also items in the list of things to do.
data IO₀ = TxtOut String
| TxtIn (String -> [Output])
| FileWrite FilePath String
| FileRead FilePath (String -> [Output])
| Beep Double
main₂ :: String -> [IO₀]
main₂ _ = [ FileRead "/dev/null" $ \_ ->
[TxtOutput "Hello World"]
]
Ok, now you may spot an imbalance: you can read a file and make output dependent on it, but you can't use the file contents to decide to e.g. also read another file. Obvious solution: make the result of the input-events also something of type IO, not just Output. That sure includes simple text output, but also allows reading additional files etc..
data IO₁ = TxtOut String
| TxtIn (String -> [IO₁])
| FileWrite FilePath String
| FileRead FilePath (String -> [IO₁])
| Beep Double
main₃ :: String -> [IO₁]
main₃ _ = [ TxtIn $ \_ ->
[TxtOut "Hello World"]
]
That would now actually allow you to express any file operation you might want in a program (though perhaps not with good performance), but it's somewhat overcomplicated:
main₃ yields a whole list of actions. Why don't we simply use the signature :: IO₁, which has this as a special case?
The lists don't really give a reliable overview of program flow anymore: most subsequent computations will only be “announced” as the result of some input operation. So we might as well ditch the list structure, and simply cons a “and then do” to each output operation.
data IO₂ = TxtOut String IO₂
| TxtIn (String -> IO₂)
| Terminate
main₄ :: IO₂
main₄ = TxtIn $ \_ ->
TxtOut "Hello World"
Terminate
Not too bad!
So what has all of this to do with monads?
In practice, you wouldn't want to use plain constructors to define all your programs. There would need to be a good couple of such fundamental constructors, yet for most higher-level stuff we would like to write a function with some nice high-level signature. It turns out most of these would look quite similar: accept some kind of meaningfully-typed value, and yield an IO action as the result.
getTime :: (UTCTime -> IO₂) -> IO₂
randomRIO :: Random r => (r,r) -> (r -> IO₂) -> IO₂
findFile :: RegEx -> (Maybe FilePath -> IO₂) -> IO₂
There's evidently a pattern here, and we'd better write it as
type IO₃ a = (a -> IO₂) -> IO₂ -- If this reminds you of continuation-passing
-- style, you're right.
getTime :: IO₃ UTCTime
randomRIO :: Random r => (r,r) -> IO₃ r
findFile :: RegEx -> IO₃ (Maybe FilePath)
Now that starts to look familiar, but we're still only dealing with thinly-disguised plain functions under the hood, and that's risky: each “value-action” has the responsibility of actually passing on the resulting action of any contained function (else the control flow of the entire program is easily disrupted by one ill-behaved action in the middle). We'd better make that requirement explicit. Well, it turns out those are the monad laws, though I'm not sure we can really formulate them without the standard bind/join operators.
At any rate, we've now reached a formulation of IO that has a proper monad instance:
data IO₄ a = TxtOut String (IO₄ a)
| TxtIn (String -> IO₄ a)
| TerminateWith a
txtOut :: String -> IO₄ ()
txtOut s = TxtOut s $ TerminateWith ()
txtIn :: IO₄ String
txtIn = TxtIn $ TerminateWith
instance Functor IO₄ where
fmap f (TerminateWith a) = TerminateWith $ f a
fmap f (TxtIn g) = TxtIn $ fmap f . g
fmap f (TxtOut s c) = TxtOut s $ fmap f c
instance Applicative IO₄ where
pure = TerminateWith
(<*>) = ap
instance Monad IO₄ where
TerminateWith x >>= f = f x
TxtOut s c >>= f = TxtOut s $ c >>= f
TxtIn g >>= f = TxtIn $ (>>=f) . g
Obviously this is not an efficient implementation of IO, but it's in principle usable.
Monads serve basically to compose functions together in a chain. Period.
Now the way they compose differs across the existing monads, thus resulting in different behaviors (e.g., to simulate mutable state in the state monad).
The confusion about monads is that being so general, i.e., a mechanism to compose functions, they can be used for many things, thus leading people to believe that monads are about state, about IO, etc, when they are only about "composing functions".
Now, one interesting thing about monads, is that the result of the composition is always of type "M a", that is, a value inside an envelope tagged with "M". This feature happens to be really nice to implement, for example, a clear separation between pure from impure code: declare all impure actions as functions of type "IO a" and provide no function, when defining the IO monad, to take out the "a" value from inside the "IO a". The result is that no function can be pure and at the same time take out a value from an "IO a", because there is no way to take such value while staying pure (the function must be inside the "IO" monad to use such value). (NOTE: well, nothing is perfect, so the "IO straitjacket" can be broken using "unsafePerformIO : IO a -> a" thus polluting what was supposed to be a pure function, but this should be used very sparingly and when you really know to be not introducing any impure code with side-effects.
Monads are just a convenient framework for solving a class of recurring problems. First, monads must be functors (i.e. must support mapping without looking at the elements (or their type)), they must also bring a binding (or chaining) operation and a way to create a monadic value from an element type (return). Finally, bind and return must satisfy two equations (left and right identities), also called the monad laws. (Alternatively one could define monads to have a flattening operation instead of binding.)
The list monad is commonly used to deal with non-determinism. The bind operation selects one element of the list (intuitively all of them in parallel worlds), lets the programmer to do some computation with them, and then combines the results in all worlds to single list (by concatenating, or flattening, a nested list). Here is how one would define a permutation function in the monadic framework of Haskell:
perm [e] = [[e]]
perm l = do (leader, index) <- zip l [0 :: Int ..]
let shortened = take index l ++ drop (index + 1) l
trailer <- perm shortened
return (leader : trailer)
Here is an example repl session:
*Main> perm "a"
["a"]
*Main> perm "ab"
["ab","ba"]
*Main> perm ""
[]
*Main> perm "abc"
["abc","acb","bac","bca","cab","cba"]
It should be noted that the list monad is in no way a side effecting computation. A mathematical structure being a monad (i.e. conforming to the above mentioned interfaces and laws) does not imply side effects, though side-effecting phenomena often nicely fit into the monadic framework.
You need monads if you have a type constructor and functions that returns values of that type family. Eventually, you would like to combine these kind of functions together. These are the three key elements to answer why.
Let me elaborate. You have Int, String and Real and functions of type Int -> String, String -> Real and so on. You can combine these functions easily, ending with Int -> Real. Life is good.
Then, one day, you need to create a new family of types. It could be because you need to consider the possibility of returning no value (Maybe), returning an error (Either), multiple results (List) and so on.
Notice that Maybe is a type constructor. It takes a type, like Int and returns a new type Maybe Int. First thing to remember, no type constructor, no monad.
Of course, you want to use your type constructor in your code, and soon you end with functions like Int -> Maybe String and String -> Maybe Float. Now, you can't easily combine your functions. Life is not good anymore.
And here's when monads come to the rescue. They allow you to combine that kind of functions again. You just need to change the composition . for >==.
Why do we need monadic types?
Since it was the quandary of I/O and its observable effects in nonstrict languages like Haskell that brought the monadic interface to such prominence:
[...] monads are used to address the more general problem of computations (involving state, input/output, backtracking, ...) returning values: they do not solve any input/output-problems directly but rather provide an elegant and flexible abstraction of many solutions to related problems. [...] For instance, no less than three different input/output-schemes are used to solve these basic problems in Imperative functional programming, the paper which originally proposed `a new model, based on monads, for performing input/output in a non-strict, purely functional language'. [...]
[Such] input/output-schemes merely provide frameworks in which side-effecting operations can safely be used with a guaranteed order of execution and without affecting the properties of the purely functional parts of the language.
Claus Reinke (pages 96-97 of 210).
(emphasis by me.)
[...] When we write effectful code – monads or no monads – we have to constantly keep in mind the context of expressions we pass around.
The fact that monadic code ‘desugars’ (is implementable in terms of) side-effect-free code is irrelevant. When we use monadic notation, we program within that notation – without considering what this notation desugars into. Thinking of the desugared code breaks the monadic abstraction. A side-effect-free, applicative code is normally compiled to (that is, desugars into) C or machine code. If the desugaring argument has any force, it may be applied just as well to the applicative code, leading to the conclusion that it all boils down to the machine code and hence all programming is imperative.
[...] From the personal experience, I have noticed that the mistakes I make when writing monadic code are exactly the mistakes I made when programming in C. Actually, monadic mistakes tend to be worse, because monadic notation (compared to that of a typical imperative language) is ungainly and obscuring.
Oleg Kiselyov (page 21 of 26).
The most difficult construct for students to understand is the monad. I introduce IO without mentioning monads.
Olaf Chitil.
More generally:
Still, today, over 25 years after the introduction of the concept of monads to the world of functional programming, beginning functional programmers struggle to grasp the concept of monads. This struggle is exemplified by the numerous blog posts about the effort of trying to learn about monads. From our own experience we notice that even at university level, bachelor level students often struggle to comprehend monads and consistently score poorly on monad-related exam questions.
Considering that the concept of monads is not likely to disappear from the functional programming landscape any time soon, it is vital that we, as the functional programming community, somehow overcome the problems novices encounter when first studying monads.
Tim Steenvoorden, Jurriën Stutterheim, Erik Barendsen and Rinus Plasmeijer.
If only there was another way to specify "a guaranteed order of execution" in Haskell, while keeping the ability to separate regular Haskell definitions from those involved in I/O (and its observable effects) - translating this variation of Philip Wadler's echo:
val echoML : unit -> unit
fun echoML () = let val c = getcML () in
if c = #"\n" then
()
else
let val _ = putcML c in
echoML ()
end
fun putcML c = TextIO.output1(TextIO.stdOut,c);
fun getcML () = valOf(TextIO.input1(TextIO.stdIn));
...could then be as simple as:
echo :: OI -> ()
echo u = let !(u1:u2:u3:_) = partsOI u in
let !c = getChar u1 in
if c == '\n' then
()
else
let !_ = putChar c u2 in
echo u3
where:
data OI -- abstract
foreign import ccall "primPartOI" partOI :: OI -> (OI, OI)
⋮
foreign import ccall "primGetCharOI" getChar :: OI -> Char
foreign import ccall "primPutCharOI" putChar :: Char -> OI -> ()
⋮
and:
partsOI :: OI -> [OI]
partsOI u = let !(u1, u2) = partOI u in u1 : partsOI u2
How would this work? At run-time, Main.main receives an initial OI pseudo-data value as an argument:
module Main(main) where
main :: OI -> ()
⋮
...from which other OI values are produced, using partOI or partsOI. All you have to do is ensure each new OI value is used at most once, in each call to an OI-based definition, foreign or otherwise. In return, you get back a plain ordinary result - it isn't e.g. paired with some odd abstract state, or requires the use of a callback continuation, etc.
Using OI, instead of the unit type () like Standard ML does, means we can avoid always having to use the monadic interface:
Once you're in the IO monad, you're stuck there forever, and are reduced to Algol-style imperative programming.
Robert Harper.
But if you really do need it:
type IO a = OI -> a
unitIO :: a -> IO a
unitIO x = \ u -> let !_ = partOI u in x
bindIO :: IO a -> (a -> IO b) -> IO b
bindIO m k = \ u -> let !(u1, u2) = partOI u in
let !x = m u1 in
let !y = k x u2 in
y
⋮
So, monadic types aren't always needed - there are other interfaces out there:
LML had a fully fledged implementation of oracles running of a multi-processor (a Sequent Symmetry) back in ca 1989. The description in the Fudgets thesis refers to this implementation. It was fairly pleasant to work with and quite practical.
[...]
These days everything is done with monads so other solutions are sometimes forgotten.
Lennart Augustsson (2006).
Wait a moment: since it so closely resembles Standard ML's direct use of effects, is this approach and its use of pseudo-data referentially transparent?
Absolutely - just find a suitable definition of "referential transparency"; there's plenty to choose from...
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"
Okay, so I am not a Haskell programmer, but I am absolutely intrigued by a lot of the ideas behind Haskell and am looking into learning it. But I'm stuck at square one: I can't seem to wrap my head around Monads, which seem to be fairly fundamental. I know there are a million questions on SO asking to explain Monads, so I'm going to be a little more specific about what's bugging me:
I read this excellent article (an introduction in Javascript), and thought that I understood Monads completely. Then I read the Wikipedia entry on Monads, and saw this:
A binding operation of polymorphic type (M t)→(t→M u)→(M u), which Haskell represents by the infix operator >>=. Its first argument is a value in a monadic type, its second argument is a function that maps from the underlying type of the first argument to another monadic type, and its result is in that other monadic type.
Okay, in the article that I cited, bind was a function which took only one argument. Wikipedia says two. What I thought I understood about Monads was the following:
A Monad's purpose is to take a function with different input and output types and to make it composable. It does this by wrapping the input and output types with a single monadic type.
A Monad consists of two interrelated functions: bind and unit. Bind takes a non-composable function f and returns a new function g that accepts the monadic type as input and returns the monadic type. g is composable. The unit function takes an argument of the type that f expected, and wraps it in the monadic type. This can then be passed to g, or to any composition of functions like g.
But there must be something wrong, because my concept of bind takes one argument: a function. But (according to Wikipedia) Haskell's bind actually takes two arguments! Where is my mistake?
You are not making a mistake. The key idea to understand here is currying - that a Haskell function of two arguments can be seen in two ways. The first is as simply a function of two arguments. If you have, for example, (+), this is usually seen as taking two arguments and adding them. The other way to see it is as a addition machine producer. (+) is a function that takes a number, say x, and makes a function that will add x.
(+) x = \y -> x + y
(+) x y = (\y -> x + y) y = x + y
When dealing with monads, sometimes it is probably better, as ephemient mentioned above, to think of =<<, the flipped version of >>=. There are two ways to look at this:
(=<<) :: (a -> m b) -> m a -> m b
which is a function of two arguments, and
(=<<) :: (a -> m b) -> (m a -> m b)
which transforms the input function to an easily composed version as the article mentioned. These are equivalent just like (+) as I explained before.
Allow me to tear down your beliefs about Monads. I sincerely hope you realize that I am not trying to be rude; I'm simply trying to avoid mincing words.
A Monad's purpose is to take a function with different input and output types and to make it composable. It does this by wrapping the input and output types with a single monadic type.
Not exactly. When you start a sentence with "A Monad's purpose", you're already on the wrong foot. Monads don't necessarily have a "purpose". Monad is simply an abstraction, a classification which applies to certain types and not to others. The purpose of the Monad abstraction is simply that, abstraction.
A Monad consists of two interrelated functions: bind and unit.
Yes and no. The combination of bind and unit are sufficient to define a Monad, but the combination of join, fmap, and unit is equally sufficient. The latter is, in fact, the way that Monads are typically described in Category Theory.
Bind takes a non-composable function f and returns a new function g that accepts the monadic type as input and returns the monadic type.
Again, not exactly. A monadic function f :: a -> m b is perfectly composable, with certain types. I can post-compose it with a function g :: m b -> c to get g . f :: a -> c, or I can pre-compose it with a function h :: c -> a to get f . h :: c -> m b.
But you got the second part absolutely right: (>>= f) :: m a -> m b. As others have noted, Haskell's bind function takes the arguments in the opposite order.
g is composable.
Well, yes. If g :: m a -> m b, then you can pre-compose it with a function f :: c -> m a to get g . f :: c -> m b, or you can post-compose it with a function h :: m b -> c to get h . g :: m a -> c. Note that c could be of the form m v where m is a Monad. I suppose when you say "composable" you mean to say "you can compose arbitrarily long chains of functions of this form", which is sort of true.
The unit function takes an argument of the type that f expected, and wraps it in the monadic type.
A roundabout way of saying it, but yes, that's about right.
This [the result of applying unit to some value] can then be passed to g, or to any composition of functions like g.
Again, yes. Although it is generally not idiomatic Haskell to call unit (or in Haskell, return) and then pass that to (>>= f).
-- instead of
return x >>= f >>= g
-- simply go with
f x >>= g
-- instead of
\x -> return x >>= f >>= g
-- simply go with
f >=> g
-- or
g <=< f
The article you link is based on sigfpe's article, which uses a flipped definition of bind:
The first thing is that I've flipped the definition of bind and written it as the word 'bind' whereas it's normally written as the operator >>=. So bind f x is normally written as x >>= f.
So, the Haskell bind takes a value enclosed in a monad, and returns a function, which takes a function and then calls it with the extracted value. I might be using non-precise terminology, so maybe better with code.
You have:
sine x = (sin x, "sine was called.")
cube x = (x * x * x, "cube was called.")
Now, translating your JS bind (Haskell does automatic currying, so calling bind f returns a function that takes a tuple, and then pattern matching takes care of unpacking it into x and s, I hope that's understandable):
bind f (x, s) = (y, s ++ t)
where (y, t) = f x
You can see it working:
*Main> :t sine
sine :: Floating t => t -> (t, [Char])
*Main> :t bind sine
bind sine :: Floating t1 => (t1, [Char]) -> (t1, [Char])
*Main> (bind sine . bind cube) (3, "")
(0.956375928404503,"cube was called.sine was called.")
Now, let's reverse arguments of bind:
bind' (x, s) f = (y, s ++ t)
where (y, t) = f x
You can clearly see it's still doing the same thing, but with a bit different syntax:
*Main> bind' (bind' (3, "") cube) sine
(0.956375928404503,"cube was called.sine was called.")
Now, Haskell has a syntax trick that allows you to use any function as an infix operator. So you can write:
*Main> (3, "") `bind'` cube `bind'` sine
(0.956375928404503,"cube was called.sine was called.")
Now rename bind' to >>= ((3, "") >>= cube >>= sine) and you've got what you were looking for. As you can see, with this definition, you can effectively get rid of the separate composition operator.
Translating the new thing back into JavaScript would yield something like this (notice that again, I only reverse the argument order):
var bind = function(tuple) {
return function(f) {
var x = tuple[0],
s = tuple[1],
fx = f(x),
y = fx[0],
t = fx[1];
return [y, s + t];
};
};
// ugly, but it's JS, after all
var f = function(x) { return bind(bind(x)(cube))(sine); }
f([3, ""]); // [0.956375928404503, "cube was called.sine was called."]
Hope this helps, and not introduces more confusion — the point is that those two bind definitions are equivalent, only differing in call syntax.