Can we create a function with void (i.e with no return value) in functional languages? Like Haskell or Scheme
In Haskell, you can write a whole family of functions that return () (unit), which is equivalent to void in C languages:
foo :: a -> ()
foo _ = ()
bar :: a -> b -> ()
bar _ _ = ()
You'll notice, however, that my implementations ignore their input, and simply return (), so they don't do anything.
You can call them like this:
Prelude> foo 42
()
Prelude> bar 42 "foo"
()
but this still doesn't accomplish anything.
You can, on the other hand, write functions that return IO (), like this:
main :: IO ()
main = putStrLn "Hello, world!"
but this is now impure. While this produces a side-effect, you could argue whether or not it's functional. At the very least, in Haskell, you can't call impure code from pure code (this is by design), so it doesn't compose.
What would the purpose be of such a function? Also if you think of a mathematical function (the base of functional programming) what would be the codomain of such a function?
So the short answer is no in haskell (i don't know enough scheme to give an informed answer regarding that).
The closest thing to void in java in haskell would be IO ()
In Racket you can:
Welcome to DrRacket, version 6.11 [3m].
Language: racket, with debugging; memory limit: 128 MB.
> (define (f) (displayln "hello") (void))
> (f)
hello
> (void? (f))
hello
#t
First, you need to remember that programming languages tend to use the word "function" in a sense that is different from the mathematical meaning. A programming-language function is just a named subroutine that may or may not produce a value that can be assigned or passed around.
Pascal made somewhat of a distinction between a "real" function and one that returned no value; the keyword procedure creates a subroutine that returns nothing, the keyword function has to return a value of some type.
C is an example that blurs this distinction; a function can have a return value of type void which in reality is type with exactly one value (although that value isn't actually represented in code; you have to pretend it exists). Every function with this return type always returns that same value. Python makes it a little more explicit; a function with no return statement or a return statement with no value actually returns the singleton value None.
Haskell has a similar type, (), which is inhabited by a single value with the same name. You can of course define a family of functions, one function per type, foo :: a -> () which ignores its argument and returns (). It doesn't have any practical value (as a pure language, a function can't do anything except return a value of its declared return type), but it does indeed exist. foo _ = ().
Incidentally, the functions of type a -> () establish () as the terminal object in the (pseudo)category Hask, which is necessary for establishing Hask as a cartesian-closed category, making it suitable for defining the semantics of Haskell.
However, Haskell also does have a type that contains no values, appropriately called Void:
data Void
Since Void is a valid type, you can imagine a type that contains functions from Void to any other type:
absurd :: Void -> a
However, since there are no values of type Void, you can't really call such a function. The unique function of type Void -> a can be defined as
absurd x = case x of {} -- There's nothing x *can* match
This is not to say that absurd has no use at all, just no practical use. Just as functions of type a -> () define () as the terminal object in Hask, absurd defines Void as the initial object in Hask.
(The latter half of this answer is a (bad) synopsis of information found in Bartosz Milewski's fantastic series of posts Category Theory for Programmers.)
In Haskell the only thing a function does is to return a value.
If a function doesn't return anything, then what the heck does it actually do?
You're probably thinking about something like a print "function", which prints something out and returns nothing. But that's not a mathematical function, that's an action. Haskell models those in a completely different way (i.e., monads). If you want to know what that is, there's elevanty billion discussions about it here on Stack Overflow, and littered across the face of the Internet.
Related
This question already has answers here:
What's so special about 'return' keyword
(3 answers)
Closed 5 years ago.
Consider these functions
f1 :: Maybe Int
f1 = return 1
f2 :: [Int]
f2 = return 1
Both have the same statement return 1. But the results are different. f1 gives value Just 1 and f2 gives value [1]
Looks like Haskell invokes two different versions of return based on return type. I like to know more about this kind of function invocation. Is there a name for this feature in programming languages?
This is a long meandering answer!
As you've probably seen from the comments and Thomas's excellent (but very technical) answer You've asked a very hard question. Well done!
Rather than try to explain the technical answer I've tried to give you a broad overview of what Haskell does behind the scenes without diving into technical detail. Hopefully it will help you to get a big picture view of what's going on.
return is an example of type inference.
Most modern languages have some notion of polymorphism. For example var x = 1 + 1 will set x equal to 2. In a statically typed language 2 will usually be an int. If you say var y = 1.0 + 1.0 then y will be a float. The operator + (which is just a function with a special syntax)
Most imperative languages, especially object oriented languages, can only do type inference one way. Every variable has a fixed type. When you call a function it looks at the types of the argument and chooses a version of that function that fits the types (or complains if it can't find one).
When you assign the result of a function to a variable the variable already has a type and if it doesn't agree with the type of the return value you get an error.
So in an imperative language the "flow" of type deduction follows time in your program Deduce the type of a variable, do something with it and deduce the type of the result. In a dynamically typed language (such as Python or javascript) the type of a variable is not assigned until the value of the variable is computed (which is why there don't seem to be types). In a statically typed language the types are worked out ahead of time (by the compiler) but the logic is the same. The compiler works out what the types of variables are going to be, but it does so by following the logic of the program in the same way as the program runs.
In Haskell the type inference also follows the logic of the program. Being Haskell it does so in a very mathematically pure way (called System F). The language of types (that is the rules by which types are deduced) are similar to Haskell itself.
Now remember Haskell is a lazy language. It doesn't work out the value of anything until it needs it. That's why it makes sense in Haskell to have infinite data structures. It never occurs to Haskell that a data structure is infinite because it doesn't bother to work it out until it needs to.
Now all that lazy magic happens at the type level too. In the same way that Haskell doesn't work out what the value of an expression is until it really needs to, Haskell doesn't work out what the type of an expression is until it really needs to.
Consider this function
func (x : y : rest) = (x,y) : func rest
func _ = []
If you ask Haskell for the type of this function it has a look at the definition, sees [] and : and deduces that it's working with lists. But it never needs to look at the types of x and y, it just knows that they have to be the same because they end up in the same list. So it deduces the type of the function as [a] -> [a] where a is a type that it hasn't bothered to work out yet.
So far no magic. But it's useful to notice the difference between this idea and how it would be done in an OO language. Haskell doesn't convert the arguments to Object, do it's thing and then convert back. Haskell just hasn't been asked explicitly what the type of the list is. So it doesn't care.
Now try typing the following into ghci
maxBound - length ""
maxBound : "Hello"
Now what just happened !? minBound bust be a Char because I put it on the front of a string and it must be an integer because I added it to 0 and got a number. Plus the two values are clearly very different.
So what is the type of minBound? Let's ask ghci!
:type minBound
minBound :: Bounded a => a
AAargh! what does that mean? Basically it means that it hasn't bothered to work out exactly what a is, but is has to be Bounded if you type :info Bounded you get three useful lines
class Bounded a where
minBound :: a
maxBound :: a
and a lot of less useful lines
So if a is Bounded there are values minBound and maxBound of type a.
In fact under the hood Bounded is just a value, it's "type" is a record with fields minBound and maxBound. Because it's a value Haskell doesn't look at it until it really needs to.
So I appear to have meandered somewhere in the region of the answer to your question. Before we move onto return (which you may have noticed from the comments is a wonderfully complex beast.) let's look at read.
ghci again
read "42" + 7
read "'H'" : "ello"
length (read "[1,2,3]")
and hopefully you won't be too surprised to find that there are definitions
read :: Read a => String -> a
class Read where
read :: String -> a
so Read a is just a record containing a single value which is a function String -> a. Its very tempting to assume that there is one read function which looks at a string, works out what type is contained in the string and returns that type. But it does the opposite. It completely ignores the string until it's needed. When the value is needed, Haskell first works out what type it's expecting, once it's done that it goes and gets the appropriate version of the read function and combines it with the string.
now consider something slightly more complex
readList :: Read a => [String] -> a
readList strs = map read strs
under the hood readList actually takes two arguments
readList' (Read a) -> [String] -> [a]
readList' {read = f} strs = map f strs
Again as Haskell is lazy it only bothers looking at the arguments when it's needs to find out the return value, at that point it knows what a is, so the compiler can go and fine the right version of Read. Until then it doesn't care.
Hopefully that's given you a bit of an idea of what's happening and why Haskell can "overload" on the return type. But it's important to remember it's not overloading in the conventional sense. Every function has only one definition. It's just that one of the arguments is a bag of functions. read_str doesn't ever know what types it's dealing with. It just knows it gets a function String -> a and some Strings, to do the application it just passes the arguments to map. map in turn doesn't even know it gets strings. When you get deeper into Haskell it becomes very important that functions don't know very much about the types they're dealing with.
Now let's look at return.
Remember how I said that the type system in Haskell was very similar to Haskell itself. Remember that in Haskell functions are just ordinary values.
Does this mean I can have a type that takes a type as an argument and returns another type? Of course it does!
You've seen some type functions Maybe takes a type a and returns another type which can either be Just a or Nothing. [] takes a type a and returns a list of as. Type functions in Haskell are usually containers. For example I could define a type function BinaryTree which stores a load of a's in a tree like structure. There are of course lots of much stranger ones.
So, if these type functions are similar to ordinary types I can have a typeclass that contains type functions. One such typeclass is Monad
class Monad m where
return a -> m a
(>>=) m a (a -> m b) -> m b
so here m is some type function. If I want to define Monad for m I need to define return and the scary looking operator below it (which is called bind)
As others have pointed out the return is a really misleading name for a fairly boring function. The team that designed Haskell have since realised their mistake and they're genuinely sorry about it. return is just an ordinary function that takes an argument and returns a Monad with that type in it. (You never asked what a Monad actually is so I'm not going to tell you)
Let's define Monad for m = Maybe!
First I need to define return. What should return x be? Remember I'm only allowed to define the function once, so I can't look at x because I don't know what type it is. I could always return Nothing, but that seems a waste of a perfectly good function. Let's define return x = Just x because that's literally the only other thing I can do.
What about the scary bind thing? what can we say about x >>= f? well x is a Maybe a of some unknown type a and f is a function that takes an a and returns a Maybe b. Somehow I need to combine these to get a Maybe b`
So I need to define Nothing >== f. I can't call f because it needs an argument of type a and I don't have a value of type a I don't even know what 'a' is. I've only got one choice which is to define
Nothing >== f = Nothing
What about Just x >>= f? Well I know x is of type a and f takes a as an argument, so I can set y = f a and deduce that y is of type b. Now I need to make a Maybe b and I've got a b so ...
Just x >>= f = Just (f x)
So I've got a Monad! what if m is List? well I can follow a similar sort of logic and define
return x = [x]
[] >>= f = []
(x : xs) >>= a = f x ++ (xs >>= f)
Hooray another Monad! It's a nice exercise to go through the steps and convince yourself that there's no other sensible way of defining this.
So what happens when I call return 1?
Nothing!
Haskell's Lazy remember. The thunk return 1 (technical term) just sits there until someone needs the value. As soon as Haskell needs the value it know what type the value should be. In particular it can deduce that m is List. Now that it knows that Haskell can find the instance of Monad for List. As soon as it does that it has access to the correct version of return.
So finally Haskell is ready To call return, which in this case returns [1]!
The return function is from the Monad class:
class Applicative m => Monad (m :: * -> *) where
...
return :: a -> m a
So return takes any value of type a and results in a value of type m a. The monad, m, as you've observed is polymorphic using the Haskell type class Monad for ad hoc polymorphism.
At this point you probably realize return is not an good, intuitive, name. It's not even a built in function or a statement like in many other languages. In fact a better-named and identically-operating function exists - pure. In almost all cases return = pure.
That is, the function return is the same as the function pure (from the Applicative class) - I often think to myself "this monadic value is purely the underlying a" and I try to use pure instead of return if there isn't already a convention in the codebase.
You can use return (or pure) for any type that is a class of Monad. This includes the Maybe monad to get a value of type Maybe a:
instance Monad Maybe where
...
return = pure -- which is from Applicative
...
instance Applicative Maybe where
pure = Just
Or for the list monad to get a value of [a]:
instance Applicative [] where
{-# INLINE pure #-}
pure x = [x]
Or, as a more complex example, Aeson's parse monad to get a value of type Parser a:
instance Applicative Parser where
pure a = Parser $ \_path _kf ks -> ks a
Is print in Haskell a pure function; why or why not? I'm thinking it's not, because it does not always return the same value as pure functions should.
A value of type IO Int is not really an Int. It's more like a piece of paper which reads "hey Haskell runtime, please produce an Int value in such and such way". The piece of paper is inert and remains the same, even if the Ints eventually produced by the runtime are different.
You send the piece of paper to the runtime by assigning it to main. If the IO action never comes in the way of main and instead languishes inside some container, it will never get executed.
Functions that return IO actions are pure like the others. They always return the same piece of paper. What the runtime does with those instructions is another matter.
If they weren't pure, we would have to think twice before changing
foo :: (Int -> IO Int) -> IO Int
foo f = liftA2 (+) (f 0) (f 0)
to:
foo :: (Int -> IO Int) -> IO Int
foo f = let x = f 0 in liftA2 (+) x x
Yes, print is a pure function. The value it returns has type IO (), which you can think of as a bunch of code that outputs the string you passed in. For each string you pass in, it always returns the same code.
If you just read the Tag of pure-function (A function that always evaluates to the same result value given the same argument value(s) and that does not cause any semantically observable side effect or output, such as mutation of mutable objects or output to I/O devices.) and then Think in the type of print:
putStrLn :: String -> IO ()
You will find a trick there, it always returns IO (), so... No, it produces effects. So in terms of Referential Transparency is not pure
For example, getLine returns IO String but it is also a pure function. (#interjay contribution), What I'm trying to say, is that the answer depends very close of the question:
On matter of value, IO () will always be the same IO () value for the same input.
And
On matter of execution, it is not pure because the execution of that
IO () could have side effects (put an string in the screen, in this
case looks so innocent, but some IO could lunch nuclear bombs, and
then return the Int 42)
You could understand better with the nice approach of #Ben here:
"There are several ways to explain how you're "purely" manipulating
the real world. One is to say that IO is just like a state monad, only
the state being threaded through is the entire world outside your
program;= (so your Stuff -> IO DBThing function really has an extra
hidden argument that receives the world, and actually returns a
DBThing along with another world; it's always called with different
worlds, and that's why it can return different DBThing values even
when called with the same Stuff). Another explanation is that an IO
DBThing value is itself an imperative program; your Haskell program is
a totally pure function doing no IO, which returns an impure program
that does IO, and the Haskell runtime system (impurely) executes the
program it returns."
And #Erik Allik:
So Haskell functions that return values of type IO a, are actually not
the functions that are being executed at runtime — what gets executed
is the IO a value itself. So these functions actually are pure but
their return values represent non-pure computations.
You can found them here Understanding pure functions in Haskell with IO
How to write the type declaration of an haskell function without arguments?
There is no such thing as a function without arguments, that would be just a value. Sure, you can write such a declaration:
five :: Int
five = 5
It might look more like what you asked for if I make it
five' :: () -> Int
five' () = 5
but that's completely equivalent (unless you write something ridiculous like five' undefined) and superfluent1.
If what you mean is something like, in C
void scream() {
printf("Aaaah!\n");
}
then that's again not a function but an action. (C programmers do call it function, but you might better say procedure, everybody would understand.) What I said above holds pretty much the same way, you'd use
scream :: IO()
scream = putStrLn "Aaaah!"
Note that the empty () do in this case not have anything to do with not having arguments (that follows already from the absence of -> arrows), instead it means there is also no return value, it's just a "side-effect-only" action.
1Actually, it differs in one relevant way: five is a constant applicative form, which sort of means it's memoised. If I had defined such a constant in some roundabout way (e.g. sum $ 5 : replicate 1000000 0) then the lengthy calculation would be carried out only once, even if five is evaluated multiple times during a program run. OTOH, wherever you would have written out five' (), the calculation would have been done anew.
Since functions in Haskell are pure (their result only depends on their arguments), the equivalent of a function with no arguments is just a value. For example, one = 1.
I read in learn you haskell that
Enum members are sequentially ordered types ... Types in this class:
(), Bool, Char ...
Also it appears in some signatures:
putChar :: Char -> IO ()
It is very difficult to find info about it in Google as the answers refer to problems of the "common parentheses" (use in function calls, precedence problems and the like).
Therefore, what does the expression () means? Is it a type? What are variables of type ()? What is used for and when is it needed?
It is both a type and a value. () is a special type "pronounced" unit, and it has one value: (), also pronounced unit. It is essentially the same as the type void in Java or C/C++. If you're familiar with Python, think of it as the NoneType which has the singleton None.
It is useful when you want to denote an action that doesn't return anything. It is most commonly used in the context of Monads, such as the IO monad. For example, if you had the following function:
getVal :: IO Int
getVal = do
putStrLn "Enter an integer value:"
n <- getLine
return $ read n
And for some reason you decided that you just wanted to annoy the user and throw away the number they just passed in:
getValAnnoy :: IO ()
getValAnnoy = do
_ <- getVal
return () -- Returns nothing
However, return is just a Monad function, so we could abstract this a bit further
throwAwayResult :: Monad m => m a -> m ()
throwAwayResult action = do
_ <- action
return ()
Then
getValAnnoy = throwAwayResult getVal
However, you don't have to write this function yourself, it already exists in Control.Monad as the function void that is even less constraining and works on Functors:
void :: Functor f => f a -> f ()
void fa = fmap (const ()) fa
Why does it work on Functor instead of Monad? Well, for each Monad instance, you can write the Functor instance as
instance Monad m => Functor m where
fmap f m = m >>= return . f
But you can't make a Monad out of every Functor. It's like how a square is a rectangle but a rectangle isn't always a square, Monads form a subset of Functors.
As others have said, it's the unit type which has one value called unit. In Haskell syntax this is easily, if confusingly, expressed as () :: (). We can make our own quite easily as well.
data Unit = Unit
>>> :t Unit :: Unit
Unit :: Unit
>>> :t () :: ()
() :: ()
It's written as () because it behaves very much like an "empty tuple" would. There are theoretical reasons why this holds, but honestly it makes a lot of simple intuitive sense too.
It's often used as the argument to a type constructor like IO or ST when its the context of the value that's interesting, not the value itself. This is intuitively true because if I tell you have I have a value of type () then you don't need to know anything more---there's only one of them!
putStrLn :: String -> IO () -- the return type is unimportant,
-- we just want the *effect*
map (const ()) :: [a] -> [()] -- this destroys all information about a list
-- keeping only the *length*
>>> [ (), (), () ] :: [()] -- this might as well just be the number `3`
-- there is nothing else interesting about it
forward :: [()] -> Int -- we have a pair of isomorphisms even
forward = length
backward :: Int -> [()]
backward n = replicate n ()
It is both a type and a value.
It is unit type, the type that has only one value. In Haskell its name and only value looks like empty tuple : ().
As others have said, () in Haskell is both the name of the "unit" type, and the only value of said type.
One of the confusing things in moving from imperative programming to Haskell is that the way the languages deal with the concept of "nothing" is different. What's more confusing is the vocabulary, because imperative languages and Haskell use the term "void" to mean diametrically different things.
In an imperative language, a "function" (which may not be a true mathematical function) may have "void" as its return type, as in this pseudocode example:
void sayHello() {
printLn("Hello!");
}
In this case, void means that the "function," if it returns, will not produce a result value. (The other possibility is that they function may not return—it may loop forever, or fail with an error or exception.)
In Haskell, however, all functions (and IO actions) must must produce a result. So when we write an IO action that doesn't produce any interesting return value, we make it return ():
sayHello :: IO ()
sayHello = putStrLn "Hello!"
Later actions will just ignore the () result value.
Now, you probably don't need to worry too much about this, but there is one place where this gets confusing, which is that in Haskell there is a type called Void, but it means something completely different from the imperative programming void. Because of this, the word "void" becomes a minefield when comparing Haskell and imperative languages, because the meaning is completely different when you switch paradigms.
In Haskell, Void is a type that doesn't have any values. The largest consequence of this is that in Haskell a function with return type Void can never return, it can only fail with an error or loop forever. Why? Because the function would have produce a value of type Void in order to return, but there isn't such a value.
This is however not relevant until you're working with some more advanced techniques, so you don't need to worry about it other than to beware of the word "void."
But the bigger lesson is that the imperative and the Haskell concepts of "no return value" are different. Haskell distinguishes between:
Things that may return but whose result won't have any information (the () type);
Things that cannot return, no matter what (the Void type).
The imperative void corresponds to the former, and not the latter.
I'm reading Learn You a Haskell, and in the monad chapters, it seems to me that () is being treated as a sort of "null" for every type. When I check the type of () in GHCi, I get
>> :t ()
() :: ()
which is an extremely confusing statement. It seems that () is a type all to itself. I'm confused as to how it fits into the language, and how it seems to be able to stand for any type.
tl;dr () does not add a "null" value to every type, hell no; () is a "dull" value in a type of its own: ().
Let me step back from the question a moment and address a common source of confusion. A key thing to absorb when learning Haskell is the distinction between its expression language and its type language. You're probably aware that the two are kept separate. But that allows the same symbol to be used in both, and that is what is going on here. There are simple textual cues to tell you which language you're looking at. You don't need to parse the whole language to detect these cues.
The top level of a Haskell module lives, by default, in the expression language. You define functions by writing equations between expressions. But when you see foo :: bar in the expression language, it means that foo is an expression and bar is its type. So when you read () :: (), you're seeing a statement which relates the () in the expression language with the () in the type language. The two () symbols mean different things, because they are not in the same language. This replication often causes confusion for beginners, until the expression/type language separation installs itself in their subconscious, at which point it becomes helpfully mnemonic.
The keyword data introduces a new datatype declaration, involving a careful mixture of the expression and type languages, as it says first what the new type is, and secondly what its values are.
data TyCon tyvar ... tyvar = ValCon1 type ... type | ... | ValConn type ... type
In such a declaration, type constructor TyCon is being added to the type language and the ValCon value constructors are being added to the expression language (and its pattern sublanguage). In a data declaration, the things which stand in argument places for the ValCons tell you the types given to the arguments when that ValCon is used in expressions. For example,
data Tree a = Leaf | Node (Tree a) a (Tree a)
declares a type constructor Tree for binary tree types storing a elements at nodes, whose values are given by value constructors Leaf and Node. I like to colour type constructors (Tree) blue and value constructors (Leaf, Node) red. There should be no blue in expressions and (unless you're using advanced features) no red in types. The built-in type Bool could be declared,
data Bool = True | False
adding blue Bool to the type language, and red True and False to the expression language. Sadly, my markdown-fu is inadequate to the task of adding the colours to this post, so you'll just have to learn to add the colours in your head.
The "unit" type uses () as a special symbol, but it works as if declared
data () = () -- the left () is blue; the right () is red
meaning that a notionally blue () is a type constructor in the type language, but that a notionally red () is a value constructor in the expression language, and indeed () :: (). [ It is not the only example of such a pun. The types of larger tuples follow the same pattern: pair syntax is as if given by
data (a, b) = (a, b)
adding (,) to both type and expression languages. But I digress.]
So the type (), often pronounced "Unit", is a type containing one value worth speaking of: that value is also written () but in the expression language, and is sometimes pronounced "void". A type with only one value is not very interesting. A value of type () contributes zero bits of information: you already know what it must be. So, while there is nothing special about type () to indicate side effects, it often shows up as the value component in a monadic type. Monadic operations tend to have types which look like
val-in-type-1 -> ... -> val-in-type-n -> effect-monad val-out-type
where the return type is a type application: the (type) function tells you which effects are possible and the (type) argument tells you what sort of value is produced by the operation. For example
put :: s -> State s ()
which is read (because application associates to the left ["as we all did in the sixties", Roger Hindley]) as
put :: s -> (State s) ()
has one value input type s, the effect-monad State s, and the value output type (). When you see () as a value output type, that just means "this operation is used only for its effect; the value delivered is uninteresting". Similarly
putStr :: String -> IO ()
delivers a string to stdout but does not return anything exciting.
The () type is also useful as an element type for container-like structures, where it indicates that the data consists just of a shape, with no interesting payload. For example, if Tree is declared as above, then Tree () is the type of binary tree shapes, storing nothing of interest at nodes. Similarly [()] is the type of lists of dull elements, and if there is nothing of interest in a list's elements, then the only information it contributes is its length.
To sum up, () is a type. Its one value, (), happens to have the same name, but that's ok because the type and expression languages are separate. It's useful to have a type representing "no information" because, in context (e.g., of a monad or a container), it tells you that only the context is interesting.
The () type can be thought of as a zero-element tuple. It's a type that can only have one value, and thus it's used where you need to have a type, but you don't actually need to convey any information. Here's a couple of uses for this.
Monadic things like IO and State have a return value, as well as performing side-effects. Sometimes the only point of the operation is to perform a side-effect, like writing to the screen or storing some state. For writing to the screen, putStrLn must have type String -> IO ? -- IO always has to have some return type, but here there's nothing useful to return. So what type should we return? We could say Int, and always return 0, but that's misleading. So we return (), the type that has only one value (and thus no useful information), to indicate that there's nothing useful coming back.
It's sometimes useful to have a type which can have no useful values. Consider if you'd implemented a type Map k v which maps keys of type k to values of type v. Then you want to implement a Set, which is really similar to a map except that you don't need the value part, just the keys. In a language like Java you might use booleans as the dummy value type, but really you just want a type that has no useful values. So you could say type Set k = Map k ()
It should be noted that () is not particularly magic. If you want you can store it in a variable and do a pattern match on it (although there's not much point):
main = do
x <- putStrLn "Hello"
case x of
() -> putStrLn "The only value..."
It is called the Unit type, usually used to represent side effects. You can think of it vaguely as Void in Java. Read more here and here etc. What can be confusing is that () syntactically represents both the type and its only value literal. Also note that it is not similar to null in Java which means an undefined reference - () is just effectively a 0-sized tuple.
I really like to think of () by analogy with tuples.
(Int, Char) is the type of all pairs of an Int and a Char, so it's values are all possible values of Int crossed with all possible values of Char. (Int, Char, String) is similarly the type of all triples of an Int, a Char, and a String.
It's easy to see how to keep extending this pattern upwards, but what about downwards?
(Int) would be the "1-tuple" type, consisting of all possible values of Int. But that would be parsed by Haskell as just putting parentheses around Int, and thus being just the type Int. And values in this type would be (1), (2), (3), etc, which also would just get parsed as ordinary Int values in parentheses. But if you think about it, a "1-tuple" is exactly the same as just a single value, so there's no need to actually have them exist.
Going down one step further to zero-tuples gives us (), which should be all possible combinations of values in an empty list of types. Well, there's exactly one way to do that, which is to contain no other values, so there should be only one value in the type (). And by analogy with tuple value syntax, we can write that value as (), which certainly looks like a tuple containing no values.
That's exactly how it works. There is no magic, and this type () and its value () are in no way treated specially by the language.
() is not in fact being treated as "a null value for any type" in the monads examples in the LYAH book. Whenever the type () is used the only value which can be returned is (). So it's used as a type to explicitly say that there cannot be any other return value. And likewise where another type is supposed to be returned, you cannot return ().
The thing to keep in mind is that when a bunch of monadic computations are composed together with do blocks or operators like >>=, >>, etc, they'll be building a value of type m a for some monad m. That choice of m has to stay the same throughout the component parts (there's no way to compose a Maybe Int with an IO Int in that way), but the a can and very often is different at each stage.
So when someone sticks an IO () in the middle of an IO String computation, that's not using the () as a null in the String type, it's simply using an IO () on the way to building an IO String, the same way you could use an Int on the way to building a String.
Yet another angle:
() is the name of a set which contains a single element called ().
Its indeed slightly confusing that the name of the set and the
element in it happens to be the same in this case.
Remember: in Haskell a type is a set that has its possible values as elements in it.
The confusion comes from other programming languages:
"void" means in most imperative languages that there is no structure in memory storing a value. It seems inconsistent because "boolean" has 2 values instead of 2 bits, while "void" has no bits instead of no values, but there it is about what a function returns in a practical sense. To be exact: its single value consumes no bit of storage.
Let's ignore the value bottom (written _|_) for a moment...
() is called Unit, written like a null-tuple. It has only one value. And it is not called
Void, because Void has not even any value, thus could not be returned by any function.
Observe this: Bool has 2 values (True and False), () has one value (()), and Void has no value (it doesn't exist). They are like sets with two/one/no elements. The least memory they need to store their value is 1 bit / no bit / impossible, respectively. Which means that a function that returns a () may return with a result value (the obvious one) that may be useless to you. Void on the other hand would imply that that function will never return and never give you any result, because there would not exist any result.
If you want to give "that value" a name, that a function returns which never returns (yes, this sounds like crazytalk), then call it bottom ("_|_", written like a reversed T). It could represent an exception or infinity loop or deadlock or "just wait longer". (Some functions will only then return bottom, iff one of their parameters is bottom.)
When you create the cartesian product / a tuple of these types, you will observe the same behaviour:
(Bool,Bool,Bool,(),()) has 2·2·2·1·1=6 differnt values. (Bool,Bool,Bool,(),Void) is like the set {t,f}×{t,f}×{t,f}×{u}×{} which has 2·2·2·1·0=0 elements, unless you count _|_ as a value.