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Tutorial: `forall` in Haskell

Β· 9 min read
Martin Sosic

Find out what Haskell's forall is all about.

You might have seen forall being used in Haskell like this:

f :: forall a. [a] -> [a]
f xs = ys ++ ys
where ys :: [a]
ys = reverse xs


liftPair :: (forall x. x -> f x) -> (a, b) -> (f a, f b)


data Showable = forall s. (Show s) => Showable s

forall is something called "type quantifier", and it gives extra meaning to polymorphic type signatures (e.g. :: a, :: a -> b, :: a -> Int, ...).

While normaly forall plays a role of the "universal quantifier", it can also play a role of the "existential quantifier" (depends on the situation).

What does all this mean and how can forall be used in Haskell? Read on to find out!

NOTE: we assume you are comfortable with basic polymorphism in Haskell.

Quick math/logic reminder​

In mathematical logic, we have

  • universal quantifier
    • symbol: βˆ€x
    • interpretation: "for all", "given any"
    • example: βˆ€x P(x) means "for all x predicate P(x) is true".
  • existential quantifier
    • symbol: βˆƒx
    • interpretation: "there exists", "there is at least one", "for some"
    • example: βˆƒx P(x) means "there is some x for which predicate P(x) is true".

Vanilla Haskell (no extensions)​

In Haskell, all polymorphic type signatures are considered to be implicitly prefixed with forall.

Therefore, if you have

f :: a -> a
g :: a -> (a -> b) -> b

it is really the same as

f :: forall a. a -> a
g :: forall a b. a -> (a -> b) -> b

What forall here does is play the role of universal quantifier. For function f, it means it is saying "for all types, this function takes that type and returns the same type.". Other way to put it would be "this funtion can be called with value of any type as its first argument, and it will return the value of that same type".

Since forall is already implicit, writing it explicitly doesn't really do anything!

Not only that, but without any extensions, you can't even write forall explicitly, you will get a syntax error, since forall is not a keyword in Haskell.

So what is the purpose of forall then? Well, obviously to be used with extensions :)!

The simplest extension is ExplicitForAll, which allows you to explicitly write forall (as we did above). This is not useful on its own though, since as we said above, explicitly writing forall doesn't change anything, it was already implicitly there.

However, there are other extensions that make use of forall keyword, like: ScopedTypeVariables, RankNTypes, ExistentialQuantification. All these extensions automatically enable ExplicitForAll extension, which means you don't need to enable it yourself when using any of these. There is also TypeApplications extension which interacts with forall and in that case you might want to use ExplicitForAll with it.

Since forall is useful only when used with extensions, let's take a look at how it is used in each one of those!

forall and extension ScopedTypeVariables​

ScopedTypeVariables enables lexical scoping of type variables by explicitly introducing them with forall.

Let's take a look at the following example:

f :: [a] -> [a]
f xs = ys ++ ys
where ys :: [a]
ys = reverse xs

This code does not compile, because compiler can't match type of ys with the return type of f. Why though, when they are both [a]? Well, that is because that is not the same a! Try changing ys :: [a] to ys :: [b] and you will get the exact same error, because it is exactly the same code -> a in ys :: [a] and a in f :: [a] -> [a] are different as and there is no connection between them. a in ys :: [a] stands for "any type", not for "that type that is reffered to with a in the type signature above".

This is where ScopedTypeVariables comes in:

{-# LANGUAGE ScopedTypeVariables #-}

f :: forall a. [a] -> [a]
f xs = ys ++ ys
where ys :: [a]
ys = reverse xs

forall now gets special powers: the type variables bound by a forall (in our case a) scope over the entire definition of the accompanying value declaration (in this case definition of f).

This means that any mention of type a in the definition of f now refers to that a from the type signature of f, which is exactly what we needed, and the code example above now compiles.

forall and extension RankNTypes​

Basically, what RankNTypes does is enable you to use forall nested in type signatures, so that it does not apply to the whole type signature but just the part of it.

This enables some cool things that you were not able to do to before, for example you can specify that your function takes a polymorphic function as an argument.

Take a look at this example:

foo :: (forall a. a -> a) -> (Char, Bool)    -- We can do this only with RankNTypes.

bar :: forall a. ((a -> a) -> (Char, Bool)) -- This is usual stuff, we don't need RankNTypes for it. Actually we can even drop `forall` since it is implicit.

In foo, forall is applied only to the first argument of foo, which is a -> a, and not to the rest of the f's type signature. This can be done only with RankNTypes extension. bar on the other hand has forall applied to the whole signature, and we could have even ommited this forall since it would be there implicitly anyway.

Now, what does this mean? If we now have specificFunc :: Int -> Int and polymorphicFunc :: a -> a, foo polymorphicFunc will compile, while foo specificFunc will not! On the other hand both bar specificFunc and bar polymorphicFunc will compile.

This is because we specified, with forall, that foo needs a polymorphic function (function that takes value of any type and returns value of that same type) as the first argument, so we can't pass it a function like specificFunc that works only for Int -> such function is too specific. On the other hand, bar needs a function that takes value of some type and returns the value of that same type, so specificFunc is completely fine since it works only with Int, while polymorphicFunc is also ok although it is more general than what is needed, since compiler can easily specialize it.

Another example is liftPair function:

liftPair :: (forall x. x -> f x) -> (a, b) -> (f a, f b)
liftPair func (y, z) = (func y, func z)
>> liftPair (:[]) (1, "a")
([1], ["a"])

liftPair takes polymorphic function and applies it to both values in the pair. There would be no way to write its type signature without using RankNTypes.

forall and extension ExistentialQuantification​

ExistentialQuantification enables us to use forall in the type signature of data constructors.

This is useful because it enables us to define heterogeneous data types, which then allows us to store different types in a single data collection (which normally you can't do in Haskell, e.g. you can't have different types in a list).

For example, if we have

data Showable = forall s. (Show s) => Showable s

now we can do

someShowables :: [Showable]
someShowables = [Showable "Hi", Showable 5, Showable (1, 2)]

printShowables :: [Showable] -> IO ()
printShowables ss = mapM_ (\(Showable s) -> print s) ss

main :: IO ()
main = printShowables someShowables

In this example this allowed us to create a heterogeneous list, but only thing we can do with the contents of it is show them.

What is interesting is that in this case, forall plays the role of an existential quantifier (therefore the name of extension, ExistentialQuantification), unlike the role of universal quantifier it normally plays.


Alternative approach to ExistentialQuantification is to use the GADTs extension, like this:

data Showable where
Showable :: (Show s) => s -> Showable

In this case forall is not needed, as it is implicit.

forall and extension TypeApplications​

TypeApplications does not change how forall works like the extensions above do, but it does have an interesting interaction with forall, so we will mention it here.

TypeApplications allows you to specify values of types variables in a type.

For example, you can do show (read @Int "5") to specify that "5" should be interpreted as an Int. read has type signature :: Read a => String -> a, so what @Int does is say that that a in the type signature is Int. Therefore, read @Int :: String -> Int.

How does forall come into play here?

Well, if an identifier’s type signature does not include an explicit forall, the type variable arguments appear in the left-to-right order in which the variables appear in the type. So, foo :: Monad m => a b -> m (a c) will have its type variables ordered as m, a, b, c, and type applications will happen in that order: if we have foo @Maybe @Either, @Maybe will apply to m while @Either will apply to a. However, if you want to force a different order, for example a, b, c, m, so that @Maybe in foo @Maybe @Either applies to a, you can refactor the signature as foo :: forall a b c m. Monad m => a b -> m (a c), and now order of type variables in forall will be used when doing type applications!

This will require you to enable ExplicitForAll extension, if it is not already enabled.


This document should give a fair idea of how forall is used and what can be done with it, but it doesn't go into much depth or cover all of the ways forall is used in Haskell.

For more in-detail explanations and further investigation, here is a couple of useful resources:

This blog post originated from the notes I wrote in wasp-lang/haskell-handbook.


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