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Concepts for higher-kinded types in C++20

🔗 Concepts for Higher-Kinded Types in C++20

🔗 Concepts

Later this year, the C++20 standard will be published. With C++20 come a great deal of new language features, including modules, ranges, consteval, and the spaceship operator. The feature that most captured my attention when I first heard about the new standard, however, was concepts.

A concept is a named set of requirements for template arguments. For an example of why this could be useful, consider the following:

template<typename In, typename Out>
void copy(In in, Out out) {
    std::copy(std::cbegin(in), std::cend(in), std::begin(out));

This just hides the iterators in a call to std::copy. Could be a handy little procedure, since explicitly dealing with iterators can be ugly. Works just fine—when In and Out have a beginning and an end, that is. If not, then the compiler issues quite an error. 300 lines of mostly unintellegible garbage in my test, as the compiler attempted every implicit conversion it know from int in order to get std::cbegin(int) to compile. No luck. Godbolt

What if there were a way to tell the compiler that copy works only for a select few types? We can write a rule for what types are valid input ranges as a concept:

template<typename T>
concept InputRange = requires(T a) {

A type T satisfies this concept (so InputRange<T> will be true) if and only if the two lines in the requires-expression compile. If we define a similar concept for output ranges, copy from before becomes:

template<typename In, typename Out>
requires InputRange<In> && OutputRange<Out>
void copy(In in, Out out) {
    std::copy(std::cbegin(in), std::cend(in), std::begin(out));

Or better yet:

template<InputRange In, OutputRange Out>
void copy(In in, Out out) {
    std::copy(std::cbegin(in), std::cend(in), std::begin(out));


With this form, when I attempt to pass an integer as the first argument to copy, the compiler tells me right away exactly what’s wrong:

<source>:17:6: note: constraints not satisfied
<source>: In instantiation of 'void copy(In, Out) [with In = int; Out = std::vector<int>]':
<source>:28:19:   required from here
<source>:6:9:   required for the satisfaction of 'InputRange<In>' [with In = int]

Very nice. Really, concepts are just a replacement for SFINAE and don’t add any new capability to the language, but they certainly do improve error messages.

🔗 Typeclasses

I’m a big fan of Haskell. Haskell is a declarative, strongly-typed language. Each value has exactly one type. Haskell has first-class functions and a concise way to represent their types: a function that takes type a as an argument and provides type b as a result has type a -> b. We run into an issue, however, when we want to overload functions—what type should (+) have? (+) :: Int -> Int -> Int would mean we need to define another function with a distinct name in order to add values of type Double! Either that, or give addition the type (+) :: a -> a -> a, which simply does not make sense for non-numeric types.

Haskell’s workaround is typeclasses, which had never been seen built into a language before Haskell 1.0 in 1990. Typeclasses work with Haskell’s parametric polymorphism to allow us to restrict the types that can act as a in that last signature for (+). A typeclass is defined as follows:

class Add a where
    (+) :: a -> a -> a

Now, the function (+) has type Add a => a -> a -> a (read “a -> a -> a, where a is of class Add”). Providing typeclass instances for Add Int and Add Double is done like so:

instance Add Int where
    x + y = {- something -}

instance Add Double where
    x + y = {- something -}

Now both can be added with the function (+), and we still need not define a nonsensical overload for non-numeric types. Good stuff.

One notable typeclass in Haskell’s standard library is Monoid, representing a type with an associative binary operator mappend and an identity element mempty and defined as follows:

class Monoid a where
    mempty :: a
    mappend :: a -> a -> a

Note here that mempty has an interesting type mostly alien to popular languages: Monoid a => a. It’s an overloaded value, almost like template<typename T> T mempty in C++. Whenever some type a is a Monoid, mempty can be of type a.

🔗 Putting it together

A concept is a set of requirements for template arguments, which can be types. A typeclass is a set of operations that can be performed on types. Very similar, huh? My first thought when learning of concepts was that I should try to implement Haskell typeclasses in C++. My first attempt:

template<typename T> T mempty;
template<typename T> T mappend(T, T);

template<typename T>
concept Monoid = requires(T a) {
    { mempty<T> } -> T;
    { mappend<T>(a, a) } -> T;

This seemed at first to work okay:

template<> int mempty<int> = 0;
template<> int mappend<int>(int a, int b) { return a + b; }

But I soon ran into an issue when trying to write a Monoid instance for my own linked-list type (identity is empty, and append is concatenation):

template<typename T> List<T> mempty<List<T>> = null<T>;

Wait a minute. This isn’t right. A full template specialization should have no template arguments, not one, but there is no way to declare typename T except in the template argument list. This was my first glimpse of the struggles to come, but I did not yet pick up on what would eventually be my takeaway from this venture.

If C++ had template templates, this approach would be very possible, like so:

template<>              // Full specialization
template<typename T>    // For all types T...
List<T> mempty<List<T>> = null<T>;

I do not yet know enough of C++ to understand why this is not allowed.

Nitash provided the following approach, similar but perhaps somewhat improved (actually working, at least):

template<typename T> T mappend(T, T) = delete;

template <> int mappend(int, int);
template<typename T> List<T> mappend(List<T>, List<T>);

template<typename T>
concept Monoid = requires (T a) {
    { T{} } -> T;
    { mappend(a, a) } -> T; 

Variable templates can’t be deleted, so we have to settle for asserting that brace-initialized {} will always result in mempty. Compared to defining a function template that takes no parameters and returns some value, this approach has the advantage of working (the other can’t express two Monoid instances for List<T> and std::vector<T>, for example), though not very well. The Bounded typeclass in Haskell has two values, minBound and maxBound, both of type Bounded a => a—these can’t both be expressed with {}!

Some struggling with templates later, I gave up on making a satisfactory Monoid typeclass and moved on to higher-kinded types, the indended topic of this post.

🔗 Kind

A kind is a type of types. A type that can be instantiated is a concrete type, we say, and a concrete type’s kind is represented by *. Examples include int, std::optional<float>, and std::pair<std::vector<std::string>, bool> in C++.

Some types cannot be instantiated. Abstract classes in C++, but let’s ignore those. OOP has no place in my attempt at elegant FP. I’m talking about class templates. std::vector is an example. What inner type would a plain vector hold? You can think of std::vector as a function on types, taking a concrete type T to the concrete type std::vector<T>. It has kind * -> *.

A higher-kinded type is a non-concrete type. Some examples are std::vector :: * -> * and std::pair :: * -> * -> * (and their Haskell approximations, [] and (,)). Concrete or not, these are types nevertheless, and can therefore belong to typeclasses. Not the same typeclasses, of course—all I’ve yet mentioned are classes of concrete types. The simplest interesting class I know of higher-kinded types is Functor:

class Functor (f :: * -> *) where
    fmap :: (a -> b) -> f a -> f b

A Functor f is a type of kind * -> * that allows one to map a function over the values it contains. There are some laws for the behavior of fmap, but I’ll skip over them. The List instance can be approximated in C++ using std::vector as follows:

template<typename A, typename Fn, typename B = std::invoke_result<Fn, A>>
std::vector<B> fmap(Fn f, std::vector<A> as) {
    std::vector<B> bs(as.size());
    std::transform(std::cbegin(as), std::cend(as), std::begin(bs), f);
    return bs;

Other simple Functors include std::optional (Maybe in Haskell) and std::pair<A> (if partial application of templates were allowed). Exercise: define fmap for each of these. A more surprising Functor is (->) x, the type of functions taking type x as an argument. Consider:

p :: a -> b    -- A function from x to y
q :: x -> a    -- Where out functor f is functions from x, q is of type f a

fmapped x = p (q x)
-- One way to combine p and q.
-- fmapped is a function taking an argument of type x
-- Apply q (of type x -> a) to x, yielding a value of type a
-- Apply p to this value, yielding a value of type b
-- So fmapped has type x -> b, or f b

In the case of functions, fmap is composition. Pretty cool.

🔗 Putting it together (reprise)

Okay, time to define Functors in C++. First attempt:

template<template<typename> typename F>       // F has kind * -> *
concept Functor = requires(F<?> fa, ? f) {    // Need to declare the inner type and the function's type somehow
    { fmap(f, fa) } -> F<?>;    // Also the resulting type.

Unlike Haskell, C++ does not allow us to use type variables without first declaring them, and this declaration must come in a very particular place. Second attempt:

template<template<typename> typename F, typename A, typename Fn, typename B = std::invoke_result<Fn, A>>
concept Functor = requires(F<A> fa, Fn f) {
    { fmap(f, fa) } -> F<B>;

Ooh, a concept that compiles! Does it do what I want? Unfortunately, it does not. Very close, yes, and it will likely work (at least somewhat) for what I want to do, but I am not interested in continuing with this fundamentally flawed typeclass.

What’s wrong with it? Recall that Functor is a class of higher-kinded types f :: * -> * admitting a suitable fmap :: (a -> b) -> f a -> f b. Look at the template arguments in the concept—what I’ve defined here is a class of sets of four types! Whether or not it behaves as a Functor class should behave, all meaning once attached to the word is now lost. I find this unacceptable.

Third attempt:

template<template<typename> typename F>
template<typename A, typename Fn, typename B = std::invoke_result<Fn, A>>
concept Functor = requires(F<A> fa, Fn f) {
    { fmap(f, fa) } -> F<B>;

Again, a template template. Again, not legal in C++. Fourth attempt:

template<template<typename> typename F>
concept Functor = template<typename A, typename Fn, typename B = std::invoke_result<Fn, A>>
                  requires(F<A> fa, Fn f) {
    { fmap(f, fa) } -> F<B>;

Okay, now I’m grasping at straws. I really want to find something that works, and that has yet to happen. A requires template is not part of the C++ grammar, and it looks hideous, but it would remedy my woes. I turned to Stack Overflow regarding this problem.

🔗 Takeaways

My question was graced with an excellent answer by user Barry. This answer begins:

C++ doesn’t have parametric polymorphism like this - you can’t do things like “for any type” in the way you want to, and in the way you can in Haskell. I think that’s fundamentally impossible in a world where overloading exists.

Fundamentally impossible, huh? Is C++ simply not expressive enough expressive in the wrong way to express such an idea? Always on the lookout for things I can claim make Haskell a better language than C++, I just had to try to come up with some reasoning as to why this might be the case.

Pretty clear after a little bit of thought. This concept approach to typeclasses does not bind methods (as they’re sometimes called in Haskell) to the concept at all. All these “methods” must be free functions declared before the concept is introduced; the concept is nothing more than a check for their existence. One of these functions could easily be overloaded like so:

template<template<typename> typename F, typename Fn>
int fmap<F, Fn, int, int>(Fn f, F<int> xs) { return 0; }

Uh-oh. If Functor is defined only in terms of one higher-kinded type, then the mere existence of this overload means our typeclass must fail for all such types, since fmap doesn’t return the correct type when mapping int to int.

In short, what I want to do is impossible.

There are other problems with concepts as typeclasses. Haskell provides the following instance:

instance Monoid a => Monoid (Maybe a) where    -- "When a is a Monoid, so is Maybe a"
    mempty = Nothing

    mappend Nothing b = b
    mappend a Nothing = a
    mappend (Just a) (Just b) = Just (mappend a b)

To use this with my C++ approach would require declaring the concept, then providing the instance for std::optional, then defining the concept. Concepts cannot be forward-declared, so this as well is impossible.

Overall, I’m quite disappointed with concepts. I wanted to use them to port Parsec to C++, but that’s tabled indefinitely until I can figure out a sensible way to deal with monads.

🔗 Comments? Questions?

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