Louis Dionne Programming and Categories, Oh My!

Laziness as a Comonad

This (first!) post will present my current understanding of how laziness can be represented as a Comonad. The interest of seeing laziness as a Comonad appeared while working on a C++ metaprogramming library called Hana as a possible generalization of the eval_if construct from the MPL library. More generally, anyone trying to implement a domain-specific language with a notion of branching inside a strict language (as opposed to non-strict like Haskell) will have to design a system to emulate laziness, in order to evaluate only the branch that was picked by the condition. This post is an attempt to describe such a system using the notion of Comonad.

To simplify this post, I’ll assume the reader to be familiar with Comonads. If that is not the case, there are several resources available online which do a better job explaining Comonad than I would. For example, see here, here and here.

It all started when I tried to define a structure that could be used to hold lazy computations in C++. This structure would hold the function and the arguments to call the function with, but it would only perform the call when explicitly evaluated. It basically went like this:

template <typename F, typename ...Args>
struct _lazy {
    F function;
    hana::_tuple<Args...> args;

template <typename F, typename ...Args>
auto eval(_lazy<F, Args...> expr) {
    return hana::unpack(expr.args, expr.function);

template <typename F>
auto lazy(F f) {
    return [=](auto ...args) -> _lazy<F, decltype(args)...> {
        return {f, {args...}};

_lazy is straightforward; it’s just a dumb holder for the function and its arguments. Likewise, eval just takes a _lazy computation and calls the function with its arguments. To do that, I use Hana’s unpack method, which satisfies

unpack(make_tuple(x1, ..., xN), f) == f(x1, ..., xN)

Finally, there’s the lazy function. It just creates a _lazy computation, but it does it in a way that gives us a nice syntax. Let’s say you have a function f. Then, lazy(f) is a lambda that expects the arguments and returns a lazy computation. Hence, lazy(f)(args...) is a lazy computation of f(args...). Achieving this (IMO) nice syntax is the reason why I’m defining lazy as above instead of the (functionally equivalent)

template <typename F, typename ...Args>
auto lazy(F f, Args ...args) {
    return _lazy<F, Args...>{f, {args...}};

Generalized types: a bit of terminology

Before we go deeper, let me quickly introduce what I’ll call generalized types for the rest of this post. Basically, the idea is that when we see the type _lazy<F, Args...>, we’re actually thinking about a container that holds a lazy value of type std::result_of_t<F(Args...)>. As a matter of fact, we could actually implement _lazy as follows:

template <typename T>
struct _lazy {
    std::function<T()> function;

    template <typename F, typename ...Args>
    explicit _lazy(F f, Args ...args)
        : function{[=]{ return f(args...); }}
    { }

We could then implement eval as

template <typename T>
auto eval(_lazy<T> expr) {
    return expr.function();

However, providing a _lazy<T> type forces us to use type erasure via std::function, which is less efficient than simply storing it like we did in our first implementation of _lazy. Also, although this is not the case here, it would make it impossible to use _lazy for metaprogramming purposes since all the type information of the F and Args... template arguments in the first implementation is erased in the second one. Providing a _lazy<T> type also restricts the implementation more, since _lazy<T> can only depend on the type of the computation’s result, not on the type of F and/or Args.... Finally, you have to know the type of the computation before creating and computing it, which is a huge problem when you’re trying to do type-level computations, since that type T is usually what you’re trying to compute! Of course, there are also advantages to using type erasure, and sometimes there’s no other option, but for our current use case it’s just not what we want.

To work around this, we could define a Lazy(T) concept that would represent a lazy computation returning something of type T. The concept could be roughly specified by saying that for any model c of Lazy(T), the eval(c) expression must be well-formed and it must represent an object of type T (or implicitly convertible to T if you’re feeling fancy). However, since there’s really only one model of that concept, I prefer saying that Lazy(T) is a generalized type instead of a concept.

Note: Obviously, there is more than a single model of Lazy(T), since I just wrote two different models, one with type erasure and one without it. However, what I mean by “only one model” is that they are all equivalent, in some way which I’m still trying to define. I think what I mean is unique up to unique isomorphism, but I don’t grok the details well enough to be sure yet.

Anyway, the important point is that I’ll refer to _lazy<F, Args...> has having a generalized type of Lazy(T) to mean that evaluating such a lazy computation yields something of type T, happily disregarding the fact that the actual type of that computation is something else entirely (_lazy<F, Args...>). In other words, I’ll consider the actual type of the lazy computation to be a simple implementation detail. In a similar vein, I’ll say that the pseudo-signature of lazy is:

\[\mathrm{lazy} : (T \to U) \to (T \to \mathrm{Lazy}(U))\]

which means that lazy is a function which takes a function from T to U, and returns a function which takes a T and returns a Lazy(U). Note that I voluntarily ignored the multi-argument case for simplicity.

Laziness as a Functor

Ok; so far, we can create lazy computations and evaluate them. This can be useful on its own, but we would like to develop a richer set of tools for manipulating lazy computations. Being guided by the goal of making Lazy a Comonad, let’s recall the definition of the Comonad type class from the Haskell Control.Comonad package:

class Functor w => Comonad w where

We see that a Comonad should also be a Functor, so we’ll start by trying to make Lazy a Functor. To do this, we must define a function named transform which has the following pseudo-signature (using generalized types instead of types):

\[\mathrm{transform} : \mathrm{Lazy}(T) \times (T \to U) \to \mathrm{Lazy}(U)\]

Note: We’re using transform instead of fmap because it’s used in Hana and it’s more familiar to C++ programmers.

Here, Lazy(T) simply represents a lazy computation which returns an object of type T, and T -> U is a function taking a T and giving back a U. Hence, transform should take a lazy computation of type T and a function transforming Ts into Us, and give back a new lazy computation of type U. An obvious way of making this happen is to simply return a new lazy computation doing the composition of the two functions:

template <typename Computation, typename F>
auto transform(Computation computation, F f) {
    auto eval_f = [=](auto x) { return f(eval(x)); };
    return lazy(eval_f)(computation);

First, we create a new function eval_f which takes a lazy computation as an argument and calls the f function with the evaluated result. Then, we lazily apply that eval_f function to our lazy computation. The result is a lazy computation which, when evaluated, will first evaluate the lazy computation and then call f with the result. Also, the types match; if computation is a lazy computation of type T (i.e. a Lazy(T)), and if f is a function of type T -> U, then transform(computation, f) is a lazy computation of type U (i.e. a Lazy(U)). We now have to make sure the laws of Functor are satisfied, i.e.

transform(computation, id) == computation

transform(transform(computation, f), g)
    == transform(computation, compose(g, f))

where compose(g, f) is simply function composition, i.e.

compose(g, f) == [](auto x) {
    return g(f(x));

As a side note, let’s observe that our definition of transform above is actually equivalent to

template <typename Computation, typename F>
auto transform(Computation computation, F f) {
    return lazy(compose(f, eval))(computation);

with the definition of compose we just gave. Now, both laws are straightforward to prove, but before we prove them we must define what it means for two computations to be equal. Intuitively, I would think of two lazy computations as being equal if their evaluated result is equal, and so we’ll use that definition of equality. In other words, for any two lazy computations a and b,

a == b    if and only if    eval(a) == eval(b)

Note: Making sure this is an equivalence relation is easy and left as an exercise.

Proving the Functor laws

Note: This part is technical and may be skipped. It’s there because proofs are important, but there’s no real insight to be gained here.

To prove the first law, we just expand transform from its definition and then use basic properties of compose.

transform(comp, id)
    // from the definition of transform
    == lazy(compose(id, eval))(comp)

    // because compose(id, anything) == anything
    == lazy(eval)(comp)

    // because eval(lazy(eval)(comp)) == eval(comp)
    // by definition of eval
    == comp

As for the second law, on the left hand side we have

transform(transform(comp, f), g)
    // definition of transform applied to transform(comp, f)
    == transform(lazy(compose(f, eval))(comp), g)

    // definition of transform applied to transform(..., g)
    == lazy(compose(g, eval))(lazy(compose(f, eval))(comp))

Let’s now evaluate this last expression:

eval(lazy(compose(g, eval))(lazy(compose(f, eval))(comp)))
    // by definition of eval
    == compose(g, eval)(lazy(compose(f, eval))(comp))

    // by definition of compose
    == g(eval(lazy(compose(f, eval))(comp)))

    // by definition of eval, again
    == g(compose(f, eval)(comp))

    // by definition of compose, again
    == g(f(eval(comp)))

On the right hand side, by definition of transform, we have

transform(comp, compose(g, f))
    == lazy(compose(compose(g, f), eval))(comp)

Evaluating this gives:

eval(lazy(compose(compose(g, f), eval))(comp))
    // by definition of eval
    == compose(compose(g, f), eval)(comp)

    // by definition of compose applied twice
    == g(f(eval(comp)))

Which tells us that the second law holds too. Phew!

Laziness as a Comonad

Now that we have a Functor, let’s look at the full definition of the Comonad type class and see what we can do with it:

class Functor w => Comonad w where
    extract :: w a -> a
    duplicate :: w a -> w (w a)
    extend :: (w a -> b) -> w a -> w b

Since our goal is to make Lazy a Comonad, the obvious first step is to substitute Lazy in that definition and see what happens:

extract :: Lazy a -> a
duplicate :: Lazy a -> Lazy (Lazy a)
extend :: (Lazy a -> b) -> Lazy a -> Lazy b

Looking at extract, it has exactly the same type as eval. duplicate does not seem to be very useful; it should take a computation that’s already lazy and give it an additional layer of laziness. It would probably be useful to compose complex lazy computations, but probably not on its own. Finally, extend looks like it should just turn a computation accepting a lazy input and a lazy input into a lazy computation of the result. At this point, it is nice to observe that the type of extend is actually a special case of lazy’s type (modulo currying):

\[\mathrm{lazy} : (T \to U) \to (T \to \mathrm{Lazy}(U))\] \[\mathrm{extend} : (\mathrm{Lazy}(T) \to U) \to (\mathrm{Lazy}(T) \to \mathrm{Lazy}(U))\]

Note: In Haskell, -> is right associative, so A -> B -> C is actually equivalent to A -> (B -> C), which is why extend in the above type class is equivalent to the extend just above.

Given a function that accepts Lazy values and a Lazy argument, lazy’s type becomes:

\[\mathrm{lazy} : (\mathrm{Lazy}(T) \to U) \to (\mathrm{Lazy}(T) \to \mathrm{Lazy}(U))\]

which is exactly the same as extend’s type. This should give us a clue that extend and lazy may very well be the same thing. Let’s go. First, extract is just eval:

template <typename Computation>
auto extract(Computation computation) {
    return eval(computation);

Second, duplicate:

template <typename Computation>
auto duplicate(Computation computation) {
    return lazy([=]{ return computation; })();

For the explanation, let’s say computation is a lazy computation of type T, i.e. computation has a generalized type of Lazy(T). What happens here is that we create a nullary function returning the lazy computation, and we then apply it lazily. evaluating that gives us back our computation, which as a generalized type of Lazy(T), and hence lazy([=]{ return computation; })() has a generalized type of Lazy(Lazy(T)), since you must eval it twice to get back the initial T.

Thirdly, let’s implement extend. Strictly speaking, this would not be required because extend can be implemented in terms of duplicate and transform, but we’ll still do it for completeness:

// F :: Lazy(T) -> U
// Computation :: Lazy(T)
// We want to return a Lazy(U)
template <typename F, typename Computation>
auto extend(F f, Computation computation) {
    return lazy(f)(computation);

We’re given a lazy computation of generalized type Lazy(T) and a function of type Lazy(T) -> U. We just use lazy to lazily apply the function to its argument, which effectively gives us a Lazy(U).

Proving the Comonad laws

Note: Again, this part is mostly technical. Feel free to skip or to skim through.

So far so good; we have definitions for an hypothetical Comonad representing laziness. The last thing we must check is whether the properties of a Comonad are respected by Lazy. There are two equivalent sets of laws that must be satisfied in order to be a Comonad, so we can pick either of those. Since I find it easier, I’ll pick the one with duplicate, transform and extract, and then I’ll make sure that my implementation of extend agrees with the default definition in terms of transform and duplicate. Translated from Haskell to C++, the laws look like:

extract(duplicate(comp)) == comp
transform(duplicate(comp), extract) == comp
duplicate(duplicate(comp)) == transform(duplicate(comp), duplicate)

Let’s prove the first law:

    // by definition of duplicate
    == extract(lazy([=]{ return comp; })())

    // by definition of extract as eval
    == [=]{ return comp; }()
    == comp

This one was easy. Let’s prove the second law:

transform(duplicate(comp), extract)
    // by definition of transform
    == lazy(compose(extract, eval))(duplicate(comp))

    // because extract == eval
    == lazy(compose(eval, eval))(duplicate(comp))

    // by definition of duplicate
    == lazy(compose(eval, eval))(lazy([=]{ return comp; })())

Evaluating this last expression gives

eval(lazy(compose(eval, eval))(lazy([=]{ return comp; })()))
    // by definition of eval
    == compose(eval, eval)(lazy([=]{ return comp; })())

    // by definition of compose
    == eval(eval(lazy([=]{ return comp; })()))

    // by definition of eval, again
    == eval([=]{ return comp; }())
    == eval(comp)


eval(transform(duplicate(comp), extract)) == eval(comp)

which proves the second law because a == b if and only if eval(a) == eval(b). This was a bit more involved. Let’s go for the third and last law. On the left hand side, we have


Evaluating this, we find

    // because eval == extract and the first law
    == duplicate(comp)

We now evaluate the right hand side:

eval(transform(duplicate(comp), duplicate))
    // by definition of transform
    == eval(lazy(compose(duplicate, eval))(duplicate(comp)))

    // by definition of eval
    == compose(duplicate, eval)(duplicate(comp))

    // by definition of compose
    == duplicate(eval(duplicate(comp)))

    // because eval == extract and the first law
    == duplicate(comp)

Hence, both expressions are equal and the third law holds. The last thing that’s left to check is that our definition of extend agrees with the default definition in terms of transform and duplicate. It should be true that:

extend(f, comp) == transform(duplicate(comp), f)

Evaluating the left hand side, we get

eval(extend(f, comp))
    // by definition of extend
    == eval(lazy(f)(comp))

    // by definition of eval
    == f(comp)

Evaluating the right hand side, we get

eval(transform(duplicate(comp), f))
    // by definition of duplicate
    == eval(transform(lazy([=]{ return comp; })(), f))

    // by definition of transform
    == eval(lazy(compose(f, eval))(lazy([=]{ return comp; })()))

    // by definition of eval
    == compose(f, eval)(lazy([=]{ return comp; })())

    // by definition of compose
    == f(eval(lazy([=]{ return comp; })()))

    // by definition of eval
    == f([=]{ return comp; }())
    == f(comp)

And hence we see that both expressions are equal. This proves that our definition of extend was correct, which completes the demontration. Phew!

So Lazy is a Comonad. What now?

My initial goal was to generalize the eval_if construct from the MPL so it could work with arbitrary Comonads. Here’s how we could now specify the type of an hypothetical eval_if working not only on types (as in the MPL). Given an arbitrary Comonad W,

\[\mathrm{eval\_if} : \mathrm{bool} \times W(T) \times W(T) \to T\]

For the special case of Lazy, this gives

\[\mathrm{eval\_if} : \mathrm{bool} \times \mathrm{Lazy}(T) \times \mathrm{Lazy}(T) \to T\]

which effectively matches our expectations. Since infinite streams can be represented as Comonads, we could, for example, peek the first element of one of two streams based on a boolean condition using eval_if. Whether this is useful is debatable; honestly I have not found an interesting use case besides Lazy so far, but at least we did not lose anything.

However, if we go a bit further down the rabbit hole, we can also see that Lazy is an Applicative Functor and a Monad. This observation is super useful because it tells us how to compose Lazy computations, which in turn can help us design a composable lazy branching system. This will be the subject of a next post, so stay tuned!


17 Mar 2015: I realized I had made an error in extend’s signature when comparing it with that of lazy. Both signatures are not only similar, they are actually the same (modulo currying). I also added a proof that my definition of extend matches the default definition in terms of duplicate and transform.