DEV Community

Matt Thornton
Matt Thornton

Posted on

Grokking Functors

In this post we're going to grok functors by "discovering" them through a worked example.

Small F# primer

Skip this if you've got basic familiarity with F#.
It should be easy enough to follow along if you've not used F# before though. You'll only need to understand the following bits.

  • F# has an option type. It represents either the presence of Some value or its absence through a None value. It is typically used instead of null to indicate missing values.
  • Pattern matching on an option type looks like:
match anOptionalValue with
| Some x -> // expression when the value exists
| None -> // expression when the value doesn't exist.
Enter fullscreen mode Exit fullscreen mode
  • F# has a pipe operator which is denoted as |>. It is an infix operator that applies the value on the left hand side to the function on the right. For example if toLower takes a string and converts it to lowercase then "ABC |> toLower would output "abc".

The scenario

Let's say we've been asked to write a function that will print a user's credit card details. The data model is straight forward, we have a CreditCard type and a User type.

type CreditCard =
    { Number: string
      Expiry: string
      Cvv: string }

type User =
    { Id: UserId
      CreditCard: CreditCard }
Enter fullscreen mode Exit fullscreen mode

Our first implementation

We want a function that takes a User and returns a string representation of their CreditCard details. This is fairly easy to implement.

let printUserCreditCard (user: User): string =
    let creditCard = user.CreditCard

    $"Number: {creditCard.Number}
    Exiry: {creditCard.Expiry}
    Cvv: {creditCard.Cvv}"
Enter fullscreen mode Exit fullscreen mode

We can even factor this out by writing a getCreditCard function and a printCreditCard function, which we can then just compose them whenever we want to print a user's credit card details.

let getCreditCard (user: User) : CreditCard = user.CreditCard

let printCreditCard (card: CreditCard) : string =
    $"Number: {card.Number}
    Exiry: {card.Expiry}
    Cvv: {card.Cvv}"

let printUserCreditCard (user: User) : string =
    user
    |> getCreditCard 
    |> printCreditCard
Enter fullscreen mode Exit fullscreen mode

Beautiful! πŸ‘Œ

A twist πŸŒͺ

All is well, until we realise that we first need to lookup the user by their id from the database. Fortunately, there's already a function implemented for this.

let lookupUser (id: UserId): User option =
    // read user from DB, if they exist return Some, else None
Enter fullscreen mode Exit fullscreen mode

Unfortunately, it returns a User option rather than a User so we can't just write

userId
|> lookupUser
|> getCreditCard
|> printCreditCard
Enter fullscreen mode Exit fullscreen mode

because getCreditCard is expecting a User, not an option User.

Let's see if we can transform getCreditCard so that it can accept an option as input instead, without changing the original getCreditCard function itself. To do this we need to write a function that will wrap it. Let's call it liftGetCreditCard, because we can think of it as "lifting" the getCreditCard function to work with option inputs.

This might seem a bit tricky to write at first, but we know that we have two inputs to liftGetCreditCard. The first is the getCreditCard function and the second is the User option. We also know that we need to return a CreditCard option. So the signature should be (User -> CreditCard) -> User option -> CreditCard option.

By following the types there's only really one thing to do, try and apply the function to the User value by pattern matching on the option. If the user doesn't exist then we can't apply the function so we have to return None.

let liftGetCreditCard getCreditCard (user: User option): CreditCard option =
    match user with
    | Some u -> u |> getCreditCard |> Some
    | None -> None
Enter fullscreen mode Exit fullscreen mode

Notice how in the Some case we have to wrap the output of getCreditCard in Some. This is because we have to make sure both branches return the same type and the only way to do that here is to make them return CreditCard option.

With this in place our pipeline is now

userId
|> lookupUser
|> liftGetCreditCard getCreditCard
|> printCreditCard
Enter fullscreen mode Exit fullscreen mode

By partially applying getCreditCard to liftGetCreditCard we created a function whose signature was User option -> CreditCard option, which is what we wanted.

Well nearly, we've got the same issue as before except on the last line now. printCreditCard only knows how to deal with CreditCard values and not CreditCard option values. So let's apply the same trick again and write liftPrintCreditCard.

let liftPrintCreditCard printCreditCard (card: CreditCard option): CreditCard option =
    match card with
    | Some cc -> cc |> printCreditCard |> Some
    | None -> None
Enter fullscreen mode Exit fullscreen mode

And our pipeline is now

userId
|> lookupUser
|> liftGetCreditCard getCreditCard
|> liftPrintCreditCard printCreditCard
Enter fullscreen mode Exit fullscreen mode

Is that a functor I see πŸ”­

If we zoom out a bit, we might notice that the two lift... functions are remarkably similar. They both perform the same basic task, which is to unwrap the option, apply a function to the value if it exists and then package this value back up as a new option. They don't really depend on what's inside the option or what the function is. The only constraint is that the function accepts as input the value that might be inside the option.

Let's see if we can write a single version called lift which defines this behaviour for any valid pair of a function, which we'll call f and an option, which we'll call x. We can erase the type definitions for now and let F# infer them for us.

let lift f x =
    match x with
    | Some y -> y |> f |> Some
    | None -> None
Enter fullscreen mode Exit fullscreen mode

Tidy, but perhaps a little bit abstract. Let's see what the inferred types tell us about this. F# has inferred it to have the type ('a -> 'b) -> 'a option -> 'b option. Where 'a and 'b are generic types. Let's place it side-by-side with liftGetCreditCard to help us make it more concrete.

(User -> CreditCard) -> User option -> CreditCard option

('a -> 'b) -> 'a option -> 'b option
Enter fullscreen mode Exit fullscreen mode

The concrete User has been replaced by the generic type 'a and the concrete CreditCard type has been replaced with the generic type 'b. That's because lift doesn't care what's inside the option box, it's just saying "give me some function f and I'll apply this to the value contained within x if it exists and repackage it as a new option".

A more intuitive name for lift would be map, because we're just mapping the contents inside the option.

So with this new map function in place let's use it in our pipeline.

userId
|> lookupUser
|> map getCreditCard
|> map printCreditCard
Enter fullscreen mode Exit fullscreen mode

Nice! This is nearly identical to the version we wrote before the option bombshell was dropped on us. So the code is still very readable.

You just discovered functors πŸ‘

That map function we wrote is what makes option values functors. Functors are just a class of things that are "mappable". Lucky for us, F# has already defined map in the Option module, so we can actually just write our code using that instead

userId
|> lookupUser
|> Option.map getCreditCard
|> Option.map printCreditCard
Enter fullscreen mode Exit fullscreen mode

Functors are just "mappable" containers πŸ“¦

Another good way to intuit functors is to think of them as value containers. For each type of container we just need to define a way to be able to map or transform its contents.

We've just discovered how to do that for options, but there are more containers that we can turn into functors too. A Result is a container which either has a value or an error and we want to be able to map the value regardless of what the error is.

The most common container though is a List or an Array. Most programmers have encountered a situation where they've needed to transform all the elements of a list before. If you've ever used Select in C# or map in JavaScript, Java etc then you've probably already grokked functors, even if you didn't realise it.

Test yourself πŸ§‘β€πŸ«

See if you can write map for the Result<'a, 'b> and List<'a> types. The answers are below, no peeking until you've had a go first!

Result solution
let map f x =
    match x with
    | Ok y -> y |> f |> Ok
    | Error e -> Error e
Enter fullscreen mode Exit fullscreen mode

This one is nearly identical to option in that we just apply the function to value of the Ok case otherwise we just propagate the Error.


List solution
let rec map f x =
    match x with
    | y:ys -> f y :: map f ys
    | [] -> []
Enter fullscreen mode Exit fullscreen mode

This is a little trickier than the others, but the basic idea is the same. If the list has some items we pick off the head of the list, apply the function to the head and add it to the front of a new list created by mapping over the tail of this one. If the list is empty we just return another empty list. By reducing the list size by one each time we call map again we guarantee that eventually we hit the base case of the empty list which terminates the recursion.


What did we learn πŸ§‘β€πŸŽ“

We learnt that functors are just types of containers that have a map function defined for them. We can use this function to transform the contents inside the container. This allows us to chain together functions that work with regular values even when those values are packaged in one of these container types.

This is useful because this pattern occurs in lots of different situations and by extracting a map function we can eradicate quite a bit of boilerplate that we'd otherwise have to do by constantly pattern matching.

Taking it further

If you enjoyed grokking functors, you'll love Grokking Monads. There we follow a similar recipe and discover a different type of function that's also very useful.

Top comments (2)

Collapse
 
tysonmn profile image
Tyson Williams

The most common container though is a List or an Array.

I disagree. I think the most common (mappable) container (in F#) is a record. See the monomorphic functors article by Mark Seemann.

Collapse
 
choc13 profile image
Matt Thornton

Thanks for pointing this out, it’s a good point. My wording was a bit sloppy here, what I was trying to convey was that List / Array are probably the most likely things that people will have already mapped over, even if they didn’t realise they were functors. I’ll improve the sentence to convey this better.