Hidden Gem in Control.Applicative

February 20, 2015

The other day I was looking through some haskell code and found a curious little combinator: <$. At first I thought it was <$>, the infix alias for fmap. I’m so used to seeing <$> because I see and write it many times a day. I decided to look it up on the official haddocks to see what was going on. The description is:

Replace all locations in the input with the same value. The default definition is fmap . const, but this may be overridden with a more efficient version.

I know I’ve read that before and didn’t understood it. Locations? What’s that supposed to mean? Why would I want to use this? “Whatever, I’ll probably never need it,” I thought. And so I never ended up using <$ until now, and that’s a real shame. Let’s take a closer look at the documentation and the type and see if we can figure out what it means.

The type is

(<$) :: Functor f => a -> f b -> f a

So we only require a Functor constraint to use it. As Bartosz Milewski notes, Functors are Containers. Remembering this analogy of “Containers” made the “locations” phrasing click for me. A list is a functor, and intuitively, locations in a list are the elements. So let’s see what it looks like to replace all locations in the input with the “same value”, or the first argument:

Prelude Control.Applicative> 3 <$ [7,8,9]

Ah ha! This makes sense too, looking at the type.

A Closer Look at the Types

The first argument is of type a, the second f b, a container of b. And that’s the last time we see b. The only thing Functor provides us is

fmap :: Functor f => (a -> b) -> f a -> f b

Given we know nothing about a and b, this only lets us apply a function to every “location” in a functor/container. There’s no breaking out early, nor adding new elements. <$ doesn’t even take a function, just a single value of a. It knows nothing about a and nothing about f b except how to map over it. It couldn’t even return an “empty” f b because functor doesn’t give it the tools to do that. The only thing <$ can do is replace each location in the functor with the given a and indeed the only implementation it can have is the one it has, fmap . const. Pretty cool!

So back to the list example. I guess this could be helpful. You could imagine a list or Map of items to validate and some condition causing you to replace all values with some invalidated value. Or maybe a list of tests and we want to replace them all with failures. This is a bit of a stretch though. There is a much more useful case…

The Killer Feature: Parsers

Haskell parsers tend to be Applicative Functors and are really nice to use so you end up using them a lot. You may be writing an Applicative parser for command line options using optparse-applicative, a parser using attoparsec, or a FromJSON instance in aeson. In any of these scenarios, if you find yourself parsing some fixed token, <$ can help!

Let’s say we’ve got the following type:

data Status = Staged | Running | Finished

I’ve been writing parsers like this for some time:

string :: Text -> Parser Text

parseStatus :: Parser Status
parseStatus = parseStaged <|> parseRunning <|> parseFinished
    parseStaged   = string "Staged" *> pure Staged
    parseRunning  = string "Running" *> pure Running
    parseFinished = string "Finished" *> pure Finished

Parser is a Functor, Applicative, and Alternative (which gives us <|>). This is where its hard to apply the description <$ of replacing all “locations” in a functor. What are locations in a parser? In this case the type is more illuminating. The f b in the type is Parser Text. f is Parser, and b is Text. We want to throw out the Text once this parser succeeds and replace it with our token. So we instead can write:

parseStatus :: Parser Status
parseStatus = parseStaged <|> parseRunning <|> parseFinished
    parseStaged   = Staged <$ string "Staged"
    parseRunning  = Running <$ string "Running"
    parseFinished = Finished <$ string "Finished"

This feels nicer. We no longer have to “lift” the value into the parser with pure. You can read this as “the result is Staged if I’m given a the string ‘Staged’”.

This little thought exercise has helped underscore the importance of reasoning with the tools that typeclasses give us. The simpler typeclasses have this great property of being extremely polymorphic, which at once makes them very powerful in their use and very constrained in their implementations.