Evan Martin (evan) wrote in evan_tech,
Evan Martin

flip id

I saw some Haskell code that used the expression flip id . What's that do?

id is the identity function -- just returns its argument, so it has type a → a .
flip converts a two-arg function to one that takes its args in the other order -- so, thinking it through, that's type (a → b → c) → (b → a → c) (the parens on the right side aren't necessary but they add clarity.

Those are straightforward enough -- but how do you flip a one-argument function? Let's just make the types work out. Start with id:
id :: a → a
Substitute in a hypothetical function of type x → y in for a. (I chose new letters to make it less confusing -- too many "a"s flying around already.)
id :: (x → y) → (x → y)
Now remove the unnecessary parens on the right (since → is right-associative):
id :: (x → y) → x → y
We now have a function that could be interpreted as having two arguments!

Now let's stick that into flip. flip's a is now x → y, its b is x, and its c is y, so:
flip :: (a → b → c) → (b → a → c)
flip id :: (x → (x → y) → y)
And remove the unnecessary parens:
flip id :: x → (x → y) → y

And now with the type we can derive the function. Let's call it flipid:
flipid x f = f x
And that is a simple function: it takes an argument and a function and calls the function with the argument. You could read it as flipping the implicit application operator; in fact, it has the same type as flip ($), which really is flipping the application operator.

But after all that I'm still not sure I would've come up with that on my own... ah, Haskell, no matter how much I learn there are always simple yet surprising corners. (Some other observations that might have helped: id f x = f x; f `id` x = f x. Or, looking back, that id is the application function, as evidenced by my two-arg form of id.)
Tags: haskell

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