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 → aSubstitute 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 → yWe 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) → yAnd now with the type we can derive the function. Let's call it
flipid:flipid x f = f xAnd 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.)