`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`

.)
## Error

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