Haskell：确定函数arity的函数？

Question

Is it possible to write a function arity :: a -> Integer to determine the arity of arbitrary functions, such that

> arity map
2
> arity foldr
3
> arity id
1
> arity "hello"
0

?

Answer 1

Yes, it can be done very, very easily:

arity :: (a -> b) -> Int
arity = const 1

Rationale: If it is a function, you can apply it to exactly 1 argument. Note that haskell syntax makes it impossible to apply to 0, 2 or more arguments as fab is really (fa) b , ie not f applied to a and b , but (f applied to a) applied to b . The result may, of course, be another function that can be applied again, and so forth.

Sounds stupid, but is nothing but the truth.

Answer 2

It's easy with OverlappingInstances :

{-# LANGUAGE FlexibleInstances, OverlappingInstances #-}

class Arity f where
  arity :: f -> Int

instance Arity x where
  arity _ = 0

instance Arity f => Arity ((->) a f) where
  arity f = 1 + arity (f undefined) 

Upd Found problem. You need to specify non-polymorphic type for polymorphic functions:

arity (foldr :: (a -> Int -> Int) -> Int -> [a] -> Int)

Don't know how to solve this yet.

Upd2 as Sjoerd Visscher commented below "you have to specify a non-polymorphic type, as the answer depends on which type you choose".

Answer 3

If id has arity 1, shouldn't id x have arity 0? But, for example, id map is identical to map , which would has arity 2 in your example.

Have the following functions the same arity?

f1 = (+)
f2 = (\x y -> x + y)
f3 x y = x + y

I think your notion of "arity" is not well defined...

Answer 4

In Haskell, every "function" takes exactly one argument. What looks like a "multi-argument" function is actually a function that takes one argument and returns another function which takes the rest of the arguments. So in that sense all functions have arity 1.

Answer 5

It's not possible with standard Haskell. It may be possible using the IncoherentInstances or similar extension.

But why do you want to do this? You can't ask a function how many arguments it expects and then use this knowledge to give it precisely that number of arguments. (Unless you're using Template Haskell, in which case, yes, I expect it is possible at compile time. Are you using Template Haskell?)

What's your actual problem that you're trying to solve?

Answer 6

How about this:

arity :: a -> Int
arity (b->c) = 1 + arity (c)
arity _ = 0