Comparing functions in Haskell

Question

Is there any way to compare two functions in Haskell?

My thought is that the answer is no since functions would not derive the Eq type class. However I\'m trying to

Answer 1

As long as the functions are over appropriate domains and ranges (Enum, Bounded, Eq, etc., in slightly different combination for domain and range), you can compare finite functions, even higher-order ones, in a straight-forward way and then automate the process using type classes.

I wrote up the idea in a post to the one of the Haskell mailing lists and then (more properly) in a paper I presented at one of the FDPE conferences (Victoria, BC). I've never seen it done before, but I wouldn't be surprised if others have also done it: it's a pretty simple idea, once you get over the bugaboo that "functions can't be compared for equality". Of course, the usual caveats about partiality and strictness apply: e.g., you can't return True for two functions that both fail to terminate for some common argument.

But for finite domains this technique works and provides a lot of practical use. (Plus it's really just fun to use :) ).

Edit: I added the code to hpaste here; it works in Hugs, but needs some tweaks for GHC.

Edit 2: I added a pragma and comment to the hpaste-d code so that it works in both GHC and Hugs. I also added this example as test3 (it returns True, as expected, and in short order):

(\x -> [(&&x), (||x), not]) == (\y -> [(y&&), (y||), not . not . not])

Answer 2

The code below may not compile, but hopefully you get the idea:

data Op = Succ | Pred deriving Eq

fromOp :: Enum a => Op -> a -> a
fromOp Succ = succ
fromOp Pred = pred

search :: Op -> Card -> [Card] -> [Card]
search op x list = if (op == Succ && rank x == King) || 
                      (op == Pred && rank x == Ace)
                   then []
                   else let c = [ n | n <- list, rank n == (fromOp op) (rank x)]
                     in if length c == 1
                     then x : search op (head c) list
                     else []

Answer 3

Well, there's no concept of "reference equality" in Haskell, and "behavioral equality" of functions isn't possible to check in the general case. Although it is possible to do it for functions operating only on small, finite types (as Rank would be), it's usually unwise because it leads to combinatorial explosion if you're not careful.

There are various ways you could accomplish your immediate goal, such as:

• Instead of using succ and pred directly, use self-guarding functions with type Enum a => a -> Maybe a that produce Nothing rather than an exception.

• Pass in a "stop" value to check against, e.g. search op end x list = if x == end then [] ....

• Forget about succ and pred entirely and just pass in a list of values to compare against, e.g. search :: [Rank] -> Card -> [Card] -> [Card], and end the search if the list of ranks is empty.

A couple other remarks on your code:

• Your list comprehension is reinventing the wheel--look for standard library functions, such as filter, that will do what you want.

• Don't compare the length of a list to a small constant--use pattern matching instead.

Answer 4

Unhelpful Answer

There is not, and will never be, a way to compare two functions for equality. There is a mathematical proof that it is not possible in general.

"Pragmatic" approaches will either depend on the internal workings of your compiler (e.g. if two functions are equal if they are represented by the same memory address internally), or be less helpful than expected. What is the expected result of succ == (+1)? How about let a == 1 in succ == (+a)?

I can't think of a way to define (==) for functions that always makes sense, and I believe neither can anyone else.

Things Irrelevant to the Question

The type signature is missing a return type.

Your function has a rank 2 type, which is not standard Haskell 98, and, more importantly, is not necessary in this case. You don't care if the passed-in function can deal with any Enum type, you only care that it works for Rank:

search :: (Rank -> Rank) -> Card -> [Card] -> [Card]   -- much simpler.

I'd try to use case more often, and if/then less often. In particular, I'd write:

case [ n | n <- list, rank n == op (rank x)] of
    [y] -> x : search op y list   -- length == 1
    _ -> []                       -- otherwise

My Real Answer

Your function basically only works with two different values for op, but its type doesn't say so; that doesn't feel right for me.

You could do it the old-fashioned way:

data Direction = Succ | Pred deriving(Eq)

search :: Direction -> Card -> [Card] -> [Card]
search dir x list = if (dir == Succ && rank x == King) ...
    ... let op = case Direction of Succ -> succ ; Pred -> pred
        in ...

Or you could make the parameter more general, and pass a function that may fail gracefully (by returning Nothing) instead:

maybeSucc x | x == maxBound = Nothing
            | otherwise = Just (succ x)

search :: (Rank -> Maybe Rank) -> Card -> [Card] -> [Card]
search op x list = case op (rank x) of
    Nothing -> []
    Just theRank -> case [ n | n <- list, rank n == theRank ] of ...

Answer 5

1) Your op is overly general. You'll only be using it for Card whatever type rank (undefined :: Card) is, so just make it RankThing -> RankThing. Also, your function type signature is missing a return value type.

2) An example intended use looks like it would be search succ Ace xs, but that's rather clumsy, how about two auxillary functions that handle the bounds:

searchUp King _ = []
searchUp x ys = search succ x ys

searchDown Ace _ = []
searchDown x ys = search pred x ys

This might read better for your users and avoid the need to check the operation

3) if you really want to check what operation is performed, and know the operation will be one of two possibilities, then you can either name the operation or test it with a known answer test (KAT). For example:

data Operation = Succ | Pred

search :: Operation -> Card -> [Card] -> []
search Succ King _ = []
search Succ x ys = search' succ x ys
search Pred Ace _ = []
search Pred x ys = search' pred x ys

And the KAT solution (rather lame, imo):

search op x ys
    | op Five == Four && rank x == Ace = []
    | op Five == Six && rank x == King = []
    | otherwise = ...