higher-order functions in Haskell

Question

I have written the following code:

hosum :: (Int->Int)->(Int->Int)     
hosum f 0 = 1
hosum f n = afunction f (-abs(n)) (abs(n))


afunction :: (Int->Int)->Int->Int->Int
afunction f a z
    |a==z
        = 0
    |otherwise
        = afunction f (a+1) z + afunction f a z

to find the sum of f(i) from -|n| to |n|.. Where is my mistake?

Answer 1

As pointed out in the comments, your code never calls the f function. There are several other things in your code that I don't understand:

1. hosum f 0 = 1. Why is it one for any f. Shouldn't it be f 0?

2. In afunction, why is the result 0 if a == z. If the range is inclusive, it should be zero only if a > z.

3. afunction in the otherwise case calls itself twice. Why doesn't it apply f to a and calls afunction f (a + 1) z only?

Now about a correct solution. The easiest(and idiomatic) way to implement it is to use standard sum and map functions. It gives a one-liner(if we don't count type signature):

hosum :: (Int -> Int) -> Int -> Int
hosum f n = sum $ map f [-abs(n)..abs(n)]

In plain English, this function takes a list of all numbers from -abs(n) to abs(n), applies f to each of them and sums them up. That's exactly what the problem statement tells us to do.