该尾递归Haskell函数的错误在哪里？

Question

I have to implement a sum function in Haskell in two ways. One function with tail recursion and the other without tail recursion.

Here is the one without tail recursion and it works perfectly

sum1 x = if x==0 then 0 else x + sum1(x-1)

Here is my attempt with tail recursion and it doesn't work:

sum2 x = help 0 y
help x y = if y==0 then x else help(x+y,y-1)

Can someone point out the mistake?

Answer 1

Your line:

help x y = if y==0 then x else 

is not the correct syntax for calling a function. Because here the Haskell compiler will interpret it as:

help x y = if y==0 then x else help 
--                                  ^ a tuple

Instead you should write:

help x y = if y==0 then x else help 
--                                  ^ two arguments

Furthermore you can also use guards, like:

helper x y | y == 0 = x
           | otherwise = help (x+y) (y-1)

Finally there is also an error in the first line of sum2 . It should be x instead of y :

sum2 x = help 0 

So in full, we get:

sum2 x = help 0 x
helper s x | x == 0 = s
           | otherwise = help (s+x) (x-1)

I also renamed y in the helper to x and x to s (as in s um) to make it less confusing (kudos to @Bergi for commenting on this).

Or use an eta reduction :

sum2 = help 0

Finally note that you do not need recursion for this. An implementation that would work faster is the following:

sum3 x = div (x*(x+1)) 2

Since:

n
---
\       (n+1) n
/   i = -------
---        2
i=1