Infinite Square Roots : Couldn't match expected type & cannot construct the infinite type

Question

Sorry for what's probably an idiot question - trying to learn Haskell at the moment;

I'm trying to build a basic function that will create an infinite list of square roots on a number, so I can get practice with the take function and how it works.

I wrote the following code;

infisqrt x = infisqrt'((x :: Float) [])

-- helper method
infisqrt' x xs = infisqrt'(sqrt(x) (xs ++ [(sqrt(x))]))

However, this is returning with two errors when trying to load the library;

:l isq

isq.hs:1:24:
Couldn't match expected type ‘[t0] -> a’ with actual type ‘Float’
Relevant bindings include
  infisqrt :: Float -> [a] -> t (bound at isq.hs:1:1)
The function ‘x :: Float’ is applied to one argument,
but its type ‘Float’ has none
In the first argument of ‘infisqrt'’, namely ‘((x :: Float) [])’
In the expression: infisqrt' ((x :: Float) [])

isq.hs:5:33:
Occurs check: cannot construct the infinite type: a ~ [a] -> a
Relevant bindings include
  xs :: [a] (bound at isq.hs:5:13)
  x :: a (bound at isq.hs:5:11)
  infisqrt' :: a -> [a] -> t (bound at isq.hs:5:1)
In the first argument of ‘sqrt’, namely ‘(x)’
In the first argument of ‘infisqrt'’, namely
  ‘(sqrt (x) (xs ++ [(sqrt (x))]))’

Can anyone let me know where I'm going wrong with this?

Answer 1

Haskell function invocation doesn't use parentheses. It looks like you're expecting this:

infisqrt x = infisqrt'((x :: Float) [])

to mean "pass x and [] as arguments to inifsqrt." However, to the compiler, it actually means "pass [] as the first argument to x, and then pass the result to infisqrt'." If you take the extra parentheses out, you should start getting traction:

infisqrt x = infisqrt' (x :: Float) []

(remember, you've got the same thing going on in infisqrt''s definition)

As a side note, it's typically preferable to put arguments' types in a function type declaration:

infisqrt :: Float -> [Float]
infisqrt x = infisqrt' x []