I have to write a simple pi approximation and I did and it works fine, but in the task it says to write a function with the header "pi_approx :: Int -> Double".

My code:

pi_approx x = sqrt (pi2_approx(x))

pi2_approx x = 
    if x/= 1
    then (6 /(x*x))   (pi2_approx(x-1))
    else (6/(1*1))

However my function works fine without "pi_approx :: Int -> Double", but when i try to use this declaration I always get the type error:

pi_approx.hs:10:14: error:

• Couldn't match expected type Double' with actual type Int'
• In the expression: ( ) (6 / (x * x)) (pi2_approx (x - 1)) In the expression: if x /= 1 then ( ) (6 / (x * x)) (pi2_approx (x - 1)) else (6 / (1 * 1)) In an equation for `pi2_approx': pi2_approx x = if x /= 1 then ( ) (6 / (x * x)) (pi2_approx (x - 1)) else (6 / (1 * 1)) | 10 | then ( ) (6 /(x*x)) (pi2_approx(x-1)) | ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

I tried to use fromIntegral in various ways and if I run the function without declaration I checked the type of the solution, which is: "(Floating a, Eq a) => a"

As I understand Double should be an instance of Floating so I don´t understand why it wont compile.

I am very new at haskell and it seems I don´t understand the core concept of data types and constraints. I can´t seem to find any simple and good documentation/explanation on the topic though. Maybe someone here can help me understand based on this example :)

CodePudding user response：

because x is an Int, hence x * x is also an Int, and you can not use an Int for (/) :: Floating a => a -> a -> a.

You need to convert this to a Double, for example with fromIntegral :: (Integral a, Num b) => a -> b:

pi2_approx :: Int -> Double
pi2_approx 1 = 6
pi2_approx x = 6 / fromIntegral (x*x)   pi2_approx (x-1)

For a large number of iterations, it gives a result close to π²:

Prelude> sqrt (pi2_approx 10000)
3.1414971639472147