Question

I am getting the following error:

exercise-2-2.hs:15:49:
    Couldn't match expected type `Double' with actual type `Int'
    In the fourth argument of `regularPolygonHelper', namely `theta'
    In the expression: regularPolygonHelper n s 0 theta r
    In an equation for `regularPolygon':
        regularPolygon n s
          = regularPolygonHelper n s 0 theta r
          where
              r = s / 2.0 * (sin (pi / n))
              theta = (2.0 * pi) / n

in the following code:

data Shape = Rectangle Side Side
           | Ellipse Radius Radius
           | RTTriangle Side Side
           | Polygon [Vertex]
    deriving Show

type Radius = Float
type Side   = Float
type Vertex = (Float, Float)

square s = Rectangle s s
circle r = Ellipse r r

regularPolygon :: Int -> Side -> Shape
regularPolygon n s = regularPolygonHelper n s 0 theta r
    where r     = s / 2.0 * (sin (pi / n))
          theta = (2.0 * pi) / n

regularPolygonHelper :: Int -> Side -> Int -> Double -> Double -> Shape
regularPolygonHelper 0 s i theta r = Polygon []
regularPolygonHelper n s i theta r = 
    (r * cos (i * theta), r * sin (i * theta)) : 
        (regularPolygonHelper (n - 1) s (i + 1) theta r)

Why is this? Isn’t (2.0 * pi) / n a double?

Answer 1

Haskell has no automatic conversion between different numeric types. You have to do this by hand. In your case, (2.0 * pi) / fromIntegral n would do the trick. (You have to add this at all the other places where you want to have a cast too) The reason for this is, that implicit conversion would make type inference much harder, IMHO it is better to have type inference than automatic conversion.