I don't understand number conversions in Haskell

haskell sqrt
David Matuszek picture David Matuszek · Aug 9, 2011 · Viewed 7.2k times · Source

Here is what I'm trying to do:

isPrime :: Int -> Bool
isPrime x = all (\y -> x `mod` y /= 0) [3, 5..floor(sqrt x)]

(I know I'm not checking for division by two--please ignore that.) Here's what I get:

No instance for (Floating Int)
  arising from a use of `sqrt'
Possible fix: add an instance declaration for (Floating Int)
In the first argument of `floor', namely `(sqrt x)'
In the expression: floor (sqrt x)
In the second argument of `all', namely `[3, 5 .. floor (sqrt x)]'

I've spent literally hours trying everything I can think of to make this list using some variant of sqrt, including nonsense like

intSqrt :: Int -> Int
intSqrt x = floor (sqrt (x + 0.0))

It seems that (sqrt 500) works fine but (sqrt x) insists on x being a Floating (why?), and there is no function I can find to convert an Int to a real (why?).

I don't want a method to test primality, I want to understand how to fix this. Why is this so hard?

Answer

hmakholm left over Monica picture hmakholm left over Monica · Aug 9, 2011

Unlike most other languages, Haskell distinguishes strictly between integral and floating-point types, and will not convert one to the other implicitly. See here for how to do the conversion explicitly. There's even a sqrt example :-)

The underlying reason for this is that the combination of implicit conversions and Haskel's (rather complex but very cool) class system would make type reconstruction very difficult -- probably it would stretch it beyond the point where it can be done by machines at all. The language designers felt that getting type classes for arithmetic was worth the cost of having to specify conversions explicitly.

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