CODEWITHSUNDEEP
haskell
Code-Apprentice
Code-Apprentice

Reputation: 83527

Testing if a Double is an Integral value in Haskell?

I have a list areas :: [Double]. Now I want to filter this list for those which are actually integral values. I want to do something like this for my predicate:

isInteger :: Double -> Bool
isInteger x = abs (fromIntegral (floor x) - x) < delta
  where delta = 0.00001

However, I would guess there is a better way to do this. Is there a Haskell idiom for checking if a real value is an integer?

Upvotes: 4

Views: 414

Answers (1)

Daniel Wagner
Daniel Wagner

Reputation: 152707

This looks fine and idiomatic to me, though you probably want to use round rather than floor. You could also consider using approxRational and checking that the denominator of the result is 1:

isInteger x = denominator (approxRational x 0.00001) == 1

Upvotes: 7

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