For rational functions that take on integer values at integer
arguments, for example n*(n+1)/2, is there a way to doctor the corresponding Haskell definition f n = n*(n+1)/2 so that the type signature becomes f :: Num a => a -> a rather than f :: Fractional a => a -> a Doug McIlroy _______________________________________________ Beginners mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners |
The problem is that the moment you divided by two, it can no longer be an instance of Num. Instances of Num include: Integers, Ints and Words. Once divided, you can no longer use that result in places where an Integer is required, because it won't be. What you can do is use the `div` function which will round down to the nearest Integer value. Then it is an instance of Integral, which includes Integers, Ints, and Words (but floating point types) You can also use `round`, `floor`, or `ceiling` to round your result to an appropriate integer after you've divided. On Fri, Dec 11, 2020 at 8:04 AM M Douglas McIlroy <[hidden email]> wrote: For rational functions that take on integer values at integer _______________________________________________ Beginners mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners |
In reply to this post by M Douglas McIlroy
[Attempting to resend] Hi - I think your logic is: I can define these two: fFrac :: Fractional a => a -> a fFrac n = n * (n+1) / 2 fInt :: Integral a => a -> a fInt n = n * (n+1) `div` 2 so that, e.g. fFrac (5.0 :: Float) == 15.0 :: Float fInt (5 :: Integer) == 15 :: Integer And all number types are either Integral or Fractional, so surely I should be able to define a single function of type: f :: Num a => a -> a This would seem reasonable, but I think there’s a problem with the last assumption. It is indeed possible for other types to be instances of Num, but not of Integral or Fractional. For example, I could define: instance Num Bool where fromInteger 0 = False fromInteger _ = True (+) = (&&) (*) = (||) abs = id signum _ = True negate = not Now this would probably be pretty dumb (and probably doesn’t comply with
expectations), but is possible. (And also pretty dumb to define it without also define an
instance Integral Bool where ..., but still possible). So I don’t think f :: Num a => a -> a could be possible, since Num by itself (& the dumb Bool instance) has no way to do the division. (At least that I can think of, but would be very interested to hear if there is). Regards, David. From: Beginners <[hidden email]> on behalf of M Douglas McIlroy <[hidden email]> For rational functions that take on integer values at integer _______________________________________________ Beginners mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners |
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