[Haskell-beginners] Ratio data constructor

[David McBride]

The Ratio type is a fraction. If a is Integer, it is an infinite precision
fraction. You can divide it arbitrarily and it will never lose any
precision (though it will take more space).

Reduce is designed to create a fraction from two numbers, not solve it. so
if you supply 5 and 10, it will return a Ratio 1/2, because 5 is the
greatest common denominator.

The % operator appears to take arbitrary numbers but it seems designed to
intend to keep the negative sign on the numerator, so as to maintain some
invariant in the data.

On Tue, Jul 20, 2021 at 4:43 PM Galaxy Being <borgauf at gmail.com> wrote:

> I'm investigating rational numbers with Haskell. This is the source I've
> found
>
> data Ratio a = !a :% !a deriving (Eq)
>
> reduce ::  (Integral a) => a -> a -> Ratio a
> {-# SPECIALISE reduce :: Integer -> Integer -> Rational #-}
> reduce _ 0              =  ratioZeroDenominatorError
> reduce x y              =  (x `quot` d) :% (y `quot` d)
>                            where d = gcd x y
> (%) :: (Integral a) => a -> a -> Ratio a
> x % y =  reduce (x * signum y) (abs y)
>
> The Ratio data type would seem to be a parameterized type with two
> parameters of the same type that must be "settled" in that they're not to
> be lazily dealt with. Then the :% is the data constructor, the : meaning
> it's a data constructor and not just an operation function. So this could
> have been
>
> data Ratio a = :% !a !a deriving (Eq)
>
> correct? But then what confuses me is in reduce, why
>
> reduce x y  =  (x `quot` d) :% (y `quot` d)
>
> and not just %? We have :% defined in the data type and then (%) defined
> as a function. What is going on here?
>
> --
> ⨽
> Lawrence Bottorff
> Grand Marais, MN, USA
> borgauf at gmail.com
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