Fwd: [Haskell-beginners] Fractional Int
wingedtachikoma at gmail.com
Fri Mar 20 18:01:22 EDT 2009
Sorry, forgot to reply to all.
---------- Forwarded message ----------
From: Sean Bartell <wingedtachikoma at gmail.com>
Date: Fri, Mar 20, 2009 at 5:58 PM
Subject: Re: [Haskell-beginners] Fractional Int
To: Zachary Turner <divisortheory at gmail.com>
For a type "a" to be Fractional requires there to be:
(/) :: a -> a -> a
You can't divide an Int by another Int and (in general) get a third
Int. You would probably want something like a "Fractionable"
(/) :: a -> a -> b
which would result in a Rational, but Haskell doesn't have this.
2009/3/20 Zachary Turner <divisortheory at gmail.com>
> Why is there no instance of Fractional Int or Fractional Integer? Obviously integers are fractions with denominator 1. I was just doing some basic stuff to get more familiar with Haskell, and was seriously racking my brain trying to figure out why the following wouldn't work:
> intToString :: Int -> [Char]
> intToString n | n<10 = chr (n + (ord '0')):
> intToString n =
> let q = truncate (n/10)
> r = n `mod` 10
> o = ord '0'
> ch = chr (r + o)
> in ch:(intToString q)
> (yes, this ends up converting the string in reverse, but that's another issue :P)
> I later realized that I could use members of the Integral typeclass such as divMod, mod, etc to make this better, but nonetheless, why should truncate(n/10) be invalid, when n is an Int? changing it to truncate((toRational n)/10) works, but I would expect Integers to already be rational.
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