[Haskell] Re: overloading of list operations

[Andrew U. Frank]

On  5 Sep 07  23 28, Brandon S. Allbery KF8NH wrote:
> 
> On Sep 5, 2007, at 21:10 , Tomi Owens wrote:
> 
> > Prelude> let f (a,b) = a * floor (100000/b)
> > Prelude> f(2,5)
> > 40000
> >
> > This function works just as I want it to.
> >
> > Now I try creating a list:
> >
> > Prelude> [(a2+b2,a)| a <- [1..4] , b<- [1..4], a2+b2<20, b<=a]
> > [(2,1),(5,2),(8,2),(10,3),(13,3),(18,3),(17,4)]
> >
> > and this works
> > So now I try to apply the function to the list:
> >
> > Prelude> map (f) [(a2+b2,a)| a <- [1..4] , b<- [1..4], a2+b2<20,
> b<=a]
> >
> > and I get this result:
> >
> > <interactive>:1:5:
> >    Ambiguous type variable `t' in the constraints:
> >      `Integral t' arising from use of `f' at <interactive>:1:5
> >      `RealFrac t' arising from use of `f' at <interactive>:1:5
> >    Probable fix: add a type signature that fixes these type 
> variable
> 
> > (s)
> >
> >
> > I'm sorry, but I don't quite get how to set the type signature and  
> > how it will apply to my function...
> 
> The problem here is that (assuming the a\sup{2} etc. are actually  
> a^2) the (^) operator expects and returns Integrals, but (/) requires 
> 
> a RealFrac.  Thus, the type of your list comprehension is inferred to 
> 
> be [(Integer,Integer)] but needs to be RealFrac a => [(Integer,a)]  
> (or, more simply, [(Integer,Double)].
> 
>    Prelude> let f (a,b) = a * floor (100000/b)
>    Prelude> :t f
>    f :: (RealFrac t1, Integral t) => (t, t1) -> t
>    Prelude> let v :: [(Integer,Double)]; v = [(a^2 + b^2,fromIntegral 
> 
> a) | a <- [1..4], b <- [1..4], a^2 + b^2 < 20, b <= a]
>    Prelude> :t v
>    v :: [(Integer, Double)]
>    Prelude> map f v
>    [200000,250000,400000,333330,433329,599994,425000]
> 
> -- 
> brandon s. allbery [solaris,freebsd,perl,pugs,haskell]
> allbery at kf8nh.com
> system administrator [openafs,heimdal,too many hats]
> allbery at ece.cmu.edu
> electrical and computer engineering, carnegie mellon university   
> KF8NH
> 
> 
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