[Haskell-cafe] noob question

[Daniel Fischer]

Am Dienstag, 26. Februar 2008 01:32 schrieb Ben:
>
> Ok, that makes sense.  There's no num k that's both Fractional and
> Integral, where as in the case where I had the number literals, those
> were two different instances. What's the usual way of working around
> this?  Something like
>
> (\k -> (1/ fromInteger k) ^ k) 3
>
> ?
Or use another exponentiation operator, in this case (**),
(\k -> (1/k)**k) :: (Floating t) => t -> t,
so in
(\k -> (1/k)**k) 3,
fromInteger 3 will have type (Floating t) => t and that will be defaulted to 
Double

Cheers,
Daniel