[Haskell-beginners] Re: question about styles of recursion

Brent Yorgey byorgey at seas.upenn.edu
Fri Mar 27 08:42:56 EDT 2009

On Fri, Mar 27, 2009 at 09:12:46AM +0000, 7stud wrote:
> Daniel Fischer <daniel.is.fischer <at> web.de> writes:
> >
> > (/) in the first equation is at type Double -> Double -> Double.
> >
> Then why don't I get an error here:
> Prelude> 2 / 4
> 0.5

There are several things going on here.  The first is that numeric
literals are overloaded; integral numeric literals (like 2) can be of
any numeric type.  What actually happens is that they get wrapped in a
call to fromIntegral.

The other thing going on is ghci's type defaulting: since it doesn't
otherwise know what type this expression should be, it picks Double.
(It would also pick Integer if that worked, but Integer doesn't work
here because of the (/).)

> Prelude> 2 / fromIntegral 4
> 0.5

>From the above discussion you can see that this is exactly the same as
the first expression.

> And why does this happen:
> Prelude> let x = 2
> Prelude> :type x
> x :: Integer

This is because of the defaulting again.  Since there are no
constraints on the type of x this time, it picks Integer.  

> Prelude> x / fromIntegral 4
> <interactive>:1:0:
>     No instance for (Fractional Integer)
>       arising from a use of `/' at <interactive>:1:0-17
>     Possible fix: add an instance declaration for (Fractional Integer)
>     In the expression: x / fromIntegral 4
>     In the definition of `it': it = x / fromIntegral 4

Of course, now that x has type Integer, this is not well-typed; you
can't divide Integers.

> And how do I read this type:
> Prelude> :type fromIntegral
> fromIntegral :: (Num b, Integral a) => a -> b
> What does the => mean?

The => indicates class constraints.  You can read this as
"fromIntegral has type a -> b, as long as b is an instance of the Num
class, and a is an instance of the Integral class."  That is,
fromIntegral can convert a value of any Integral type (Int, Integer)
to a value of any Num type.


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