[Haskell] Re: Implicit type of numeric constants
sievers at math2.nat.tu-bs.de
Thu Sep 21 13:54:52 EDT 2006
Robert Stroud wrote:
> Thanks - that's a helpful example. But why is the following not
> equivalent to the untyped "k = 2" case:
> let f :: Int -> Int -> Int ; f x y = x * y in (f 2 2, 1/2)
> Does the type of 2 effectively get decided twice, once as an Int, and
> once as a Fractional, and is this the "repeated computation" that the
> monomorphism restriction is intended to prevent?
2 has the fixed polymorphic type Num a => a, which gets resolved
at each occurance. This resolving happens at compile time, so there
is no repeated computation at run time involved. However, you need
to somehow get both a 2::Int and a 2::Double which may be seen as a
trivial case of this repeated computation.
> Otherwise, I would have expected that it wouldn't make any difference
> whether I used a named 2 or an anonymous 2, but imposing the
> monomorphism restriction on the named 2 seems to break referential
Only if you expect referential transparency for implicitly typed values.
All you have to do is say
k :: Num a => a
k = 2
and everything is fine.
More information about the Haskell