On 20 Sep 2006, at 17:28, Christian Sievers wrote:
However, if I type an apparently equivalent let expression into Hugs
directly, then I get the value 4 as expected
let k = 2 ; f :: Int -> Int -> Int ; f x y = x * y in f k k
Why is there a difference in behaviour?
Here, there is no defaulting, 'k' has the polymorphic type you
expect, and
the use of 'k' as an argument to the monomorphically typed 'f'
chooses the
right instance of 'Num'.
Well, there is no defaulting at the stage of type checking when k
is given
type Num a => a, the monomorphism restriction applies and this type
is not
generalised to forall a . (Num a) => a, then the use of k forces
the type
variable a to be Int, and then there is no longer any need for
defaulting.
So k gets a monotype which is determined by its usage, you cannot
do e.g.
let k = 2 ; f :: Int -> Int -> Int ; f x y = x * y in (f k k, 1/k)
whereas let k :: Num a => a; k = 2; ... is possible.
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?
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
transparency.
Thanks.
Robert
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