On 21 Sep 2006, at 10:46, Robert Stroud wrote:
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.
Aha - light begins to dawn... :-)
Each 2 is a different 2, so the type of 2 can be different each time
it's used, whereas there's only one k so it can only have one type.
Is that right?
Robert
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