explicit signatures and default for integer literals

Dinko Tenev dinko.tenev at gmail.com
Tue May 31 03:55:04 EDT 2005

On 5/31/05, Seth Kurtzberg <seth at cql.com> wrote:
> Because I'm not smart enough to understand it?
> ;-)

OK, sorry about the notation (I just didn't feel like doing so much
typing, you know :)

Here goes...

First we observe that, g = new . flip zip [0..], so, without the type
specification, it has the general type (New [(a, b1)] b, Num b1, Enum
b1) => [a] -> b, as reported by GHC.

Then we infer from

    (1) g :: (New [(u, v)] w, Num v, Enum v) => [u] -> w


    (2) instance New [(a, b)] (Map a b)

that in (New [(u, v)] w), w can only be (Map u v), so subst (Map u v)
for w to obtain

    (3) g :: (New [(u, v)] (Map u v), Num v, Enum v) => [u] -> Map u v

Furthermore, for g we have already specified

    (4) g :: Ord a => [a] -> Map a Int

so finally substituting Int for v should give us

    (5) g :: (New [(u, Int)] (Map u Int), Num Int, Enum Int, Ord Int)
=> [u] -> Map u Int

Is there any good reason why GHC won't do this?  Particularly, is
there any good reason not to infer (New [(u, v)] (Map u v)) from (New
[(u, v)] w) as in (1), (2) --> (3) above?


D. Tenev

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