But if you get a parameter called f, and you do (f x y), what arity is
f? If applying functions of different arities had the same syntax, you
wouldn't know. But you want to know if arity mismatch is a type error.

On Thu, Feb 12, 2015 at 2:31 AM, Keean Schupke <[email protected]> wrote:
> Not sure why you need arity inference, type inference I have done works fine
> without it. Any given function name generally has the same arity in all
> versions, so you know the required arity from the function definitions. You
> don't need the arity during type inference.
>
> Keean.
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