On Tue, Feb 21, 2006 at 01:07:51PM -, Simon Peyton-Jones wrote:
> I'm not sure this works. Consider this
>
> newContinuation (\k -> return (callContinuation k)) ...
>
> The partial application (callContinuation k) has no 's' in its type, and so
> can go anywhere.
Ah, you are right. si
I'm not sure this works. Consider this
newContinuation (\k -> return (callContinuation k)) ...
The partial application (callContinuation k) has no 's' in its type, and so can
go anywhere.
Simon
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