Fractional/negative fixity?

Ben Rudiak-Gould Benjamin.Rudiak-Gould at cl.cam.ac.uk
Fri Nov 10 12:11:58 EST 2006


apfelmus at quantentunnel.de wrote:
>I think that computable real fixity levels are useful, too.

Only finitely many operators can be declared in a given Haskell program. 
Thus the strongest property you need in your set of precedence levels is 
that given arbitrary finite sets of precedences A and B, with no precedence 
in A being higher than any precedence in B, there should exist a precedence 
higher than any in A and lower than any in B. The rationals already satisfy 
this property, so there's no need for anything bigger (in the sense of being 
a superset). The rationals/reals with terminating decimal expansions also 
satisfy this property. The integers don't, of course. Thus there's a benefit 
to extending Haskell with fractional fixities, but no additional benefit to 
using any larger totally ordered set.

-- Ben



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