Re: type conversion discussion

Peter Eisentraut <peter_e(at)gmx(dot)net> writes:
>> I don't think so. The lattice property only says that the set A has a
>> glb within the equivalence class. AFAICT it doesn't promise that the
>> glb will be >= Q, so you can't necessarily use the glb as the function
>> to call.

> Since all functions in A are >=Q by definition, Q is at least _a_ lower
> bound on A. The glb(A) is also a lower bound on A, and since it's the
> greatest it must also be >=Q.

No, you're not catching my point. glb(A) is the greatest lower bound
*within the set of available functions*. Q, the requested call
signature, is *not* in that set (if it were then we'd not have any
ambiguity to resolve, because there's an exact match). The fact that
the set of available functions forms a lattice gives you no guarantee
whatever that glb(A) >= Q, because Q is not constrained by the lattice
property.

regards, tom lane