Michael Paquier wrote:
> On Tue, Aug 4, 2015 at 2:43 PM, Alvaro Herrera <alvherre@2ndquadrant.com> wrote:
> > Maybe the solution to this is to add the concept of "addition" of two
> > lock modes, where the result is another lock mode that conflicts with
> > any lock that would conflict with either of the two operand lock modes.
> That's commutative, as this is basically looking at the conflict table
> to get the union of the bits to indicate what are all the locks
> conflicting with lock A and lock B, and then we select the lock on the
> table that includes the whole union, with a minimum number of them.
Yes.
> Now, let's take for example this case with locks A, B, C, D:
> - Lock A conflicts with ACD
> - B with BCD
> - C with itself
> - D with itself
> What would you choose as a result of add(C,D)? A or B? Or the super
> lock conflicting with all of them?
This appears to me an hypothetical case that I don't think occurs in our
conflicts table, so I wouldn't worry about it.
--
Álvaro Herrera http://www.2ndQuadrant.com/
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