>Are we guaranteed that r1 and r4 will both see the value -1?
I think the current model guarantees that r1 will be -1 but doesn't
actually guarantee that r4 will be anything in particular (except it won't
be out of thin air). The realities of implementing the r1 = -1 guarantee
will almost certainly force r4 = -1 for all implementations that I can
imagine but I don't think we need to guarantee anything.
Martin Trotter
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