On Tue, Jul 24, 2018 at 02:28:45AM +0100, Ramsay Jones wrote: > > > On 23/07/18 22:15, Luc Van Oostenryck wrote: > > Since an LSR with an in-range shift count will insert a zero in > > I'm assuming that the LSR shift count N is non-zero (otherwise > LSR(x,0) => x, which means ASR(LSR(x,0),N') != LSR(x,N')), right? Yes. This is normally guaranted at all time. But yes, known that I'm thinking about it, since a zero LSR is not simplified away at expansion time, there is a very small possibility that this simplification is done with an LSR of 0 (at the first run of the simplification loop, if the LSR and the ASR are in two different Basic Blocks and for some reasons the ASR's BB is processed before the LSR's BB (which normally doesn't occurs)). Thank for catching that this! -- Luc -- To unsubscribe from this list: send the line "unsubscribe linux-sparse" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html
