On Fri, Nov 14, 2014 at 08:35:48PM +0100, Bernd Schmidt wrote: > On 11/14/2014 08:19 PM, Segher Boessenkool wrote: > >+ /* If I2 is a PARALLEL of two SETs of REGs (and perhaps some CLOBBERs), > >+ make those two SETs separate I1 and I2 insns, and make an I0 that is > >+ the original I1. */ > >+ if (i0 == 0 > >+ && GET_CODE (PATTERN (i2)) == PARALLEL > >+ && XVECLEN (PATTERN (i2), 0) >= 2 > >+ && GET_CODE (XVECEXP (PATTERN (i2), 0, 0)) == SET > >+ && GET_CODE (XVECEXP (PATTERN (i2), 0, 1)) == SET > >+ && REG_P (SET_DEST (XVECEXP (PATTERN (i2), 0, 0))) > >+ && REG_P (SET_DEST (XVECEXP (PATTERN (i2), 0, 1))) > >+ && !reg_used_between_p (SET_DEST (XVECEXP (PATTERN (i2), 0, 0)), > >i2, i3) > >+ && !reg_used_between_p (SET_DEST (XVECEXP (PATTERN (i2), 0, 1)), > >i2, i3) > > Don't we have other code in combine checking the reg_used_between case?
It doesn't make any sense at all. What I wanted to check is whether splitting the parallel creates a conflict, but I woefully failed at that. Will fix. > >+ && (XVECLEN (PATTERN (i2), 0) == 2 > >+ || GET_CODE (XVECEXP (PATTERN (i2), 0, 2)) == CLOBBER)) > > This probably wants to test for XVECLEN == 3 for the second case. Can > then drop the earlier test. It needs to test there are exactly two SETs, any amount of clobbers, and nothing else. > I think you also need to check that !reg_overlap_mentioned_p between the > two dests and the other set's sources. Only the dest of the new I1 with the sources of the new I2, but yes. Segher
