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2017 Nov 11
2
Update control flow graph when splitting a machine basic block?
...implicit %r0, implicit %r1, implicit-def %sp, implicit-def %r0, debug-location !25 DBG_VALUE 2, 0, !17, !18, debug-location !27 DBG_VALUE debug-use %r0, debug-use _, !16, !18, debug-location !26 tCMPi8 %r0, 2, 14, _, implicit-def %cpsr, debug-location !32 t2IT 11, 28, implicit-def %itstate %r0 = tMOVi8 _, 1, 11, %cpsr, implicit %r0, implicit %itstate tPOP_RET 11, %cpsr, def %r4, def %r6, def %r7, def %pc, implicit %r0, implicit %r4, implicit killed %itstate, debug-location !44 %r1 = t2MOVi 2, 14, _, _ t2B %bb.3.for.body, 14, _ Note that a terminator tPOP_RET is befor...
2017 Nov 11
2
Update control flow graph when splitting a machine basic block?
> The right way to update the CFG very much depends on how you're > transforming it. I would like to export the CFG for control flow checking. Theoretically, it should be possible for a compiler to know every target of every control flow instruction, except for computed targets that are not known at compile-time. When a machine basic block is split between two branches, as shown below:
2015 Sep 17
2
Register Number
...The 0x01 would be the encoding used in generating the binary. The D0 has id 14 on ARM because there are 13 other registers preceding it: namespace ARM { enum { NoRegister, APSR = 1, APSR_NZCV = 2, CPSR = 3, FPEXC = 4, FPINST = 5, FPSCR = 6, FPSCR_NZCV = 7, FPSID = 8, ITSTATE = 9, LR = 10, PC = 11, SP = 12, SPSR = 13, D0 = 14, ... -Krzysztof -- Qualcomm Innovation Center, Inc. is a member of Code Aurora Forum, hosted by The Linux Foundation
2015 Sep 17
2
Register Number
Dear all, in my TestRegisterInfo.td file, I defined a register like this: class TestReg<bits<6> enc, string name> : Register<name> { let HWEncoding{5-0} = enc; let Namespace = "TEST"; } def D0 : TestReg<0x01, "d0">, DwarfRegNum<[1]>; but when I compile, the result I have in TestGenAsmMatcher.inc is this: case 'd': // 7