原始:(A)=E8H,(R0)=40H,(R1)=20H,(R4)=3AH,(40H)=2CH,(20H)=0FH
变换:
(1)MOV A,@R0 // 将40H单元数据送A, (A)=2CH , 不影响状态位
结果:(A)=2CH,(R0)=40H,(R1)=20H,(R4)=3AH,(40H)=2CH,(20H)=0FH
(2)ANL 40H,#0FH // 40H单元数据与0FH相与, (40H)=0CH, 不影响状态位
结果:(A)=E8H,(R0)=40H,(R1)=20H,(R4)=3AH,(40H)=0CH,(20H)=0FH
(3)ADD A,R4 // A中的值加上R4中的值, (A)=22H, CY=1,AC=1,OV=1
原始:(A)=22H,(R0)=40H,(R1)=20H,(R4)=3AH,(40H)=2CH,(20H)=0FH
(4)SWAP A // A中数据作半字节交换, (A)=8EH, 不影响状态位
结果:(A)=8EH,(R0)=40H,(R1)=20H,(R4)=3AH,(40H)=2CH,(20H)=0FH
(5)DEC @R1 // 20H单元数值减1, (20H)=0EH, 不影响状态位
结果:(A)=E8H,(R0)=40H,(R1)=20H,(R4)=3AH,(40H)=2CH,(20H)=0EH
(6)XCHD A,@R1 // 20H单元低半字节与累加器交换, 不影响状态位
结果:(A)=EFH,(R0)=40H,(R1)=20H,(R4)=3AH,(40H)=2CH,(20H)=08H
追问(3)中的ov是不是该是0呢啊
来自:求助得到的回答