1、(4) LDS SI,BXDI MOVSI,BXEA= BX+DI=56H, PA=DS*16+EA=91D0H+56H =9226H SI=(09226H)=00F6H, DS=(09228H)=1E40HEA= SI=00F6H, PA=DS*16+EA=1E400H+00F6H=1E4F6H(1E4F6H)= BX=0024H(5) XCHG CX,BX+32H XCHGBX+20HSI,AXEA= BX+32H=56H, PA=DS*16+EA=91D0H+56H =9226H (09226H)= CX=5678H , CX=(09226H)=00F6H EA= BX+20H+SI=5
2、6H,PA=DS*16+EA=91D0H+56H =9226H AX=(09226H)=5678H ,(09226H) = AX=1234H3.2设DS=1000H,SS=2000H,AX=1A2BH,BX=1200H,CX=339AH,BP=1200H,SP=1350H,SI=1354H,(11350H)=0A5H,(11351H)=3CH,(11352H)=0FFH,(11353H)=26H,(11354H)=52H,(11355H)=OE7H,(126A4H)=9DH,(126A5H)=16H,(21350H)=88H,(21351H)=51H下列各指令都在此环境下执行,在下列各小题的空
3、格中填入相应各指令的执行结果。 (1)MOV AX,1352H AX= 解:AX=1352H (2)MOV AX,1352H ;AX=PA=DS*16+EA=10000H+1352H=11352H (11352H)=0FFH,(11353H)=26H AX=26FFH (3)MOV 0150HBX,CH (11350H)= (11351H)= EA=BX+0150H=1350H PA=DS*16+EA=10000H+1350H=11350H, CH=33H (11350H)=33H, (11351H)的值不变 ,(11351H)=3CH(4) MOV AX,0150HBP AX=_EA= BP
4、0150H1350H PA=SS*16+EA=20000H+1350H=21350H AX=5188H5)POP AX ; AX=_,SP=_EA= SP=1350H AX=5188H, SP=1350H+2H=1352H(6)ADDSI,CX(11354H)=_,(11355H)=_,SF=_ZF=_, PF=_, CF=_, OF=_EA=SI=1354H, PA=DS*16+EA=10000H+1354H=11354HCX=339AH, (11354H)=52H,(11355H)=OE7H0E752H+339AH=11AECH-(11355H): (11354H) (11354H) =0
5、ECH, (11355H)= 1AH(11354H) =0ECH, (11355H)= 1AHCF=1,ZF=0,PF(低八位奇偶校验):0ECH= 11101100BPF=0SF(最高位状态),1H=0001BSF=0 OF(溢出标志) 0E752H1110011101010010B 339AH=11001110011010B 111001*0+ 11001110011010 10001101011101100(7)SUB BH,0150HBXSI_,(7)SUB BH,0150HBXSIBH=_,SF=_,ZF=_,PF=_,CF=_,0F=_EA=0150H+BX+SI=26A4H;PA
6、=DS*16+EA=10000H+26A4H=126A4H; (126A4H)=9DH,BH=12HBH=75H, SF=0,ZF=0,PF=0, CF=1 ,OF=0(8)INC BYTE PTR 0152HBX(11352H)=_,(11353H)=_,CF=_EA=0152H+ BX= 1352H,PA=DS*16+EA=11352, (11352H)=0FFH, (11352H)=00H, (11353H)= 26H, 不影响CF(9)INC WORD PTR 0152HBXEA=0152H+ BX= 1352H, PA=DS*16+EA=11532, (11352H)=0FFH, (
7、11353H)= 26H (11352H)=00H, (11353H)= 27H, 不影响CF (10)SAR BYTE PTR 0150HBX,1(11350H) =_, CF=_, OF=_EA=BX+0150H=1350H PADS*16+EA=11350H, (11350H)=0A5H= 10100101B 11010010B=0D2H, CF=1,OF=0(当移位数为1是,最高位不变则OF=0)(11)SAL BYTE PTR 0150HBX,1(11350H)=_,CF=_,OF=_EA=BX+0150H=1350H,PA=DS*16+EA=11350, 01001010B=4AH
8、, CF=1,OF=1 3、3 设下列各转移指令的第一字节在内存中的地址为CS=2000H和IP=016EH,且环境均为DS=6000H,BX=16C0H,(616C0H)=46H,(616C1H)=01H,(616C2H)=00H,(616C3H)=30H,(61732H)=70H,(61733H)=17H。写出下列各无条件转移指令执行后CS和IP值。个指令左首的16进制编码是该指令的机器码。指令中的目的地址用相应的标号表示。(1)EBE7 JMP SHOURT AGAIN(2)E90016 JMP NEARPTR OTHER(3)E3 JMP BX(4)EA46010010 JMP FAR
9、 PROB(5)FF67 JMP WORD PTR 0072HBX (6)FFEB JMP DWORD PTR BX(1)E7补码为-19,IP目标=IP源+2+EA(即-19) =016EH+2-19=0157H 因为段内寻址,所以cs=2000H不变(2)IP目标=IP源+3 +EA=016EH+3+1600H=1771H 因为段内寻址 所以cs=2000H不变(3) IP=16C0H, 因为段内寻址 所以cs=2000H不变(4)段间寻址,有机器码可看出IP=0146H CS=3000H(5)段内寻址,所以CS=2000H不变 DS*16+0072H+BX=61732H (61732H)
10、=70H,(61733H)=17H IP=1770H(6)PA=DS*16+BX=60000H+16C0H=616C0H (616C0H)=46H (616C1H)=01H IP=0146H (616C2H)=00H (616C3H)=30H CS=3000H34 阅读下列各小题的指令序列,在后面空格中填入 该指令序列的执行结果。(1) MOV BL,85H MOV AL,17H ADD AL,BL DAAAL=_, BL=_, CF=_17H+85H9CHALDAA 压缩的BCD码加法十进制调整指令。(AL的低4位9或AF=1,ALAL+06H,AF1;AF是辅助进位标志用以标志D3向D4的
11、进位AL的高4位9或CF=1,ALAL+60H,CF1;)AL=9CH+ 06H=0A2HAL=0A2H+60H=02H, BL=85H CF=1(2) MOV AX,BX;NOT AX; ADD AX,BX; lNC AXAX=_,CF=_INC不影响标志位 AX=0000H, CF=0(3)MOV AX,0FF60H ; STC ; MOV DX,96 XOR DH,0FFH ; SBB AX DXXOR 命令 会使 CF0,OF0 96=60H,AX=0000H, CF=0(4)MOV BX,0FFFEH ; MOV CL,2;SAR BX,CLBX=_,CF=_0FFFEH=11111
12、11111111110B 1111111111111111B,CF=0 1111 1111 1111 1111B,CF=135 阅读分析下列指令序列ADD AX,BXJNO LlJNO L2SUB AX,BXJNC L3JNO L4JMP L5 (1)AX=14C6H,BX=80DCH ADD AX,BX OF=0,CF=0; L1(2)AX=0B568H,BX=5487H OF=0,CF=1;(3)AX=42C8H,BX=608DHADD AX,BX CF=0 ,OF=1, AX=0AC55HSUB AX,BX;CF=0,OF=0;L3(4) AX=0D023H,BX=9FDOHOF=1,CF=1,AX=6FF3HSUB AX,BX ;CF=1,OF=1;L5(5)AX=9FDOH,BX=0D023H3.6 AND AL,AL JZ BRCHl RCR AL,1 JZ BRCH2 RCL AL,1 INC AL JZ BRCH3上述程序运行后,试回答:(1)当AL =00H时, 程序转向BRCHl(2)当AL =01H时, 程序转向BRCH2