指令試題和答案

1、1 MACR *AR3-, *AR4-, A, BABTFRCTAR3AR40100H0200H指令執(zhí)行前ABTFRCTAR3AR40100H0200H指令執(zhí)行后00 0000 1000H00 0000 1000H00 0000 2200H 00 000D 0000H0003H 0345H0 00100H 00FFH0200H 01FFH0345H 0345H0400H 0400H（1） RA+(AR1)*(AR2) à B R1000H+0345H*0400H=R1000H+D,1400H=RD,2400H=D,0000H à B （2） (AR1)=0345H 

2、4; T （3） AR1-1 à AR1=00FFH, AR2-1 à AR2=01FFH2 MAS *AR1+, *AR2+, A, BABTFRCTAR1AR20100H0200H指令執(zhí)行前ABTFRCTAR1AR20100H0200H指令執(zhí)行后00 002F 2850H00 002F 2850H00 0000 2200H 00 0015 0050H0003H 0345H1 10100H 0101H0200H 0201H0345H 0345H 0400H 0400H（1） A-(AR1)×(AR2)<<1à B (AR1)×(A

3、R2)=0345H×0400H=D,1400H D1400H<<1=1A,2800H 2F,2850H-1A,2800H=15,0050HàB （2） (AR1) à T=0345H （3） AR1+1àAR1=0101H，AR2+1 àAR2=0201H3 DADD *AR1-, A, BABAR1C16055EH 055FH 0560H 0561H指令執(zhí)行前ABAR1C16055EH055FH 0560H0561H指令執(zhí)行后00 8006 2090H00 8006 2090H00 0000 0000H 00 B55B 33C3H

4、055F H 055DH1 1 1333H 1333H3555H 3555H5777H 5777H 7999H 7999H A 8006 2090H + (AR1) 3555 1333H = B55B 33C3H à B（練習(xí)） LDM *AR1, ADST B, *AR2+BIT *AR1, 12ABTCAR1AR20002H0082H1882H1883H指令執(zhí)行前ABTCAR1AR20002H0082H1882H1883H指令執(zhí)行后00 0000 0000H 00 0000 0999H00 6500 F009H 00 6500 F009HX 11882H 0002H1882H 1

5、884H0999H0999H2700H 2700H 0516H 6500H0333H F009H（1） LDM的真實(shí)尋址是：(0002H)，所以(0002H)=0999H à A=00 0000 0999H，0002H à AR1=0002H （2） BH=6500H à (1882H) BL=F009H à (1883H) AR2+2 à AR2=1884H （3） (AR1)(15 - 12)=(0002H)(3)=0999H(3) = 1001B(3) = 1 à TC4 LDM *AR1, ABIT *AR1, 7MIN BS

6、TH A, 2ABDPTCAR10002H0062H0082H10E2H指令執(zhí)行前ABDPTCAR10002H0062H0082H10E2H指令執(zhí)行后00 0000 0000H 00 0000 2700H00 0050 F900H 00 0000 2700H1 1X 110E2H 0062H0999H 0999H2700H 2700H 0516H 0000H0333H 0333H(1) MMR間接尋址：AR1 = 10E2H = 0001,0000,1110,0010B à0000,0000,0|110,0010B = 0062H à AR1 (AR1) = ( 0062H

7、 ) = 2700H à A(2) (AR1)(15-7) = (2700H)(8) = (0010,0111,0000,0000B) (8) = 1 à TC(3) A2700H < B50,F900HA2700H à B(4) 即STH A, 2，所以：Hi(A) = 0000H à (0082H)1 編寫匯編程序，計(jì)算矩陣和的乘積。.title “matrix” .mmregs.bss a1,3.bss a2,3.bss b1,3.bss b2,3.bss c, 4.datatable: word 55, -2, 3, 85, -75, 5

8、word -35, -6,25, 5, 3, 1.textstart: STM #a1, AR1 RPT #11MVPD table, *AR1+STM #a1, AR2STM #a2, AR3STM #b1, AR4STM #b2, AR5STM #c, AR6RSBX FRCTRPTZ A, #2MAC *AR2+, *AR4+, A, BSTH B, *AR6+ ; c11STM #a1, AR2STM #b1, AR4RPTZ A, #2MAC *AR2+, *AR5+, A, BSTH B, *AR6+ ; c12RPTZ A, #2MAC *AR3+, *AR4+, A, BSTH

9、 B, *AR6+ ; c21STM #a2, AR3STM #b2, AR5RPTZ A, #2MAC *AR3+, *AR5+, A, BSTH B, *AR6 ; c22.end2編寫匯編程序，求數(shù)列和相應(yīng)元素的最大值，結(jié)果保存為數(shù)列。（至少需寫出關(guān)鍵偽指令和指令步驟） （附：指令“MAX ”的功能代數(shù)式為）.title “compare.asm”.mmregs.bss x,3.bss y,3.bss w,3.datatable: .word -19, 27, 5, 36, -23, -18,.textstart: STM #2, AR0STM #x, AR1STM #x, AR2STM #y, AR3RPT #5MVPD table, *AR1+ ; e