分析下列程序段: BUF DW 2456H, 7816H, 5731H, 896AH MOV BX, OFFSET BUF MOV AL,2 XLAT MOV CL, AL MOV AL,6 XLAT INT 3 程序段执行后结果:CL=();AL=()
现有(DS)=2000H,(BX)=0100H,(SI)=0002H,(20100H)=12H,(20101H)=34H,(20102H)=56H,(20103H)=78H,(21200H)=2AH,(21201H)=4CH,(21202H)=B7H,(21203H)=65H,试说明下列各条指令执行完后AX寄存器的内容。MOV AX,[BX]。
现有(DS)=2000H,(BX)=0100H,(SI)=0002H,(20100H)=12H,(20101H)=34H,(20102H)=56H,(20103H)=78H,(21200H)=2AH,(21201H)=4CH,(21202H)=B7H,(21203H)=65H,试说明下列各条指令执行完后AX寄存器的内容。MOV AX,[1200H]。
已知SP=25H,PC=4345H,(24H)=12H,(25H)=34H,(26H)=56H,当执行RET指令后,SP=(),PC=()。
现有(DS)=2000H,(BX)=0100H,(SI)=0002H,(20100H)=12H,(20101H)=34H,(20102H)=56H,(20103H)=78H,(21200H)=2AH,(21201H)=4CH,(21202H)=B7H,(21203H)=65H,试说明下列各条指令执行完后AX寄存器的内容。MOV AX,1200H。
已知DS= ES =1000H执行下列程序: MOV SI,101H MOV DI,100H MOV CX,4 LP1:MOV AL,[ SI] MOV[ DI] ,AL INC SI INC DI LOOP LP1 INT 3 DS:101H DB 69H,23H,56H,3AH 程序运行后结果为:(10100H)= ()H(10101H)= ()H(10102H)= ()H(10103H)= ()H
有如下程序: START: MOV BL,67H MOV AL,BL MOV CL,4 SHR AL,CL MOV BH,AL OR BH,30H MOV AL,BL AND AL,OFH MOV BL,AL OR BL,30H INT 3 运行上面程序后,BH=(),BL=()
有程序如下: ORG 2800H BUF DB 67,4,57,34,89,123 START:MOV SI, 04 MOV BX, OFFSET BUF MOV AL, [BX+ SI] CALL SUB1 INT 3 SUB1: PUSHF MOV AH, OOH MOV BL, OAH DIV BL POPF RET该程序运行后,AH= ()AL=()
现有(DS)=2000H,(BX)=0100H,(SI)=0002H,(20100H)=12H,(20101H)=34H,(20102H)=56H,(20103H)=78H,(21200H)=2AH,(21201H)=4CH,(21202H)=B7H,(21203H)=65H,试说明下列各条指令执行完后AX寄存器的内容。MOV AX,BX。
若NUM单元中的内容为23H,程序段如下:MOV AL, NUMOR AL, ALJZ ZERO:JS NUBYS;PLUS: MOV DL,‘+’;JMP EXITNUBYS: MOV DL.‘一’JMP EXITZERO: MOV DL.‘0’EXIT: MOV AH,2INT 21H......;说明程序执行后,屏幕上显示什么?
阅读程序,说明程序执行后AL,CL,CH,BL的值;DA1 DB83H,72H,61H,94H,5AHMOV CX,WORD PTR DA1AND CX,OFHMOV AL,DA1 +3MOV BL,DA1上述指令序列执行后,AL=(),CL=(),CH=(),BL=()
已知DS=ES=2000H执行下列程序: MOV SI,301H MOV DI,300H MOV CX,4 CLD REP MOVSB INT 3 DS:301H DB 22H,33H,44H,55H 程序运行后结果为:(20300H)=()H(20301H)=()H (20302H)=()H(20303H)=()H
若(DS)=2000H,(BX)=0050H,(20050H);12H,(20051H)=34H,执行MOV AX,[BX]指令后,AX寄存器中的内容是()。
已知 ( R0 )=20H, (20H )=36H, (21H) =17H, (36H) =34H, 执行过程如下:MOV A , @R0 ; (A)=MOV R0 , A ; (R0)=MOV A , @R0ADD A , 21H ; (A)=ORL A , #21H ; (A)=RL A ; (A)=MOV R2 , ARET
现有如下程序段include "stdio.h"main(){ int k[30]={12,324,45,6,768,98,21,34,453,456};int count=0,i=0;while(k[i]){ if(k[i]%2==0||k[i]%5==0)count++;i++; }printf("%d,%d\n",count,i);}则程序段的输出结果为
某发电厂有三台机组并列运行,其耗量特性分别为: 2F12.80.26P0.0015PG1G1(T/h)2F23.50.29P0.0015PG2G2(T/h) 2F34.00.17PG30.0015PG3(T/h) 机组功率约束条件为: 20MWPG150MW 20MWPG2100MW20MWPG3100MW 当负荷功率为200MW时机组间负荷的最优分配为()。
若有如下程序: include"stdio.h" void main() {FILE *fp; fp=fopen("test","wb"); fprintf(fp,"%d%.1f,%c%c",5,238.41,'B','h'); fclose(fp); } 则程序运行后向文件输出的结果是()。
以下程序段中,有数据2000H的字单元的偏移量分别是()。\nORG20H\nVAR1DB0,’0’,20H,0,20H\nDWVAR1
已知物理地址(371F0H)=12H,(371F1H)=34H,(371F2H)=56H,如从地址371F0H中取出一个字的内容是34
已知,(DS)=2000H,(BX)=100H, (SI)=02H, 从物理地址20100H单元开始,依次存放数据12H、34H、56H、78
有如下程序:include<stdio.h>main(){int x=23; do {Printf("%d",x--);} while(! x);}该程序的执
分析下列程序段:BUF DW 2456H, 7816H, 5731H, 896AHMOV BX, OFFSET BUFMOV AL,2XLATMOV CL, ALMOV AL,6XLATINT 3程序段执行后结果:CL=();AL=()
执行以下程序后,BCDBUF+1字节单元的内容是()H 。 DATA SEGMENT ASCBUF DB 39H, 38H, 35H, 37H,
25、已知BUF DW ‘AB’,则汇编后BUF+1单元存放的内容为 H。