思考题与习题及答案赵晓安.docVIP

思考题与习题4及答案 4－1 试编程实现将20H单元的两位BCD数相乘，结果以BCD码的形式存于21H单元当中。 4-1-1 ORG 0000H MOV 20H,#56H MOV A,20H ANL A,#0FH MOV B,A MOV A,20H ANL A,#0F0H SWAP A MOV R2,A DEC B LOOP: ADD A,R2 DA A DJNZ B,LOOP MOV 21H,A SJMP $ END 4-1-2 ORG 0000H MOV 20H,#56H MOV A,20H MOV R0,#21H MOV 21H,#0 XCHD A,@R0 MOV B,@R0 SWAP A MOV R2,A DEC B LOOP: ADD A,R2 DA A DJNZ B,LOOP MOV 21H,A SJMP $ END 4-1-3 错误做法 ORG 0000H MOV 20H,#56H MOV A,20H ANL A,#0FH MOV B,A MOV A,20H ANL A,#0F0H SWAP A MOV R2,A MUL AB; (A)=1EH DA A; (A)=24H ERROR MOV 21H,A SJMP $ END 4－2 设一个加数在50H～52H单元中，另一个加数在53H～55H中，进位存放在位地址00中， 试编程实现两数求和。 ORG 000H ONE DATA 50H TWO DATA 53H WEI BIT 00H MOV R0,#ONE MOV R1,#TWO ; MOV C,WEI CLR C MOV R2,#3 LOOP: MOV A,@R0 ADDC A,@R1 MOV @R0,A INC R0 INC R1 DJNZ R2,LOOP MOV WEI,C SJMP $ END 4-2-2 ORG 0000H ONE DATA 50H TWO DATA 53H WEI BIT 00H MOV R0,#ONE MOV R1,#TWO CLR C MOV R2,#3 SETB RS0 MOV R0,#58H CLR RS0 LOOP: MOV A,@R0 ADDC A,@R1 INC R0 INC R1 SETB RS0 MOV @R0,A INC R0 CLR RS0 DJNZ R2,LOOP MOV WEI,C SJMP $ END 4－3试用除法指令将20H单元的两位BCD码变成ASCII码后存入21H和22H单元。 4-3 ORG 0000H MOV 20H,#67H MOV A,20H MOV B,#10H DIV AB ADD A,#30H MOV 21H,A ORL B,#30H MOV 22H,B SJMP $ END 4－4从20H单元开始存放一组带符号数，其数目已存在1FH单元。要求统计出大于0、小于0和等于0的个数并存于ONE、TWO、THREE 三个单元。 4-4 ORG 0000H ONE EQU 30H TWO EQU 31H THREE EQU 32H NUM EQU 1FH MOV NUM,#0FH MOV R0,#20H;-1,2,34,0A0H,0.... MOV R1,#ONE LOOP: MOV @R1,#0 INC R1 CJNE R1,#34H,LOOP LOOP1: MOV A,@R0 JZ MTHREE JB ACC.7,MTWO INC ONE SJMP DJNZZ MTHREE: INC THREE SJMP DJNZZ MTWO: INC TWO DJNZZ: INC R0 DJNZ NUM,LOOP1;NUM=0FH SJMP $ END 4－5在用JMP指令实现128分支的程序中，试用LJMP代替AJMP指令。修改后的程序有何特点？最多能实现多少分支？ 4-5 AJMP指令表 128分支占128*