用汇编怎么写“利用查表方法把DX的内容(16进制数)转换成ASCII码，保存到100H开始的内存单元中”

Question

帮忙啊

Answer 1

changeline macro;定义一个用于换行的宏（无参数）
mov dl, 0dH
mov ah, 02H;回车
int 21H
mov dl, 0aH
mov ah, 02H;换行
int 21H
endm

data segment
x dw 0
y dw 0
da1 dw 0
da2 dw 0
string1 db 'Input incorrectly, please input again:', 0dH, 0aH, '$'
string2 db 'Please input the first number(enter for ending):', 0dH, 0aH, '$'
string3 db 'Please input the seconde number(enter for ending ) :', 0dH, 0aH, '$'
string4 db 'Please input the operation you want:(1 for add; 2 for sub; 3 for mul; 4 for div)', 0dH, 0aH, '$'
string5 db 'Please input the system you want:(1 for 2,2 for 8,3 for 10,4 for 16 ) ', 0dH, 0aH, '$'
string6 db 'The key is:', 0dH, 0aH, '$'
data ends

input macro num;定义一个用于输入的宏,并判断输入是否正确
local again, mistake, last;local伪指令防止多次调用造成错误
mov cx, 0400H
xor bx, bx;bx清零
mov dl, 10

again: mov ah, 01H;接收数据（字符）
int 21H
cmp al, 0dH;判断是否是回车,遇回车则结束输入
jz last
cmp al, '0';和字符0比较大小
jb mistake;小于等于跳转
cmp al, '9';和字符9比较大小
ja mistake
xor ah, ah
sub al, 30H;与0的ASCII码相减
xchg ax, bx;将ax中的数保存在bx
mul dl;al与dl相乘，因为用户输入时是先输高位，再输次高位，所以高位和次高位之间相差10
add bx, ax;bx中的数就是用户刚刚输入的数
dec ch
jnz again
jmp last
mistake: lea dx, string1
mov ah, 09H
int 21H
mov ch, 4
xor bx, bx
jmp again
last: mov num, bx ;将用户输入的数放入内存单元num中
endm

code segment 'code'
assume ds: data, cs: code
start:
mov ax, data
mov ds, ax
lea dx, string2
mov ah, 09H
int 21H
input x
changeline

lea dx, string3
mov ah, 09H
int 21H
input y
changeline

call proc3
call proc1

mov ax, da1
call proc2
mov ah, 4cH
int 21H

proc1 proc;该子程序用于判断该用何种进制输出
changeline;调用用于换行的宏
lea dx, string5
mov ah, 09H
int 21H

again1: mov ah, 01H
int 21H
cmp al, '1'
jnz s21
mov da2, 2
jmp last1
s21: cmp al, '2'
jnz s31
mov da2, 8
jmp last1
s31: cmp al, '3'
jnz s41
mov da2, 10
jmp last1
s41: cmp al, '4'
jnz mistake1
mov da2, 16
jmp last1
mistake1: lea dx, string1
mov ah, 09H
int 21H
jmp again1
last1:
changeline
lea dx, string6
mov ah, 09H
int 21H
ret
proc1 endp

proc2 proc
xor dx, dx
mov bx, da2
div bx
push dx;将余数放到堆栈中保存
cmp ax, 0
jz then
call proc2
then:pop dx;将结果从堆栈中取出放入dx中
or dl, 30H
cmp dl, 39H
jbe next0
add dl, 07H
next0: mov ah, 02H;用单字符显示输出
int 21H
ret
proc2 endp

proc3 proc;该子程序用于判断用户将要进行何种运算
lea dx, string4
mov ah, 09H
int 21H
again: mov ah, 01H
int 21H

cmp al, '1'
jnz s2
mov ax, x;x与y进行相加
add ax, y
jmp last
s2: cmp al, '2'
jnz s3
mov ax, x
sub ax, y
jmp last
s3: cmp al, '3'
jnz s4
mov ax, x
mov bx, y
mul bx
jmp last
s4: cmp al, '4'
jnz mistake
mov ax, x
xor dx, dx
mov bx, y
div y
jmp last
mistake: lea dx, string1
mov ah, 09H
int 21H
jmp again
last: mov da1, ax
ret
proc3 endp

code ends

end start