输入一位十六进制数，输出其十进制表示

比如输入:f
则输出：15
（这里使用小写f 不对大写F进行判断）

方法一（利用AAA）

stack segment
db 10 dup(?)
stack ends

data segment
string db 'please input a num: $'
result db 0ah,0dh,'result is : $'
data ends

code segment
assume cs:code,ds:data,ss:stack
start:
mov ax,data
mov ds,ax

lea dx,string
mov ah,09h
int 21h

mov ah,01h
int 21h

sub al,30h
cmp al,10
jb result_1
sub al,27h
xor ah,ah
AAA
push ax
push ax
lea dx,result
mov ah,09h
int 21h

pop ax

add ah,30h
mov dl,ah
mov ah,02h
int 21h
pop ax
add al,30h
mov dl,al
mov ah,02h
int 21h
jmp jieshu

result_1:
lea dx,result
mov ah,09h
int 21h

add al,30h
mov dl,al
mov ah,02h
int 21h
jmp jieshu

jieshu:
mov ax,4c00h
int 21h

code ends
end start

结果

方法二：(利用 除以进制10)

stack segment
stack ends

data segment
string_1 db 'please input a num: $'
string_2 db 0ah,0dh,'the result is : $'
result db 0,0
data ends

code segment
assume cs:code,ds:data,ss:stack
start:
mov ax,data
mov ds,ax

lea dx,string_1
mov ah,09h
int 21h

mov ah,01h
int 21h

sub al,30h
mov cx,2
mov si,offset result+1
mov ah,0
mov bl,10
cmp al,10
jb shuzi
sub al,27h



shuzi:
div bl
mov [si],ah
dec si
mov ah,0
loop shuzi

mov si,offset result
mov cx,2

lea dx,string_2
mov ah,09h
int 21h

shuchu:
mov dl,[si]
add dl,30h
mov ah,02h
int 21h
inc si
loop shuchu

mov ax,4c00h
int 21h



code ends
end start

总结

• 这里有个坑，使用dos的2号功能后，输出字符的ASCII值会被送入al中。太坑了。
• 一定记得对ax进行保护，这里保护了两次！