data segment
tital1 db 'Please input shi jin zi shu (example 13): ','$'
tital2 db 'it is transpranted into shi liu jin zi shu :','$'
tital3 db 'Do you want to have a try again (yes "y",no "n"): ','$'
tital4 db 'welcome to use this program It is made by WQ 2012/5/17','$'
kongzi db 0dh,0ah,'$'
data ends
code segment
assume cs:code,ds:data
start:
mov ax,data
mov ds,ax
retry:
lea dx,kongzi ;换行回车
mov ah,09h
int 21h
lea dx,tital1 ;输出抬头1
mov ah,09h
int 21h
mov al,0 ;初始化al
mov ah,01h ;输入并回显
int 21h
mov bh,al ;放在bx中保存
mov ah,01h
int 21h
mov bl,al
lea dx,kongzi ;换行回车
mov ah,09h
int 21h
lea dx,tital2 ;输出抬头2
mov ah,09h
int 21h
[page]
mov ax,0 ;初始化ax
sub bh,30h ;把ascII码转化为十进制数
mov al,bh ;输入的第一个数(十位数)
mov cl,0ah
mul cl ;al被乘十
sub bl,30h
add al,bl ;彻底转化为十进制数
mov dl,al ;dl 中为十进制数
mov bl,10h ;16
mov cl,al
mov dl,0
l1:cmp cl,bl ;和十六比较看看是否比十六大
jnae l2
inc dl ;看该数里面有几个十六
sub cl,bl
jmp l1
l2: add dl,30h ;比十六小的话第一位数应该为0,
;因为99化成十六进制数为63h第二位不会大于六
mov ah,02h
int 21h
cmp cl,0ah ;cl和十进行比较
jae l3
mov dl,cl
add dl,30h ;十以内的数加30
mov ah,02h
int 21h
jmp tishi
l3: mov dl,cl ;比十大的要变成字母
add dl,37h ;十进制二位数和十六进制abcdef差37h
mov ah,02h
int 21h
tishi:
lea dx,kongzi;控制换行回车
mov ah,09h
int 21h
lea dx,tital3;输出抬头3
mov ah,09h
int 21h
mov ah,01h ;判断是否继续
int 21h
cmp al,'y'
je retry
lea dx,kongzi;控制换行回车
mov ah,09h
int 21h
lea dx,tital4
mov ah,09h
int 21h
over: ;结束
mov ah,4ch
int 21h
code ends
end start
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