CODE SEGMENT
ASSUME CS:CODE
MAIN: MOV AH,1
INT 21H
MOV BL,AL
INT 21H
MOV CL,AL
INT 21H
&#[……]
code segment
assume cs:code
main:mov ah,1
int 21h
MOV BL,AL
MOV AH,2
MOV DL,13
INT 21H
MOV DL,[……]
data segment
buf db 100,101 dup(0)
buf1 db ‘qing shu ru’,13,10,'(ge shu zai 100 yi nei)=’,’$’
data ends
code segment
assume ds:data,cs:code
main:mov dx,data
mov ds,dx
lea dx,[buf1]
[……]
data segment
int1 db ‘935797238’,13,10,24h
int2 db ‘843758912’,13,10,24h data ends
code segment
assume cs:code,ds:data
main: mov ax,data
mov ds,ax
lea dx,[int1][……]
；设变量int1和int2的定义如下 ；int1 ‘935797238’ ；int2 ‘843758912’ ；编写程序段把int1和int2中以字符串形式存放的两个长整数相加，结果仍然以同样的形式放会变量int1中，不计溢出！
data segment
int1 db ‘935797238’,13,10,24h
int2 db ‘843758912’,13,10,24h
data ends
code segment
assume cs:code,ds:data[……]
data segment
buf db ‘THE ASCII CODE OF[‘,’$’
buf1 db ‘] IS ‘,’$’
buf2 db 13,10,’$’
data ends
code segment
assume cs:code,ds:data
main:mov ax,data
mov ds,ax
lea dx,[buf]
mov ah,9;输入提示
int 21h[……]
data segment
int1 db 30,31 dup(0),’$’
int2 db 30,31 dup(0),’$’
space db 13,10,’$’
data ends
code segment
assume cs:code,ds:data
main:mov dx,data
mov ds,dx
lea dx,[i[……]
data segment
buf1 db ‘shu ru yi ge dan ci,jieguo:huo ba kong ge shan diao’,13,10,’$’
int1 db 100,101 dup(0),’$’
buf db 100,101 dup(0),’$’
data ends
code segment
assume cs:code,ds:data
main:mov dx,data
mov ds,dx[……]
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