下面的汇编代码出了什么问题？

Question

获取输入字符串的代码

在Windows机器上使用nasm进行汇编：nasm file.asm -o file.com

;read the string
mov ah,0x0A ;read 
mov dx,buffer   ;pointer to buffer
int 0x21

;newline
mov ah,2
mov dl,10
int 0x21
mov dl,13
int 0x21

;put $ sign at end of string
mov bx,buffer+1
mov dx,buffer+2
add dl,byte[bx]
mov bx,dx
mov byte[bx],'$'

;output
mov dx,buffer+2
mov ah,9
int 0x21

;exit
mov ah,0x4c
int 0x21

;buffer

buffer:
db 255  ;len of buffer
db 0    ;num of char read
db 255  ;actual string

;############################

它是长度为2个字符串的字符串的工作文件，之后它发出垃圾。

Answer 1

这是应该如何解决的。查看;;的行。

bits 16 ;; you have to tell NASM to generate 16-bit code
org 0x100 ;; you have to start generating addresses at 0x100 for .COM programs

;read the string
mov ah,0x0A ;read 
mov dx,buffer   ;pointer to buffer
int 0x21

;newline
mov ah,2
mov dl,10
int 0x21
mov dl,13
int 0x21

;put $ sign at end of string
mov bx,buffer+1
mov dx,buffer+2
add dl,byte[bx]
adc dh, 0 ;; we're doing 16-bit+8-bit=16-bit addition
mov bx,dx
mov byte[bx],'$'

;output
mov dx,buffer+2
mov ah,9
int 0x21

;exit
mov ah,0x4c
int 0x21

;buffer

buffer:
db 254; 255  ;len of buffer ;; my doc says the range is 1 to 254
; db 0    ;num of char read ;; you only need to reserve memory
; db 255  ;actual string ;; you only need to reserve memory
resb 255 ;; reserve memory (1 for length, 254 for text)
resb 1 ;; reserve memory for "$"

如果缓冲区是.COM程序中的最后一项，您实际上可以删除db 254之后的所有行。