Assembly 什么';下面的汇编代码有什么问题?

Assembly 什么';下面的汇编代码有什么问题?,assembly,x86,dos,nasm,Assembly,X86,Dos,Nasm,获取输入字符串的代码 在windows计算机上使用nasm进行组装: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,b

获取输入字符串的代码

在windows计算机上使用nasm进行组装:
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

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

它是长度不超过2个字符的字符串的工作文件,之后它将发出垃圾。

下面介绍如何修复它。请参见带有
;;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程序中的最后一个内容,那么实际上可以删除
db 254
之后的所有行。请参见带有
;;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程序中的最后一项内容,那么实际上可以删除
db 254

之后的所有行。如果您能用输入和输出的真实示例解释代码应该做什么,以及它实际做什么,那就太好了。定义什么“没有错”如果你能用输入和输出的真实例子来解释代码应该做什么,以及它实际做什么,那就太好了。定义“没有错”的含义。