string 使用 DOS 中断 MASM 获取字符串输入和显示输入
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Getting string input and displaying input with DOS interrupts MASM
提问by Jumpman
In MASM, I created a buffer variable to hold the user string input from keyboard. I am stuck on how to hold the string input into that buffer variable. I don't have any libraries linked like the irvine ones and want to do this with DOS interrupts. So far I have something along the lines of
在 MASM 中，我创建了一个缓冲区变量来保存从键盘输入的用户字符串。我被困在如何将字符串输入保存到该缓冲区变量中。我没有像 irvine 那样链接的任何库，并且想用 DOS 中断来做到这一点。到目前为止，我有一些类似的东西
.model small
.stack 100h
.data
buff db 25 dup(0), 10, 13
lbuff EQU ($ - buff) ; bytes in a string
.code
main:
mov ax, @data
mov ds, ax
mov ah, 0Ah ; doesn't work
mov buff, ah ; doesn't seem right
int 21h
mov ax, 4000h ; display to screen
mov bx, 1
mov cx, lbuff
mov dx, OFFSET buff
int 21h
mov ah, 4ch
int 21h
end main
I assume using 0Ah is correct as it is for reading array of input of buffered characters.
我认为使用 0Ah 是正确的，因为它用于读取缓冲字符的输入数组。
回答by Jose Manuel Abarca Rodríguez
I made some changes to your code. First, the "buff" variable needs the three level format (max number of characters allowed, another byte for the number of characteres entered, and the buffer itself) because that's what service 0AH requires. To use service 0AH I added "offset buff" (as Wolfgang said). Here it is:
我对您的代码进行了一些更改。首先，“buff”变量需要三级格式（允许的最大字符数、输入字符数的另一个字节和缓冲区本身），因为这是服务 0AH 所需要的。为了使用服务 0AH，我添加了“偏移增益”（正如沃尔夫冈所说）。这里是：
.model small
.stack 100h
.data
buff db 26 ;MAX NUMBER OF CHARACTERS ALLOWED (25).
db ? ;NUMBER OF CHARACTERS ENTERED BY USER.
db 26 dup(0) ;CHARACTERS ENTERED BY USER.
.code
main:
mov ax, @data
mov ds, ax
;CAPTURE STRING FROM KEYBOARD.
mov ah, 0Ah ;SERVICE TO CAPTURE STRING FROM KEYBOARD.
mov dx, offset buff
int 21h
;CHANGE CHR(13) BY '$'.
mov si, offset buff + 1 ;NUMBER OF CHARACTERS ENTERED.
mov cl, [ si ] ;MOVE LENGTH TO CL.
mov ch, 0 ;CLEAR CH TO USE CX.
inc cx ;TO REACH CHR(13).
add si, cx ;NOW SI POINTS TO CHR(13).
mov al, '$'
mov [ si ], al ;REPLACE CHR(13) BY '$'.
;DISPLAY STRING.
mov ah, 9 ;SERVICE TO DISPLAY STRING.
mov dx, offset buff + 2 ;MUST END WITH '$'.
int 21h
mov ah, 4ch
int 21h
end main
When 0AH captures the string from keyboard, it ends with ENTER (character 13), that's why, if you want to capture 25 characters, you must specify 26.
当0AH从键盘捕获字符串时，它以ENTER（字符13）结尾，这就是为什么如果要捕获25个字符，则必须指定26。
To know how many characters the user entered (length), access the second byte (offset buff + 1). The ENTER is not included, so, if user types 8 characters and ENTER, this second byte will contain the number 8, not 9.
要知道用户输入了多少个字符（长度），请访问第二个字节（偏移 buff + 1）。不包括 ENTER，因此，如果用户键入 8 个字符并按 ENTER，则第二个字节将包含数字 8，而不是 9。
The entered characters start at offset buff + 2, and they end when character 13 appears. We use this to add the length to buff+2 + 1 to replace chr(13) by '$'. Now we can display the string.
输入的字符从偏移 buff + 2 开始，并在字符 13 出现时结束。我们使用它来将长度添加到 buff+2 + 1 以将 chr(13) 替换为 '$'。现在我们可以显示字符串了。
回答by Q10Viking
This is my code,maybe can help you.
这是我的代码，也许可以帮助你。
;Input String Copy output
dataarea segment
BUFFER db 81
db ?
STRING DB 81 DUP(?)
STR1 DB 10,13,'$'
dataarea ends
extra segment
MESS1 DB 'After Copy',10,13,'$'
MESS2 DB 81 DUP(?)
extra ends
code segment
main proc far
assume cs:code,ds:dataarea,es:extra
start:
push ds
sub ax,ax
push ax
mov ax,dataarea
mov ds,ax
mov ax,extra
mov es,ax
lea dx,BUFFER
mov ah,0ah
int 21h
lea si,STRING
lea di,MESS2
mov ch,0
mov cl,BUFFER+1
cld
rep movsb
mov al,'$'
mov es:[di],al
lea dx,STR1 ;to next line
mov ah,09h
int 21h
push es
pop ds
lea dx,MESS1 ;output:after copy
mov ah,09h
int 21h
lea dx,MESS2
mov ah,09h
int 21h
ret
main endp
code ends
end start
And the result is:
结果是：
c:\demo.exe
Hello World!
After Copy
Hello World!
回答by rashedcs
You may follow this code :
您可以遵循以下代码：
; Problem : input array from user
.MODEL SMALL
.STACK
.DATA
ARR DB 10 DUB (?)
.CODE
MAIN PROC
MOV AX, @DATA
MOV DS, AX
XOR BX, BX
MOV CX, 5
FOR:
MOV AH, 1
INT 21H
MOV ARR[BX], AL
INC BX
LOOP FOR
XOR BX, BX
MOV CX, 5
PRINT:
MOV AX, ARR[BX] ;point to the current index
MOV AH, 2 ;output
MOV DL, AX
INT 21H
INC BX ;move pointer to the next element
LOOP PRINT ;loop until done
MAIN ENDP
回答by AbbieRojo
;try this one, it takes a 10 character string input from user and displays it after in this manner, "Hello *10character string input"
;试试这个，它需要用户输入的10个字符串，然后以这种方式显示，“你好*10个字符串输入”
.MODEL TINY
.CODE
.286
ORG 100h
START:
MOV DX, OFFSET BUFFER
MOV AH, 0ah
INT 21h
JMP PRINT
BUFFER DB 10,?, 10 dup(' ')
PRINT:
MOV AH, 02
MOV DL, 0ah
INT 21h
MOV AH, 9
MOV DX, OFFSET M1
INT 21h
XOR BX, BX
MOV BL, BUFFER[1]
MOV BUFFER [BX+2], '$'
MOV DX, OFFSET BUFFER +2
MOV AH, 9
INT 21h
M1: db 'Hello $'
END START
END
