此代码将1位十六进制转换为十进制，它正在工作，复制互联网，但我想了解它如何工作，并添加一个循环，我可以达到4位数。我想要一个帮助如何执行此循环。
我正在使用masm16和DosBox。
.model small
.stack 100h
.data
msg1 db 10,13,'ENTER A HEX DIGIT:$'
msg2 db 10,13,'IN DECIMAL IS IT:$'
msg3 db 10,13,'DO YOU WANT TO DO IT AGAIN?$'
msg4 db 10,13,'ILLEGAL CHARACTER- ENTER 0-9 OR A-F:$'
.code
again:
mov ax,@data
mov ds,ax
lea dx,msg1
mov ah,9
int 21h
mov ah,1
int 21h
mov bl,al
jmp go
go:
cmp bl,'9'
ja hex
jb num
je num
hex:
cmp bl,'F'
ja illegal
lea dx,msg2
mov ah,9
int 21h
mov dl,49d
mov ah,2
int 21h
sub bl,17d
mov dl,bl
mov ah,2
int 21h
jmp inp
inp:
lea dx,msg3
mov ah,9
int 21h
mov ah,1
int 21h
mov cl,al
cmp cl,'y'
je again
cmp cl,'Y'
je again
jmp exit
num:
cmp bl,'0'
jb illegal
lea dx,msg2
mov ah,9
int 21h
mov dl,bl
mov ah,2
int 21h
jmp inp
illegal:
lea dx,msg4
mov ah,9
int 21h
mov ah,1
int 21h
mov bl,al
jmp go
exit:
end
答案 0 :(得分：0)
我发现了一个可以满足您需求的旧程序，但它与您的程序非常不同。我会在这里粘贴它，因为它可以帮助你。我添加了一些你的代码并进行了一些其他更改。这是它的工作原理：
以下是代码，只需复制粘贴并运行：
.model small
.stack 100h
.data
msg1 db 10,13,10,13,'ENTER 1 TO 4 HEX DIGITS:$'
msg2 db 10,13,'IN DECIMAL IS IT:$'
msg3 db 10,13,10,13,'DO YOU WANT TO DO IT AGAIN (Y/N)?$'
msg4 db 10,13,'ILLEGAL CHARACTER- ENTER 0-9 OR A-F:$'
hex db 5,?,5 dup(?) ;VARIABLE WITH 3 SECTIONS.
buffer db 6 dup('$') ;RESULT COULD HAVE 5 DIGITS.
.code
mov ax, @data
mov ds, ax
again:
;CLEAR BUFFER (IN CASE IT HOLDS PREVIOUS RESULT).
call clear_buffer
;DISPLAY 'ENTER 1 TO 4 HEX DIGITS:$'
mov ah, 9
lea dx, msg1
int 21h
;CAPTURE HEX NUMBER AS STRING.
mov ah, 0ah
lea dx, hex
int 21h
;CONVERT HEX-STRING TO NUMBER.
lea si, hex+2 ;CHARS OF THE HEX-STRING.
mov bh, [si-1] ;SECOND BYTE IS LENGTH.
call hex2number ;NUMBER RETURNS IN AX.
;CONVERT NUMBER TO DECIMAL-STRING TO DISPLAY.
lea si, buffer
call number2string ;STRING RETURNS IN SI (BUFFER).
;DISPLAY 'IN DECIMAL IS IT:$'
mov ah, 9
lea dx, msg2
int 21h
;DISPLAY NUMBER AS STRING.
mov ah, 9
lea dx, buffer
int 21h
illegal: ;JUMP HERE WHEN INVALID CHARACTER FOUND.
;DISPLAY 'DO YOU WANT TO DO IT AGAIN (Y/N)?$'
mov ah, 9
lea dx, msg3
int 21h
;CAPTURE KEY.
mov ah, 1
int 21h
cmp al,'y'
je again
cmp al,'Y'
je again
;TERMINATE PROGRAM.
mov ax, 4c00h
int 21h
;---------------------------------------------
;FILL VARIABLE "BUFFER" WITH "$".
;EVERYTIME THE USER WANTS TO DO IT AGAIN, THE
;PREVIOUS RESULT MUST BE CLEARED.
clear_buffer proc
lea si, buffer
mov al, '$'
mov cx, 5
clearing:
mov [si], al
inc si
loop clearing
ret
clear_buffer endp
;---------------------------------------------
;INPUT : BH = STRING LENGTH (1..4).
; SI = OFFSET HEX-STRING.
;OUTPUT : AX = NUMBER.
hex2number proc
MOV AX, 0 ;THE NUMBER.
Ciclo:
;■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■
; SHL AX, 4 ;SHIFT LEFT LOWER 4 BITS.
;SHIFT LEFT AL AND AH MANUALLY 4 TIMES TO SIMULATE SHL AX,4.
shl al, 1
rcl ah, 1
shl al, 1
rcl ah, 1
shl al, 1
rcl ah, 1
shl al, 1
rcl ah, 1
;■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■■
MOV BL, [ SI ] ;GET ONE HEX CHAR FROM STRING.
call validate
CMP BL, 'A' ;BL = 'A'..'F' : LETTER.
JAE letterAF ;BL = '0'..'9' : DIGIT.
;CharIsDigit09.
SUB BL, 48 ;CONVERT DIGIT TO NUMBER.
JMP continue
letterAF:
SUB BL, 55 ;CONVERT LETTER TO NUMBER.
continue:
OR AL, BL ;CLEAR UPPER 4 BITS.
INC SI ;NEXT HEX CHAR.
DEC BH ;BH == 0 : FINISH.
JNZ Ciclo ;BH != 0 : REPEAT.
Fin:
RET
hex2number endp
;---------------------------------------------
;INPUT : BL = HEX CHAR TO VALIDATE.
validate proc
cmp bl, '0'
jb error ;IF BL < '0'
cmp bl, 'F'
ja error ;IF BL > 'F'
cmp bl, '9'
jbe ok ;IF BL <= '9'
cmp bl, 'A'
jae ok ;IF BL >= 'A'
error:
pop ax ;REMOVE CALL VALIDATE.
pop ax ;REMOVE CALL HEX2NUMBER.
;DISPLAY 'ILLEGAL CHARACTER- ENTER 0-9 OR A-F$'
mov ah, 9
lea dx, msg4
int 21h
jmp illegal ;GO TO 'DO YOU WANT TO DO IT AGAIN (Y/N)?$'
ok:
ret
validate endp
;---------------------------------------------
;INPUT : AX = NUMBER TO CONVERT TO DECIMAL.
; SI = OFFSET STRING.
;ALGORITHM : EXTRACT DIGITS ONE BY ONE, STORE
;THEM IN STACK, THEN EXTRACT THEM IN REVERSE
;ORDER TO CONSTRUCT STRING.
number2string proc
mov bx, 10 ;DIGITS ARE EXTRACTED DIVIDING BY 10.
mov cx, 0 ;COUNTER FOR EXTRACTED DIGITS.
cycle1:
mov dx, 0 ;NECESSARY TO DIVIDE BY BX.
div bx ;DX:AX / 10 = AX:QUOTIENT DX:REMAINDER.
push dx ;PRESERVE DIGIT EXTRACTED FOR LATER.
inc cx ;INCREASE COUNTER FOR EVERY DIGIT EXTRACTED.
cmp ax, 0 ;IF NUMBER IS
jne cycle1 ;NOT ZERO, LOOP.
;NOW RETRIEVE PUSHED DIGITS.
lea si, buffer
cycle2:
pop dx
add dl, 48 ;CONVERT DIGIT TO CHARACTER.
mov [ si ], dl
inc si
loop cycle2
ret
number2string endp
;------------------------------------------
end
有关int 21h的一些解释啊= 0Ah：此服务需要一个包含三个部分的变量：
hex db 5 ;MAX LENGTH ALLOWED (4 CHARS).
db ? ;LENGTH ENTERED BY USER.
db 5 dup(?) ;STRING CHARS (PLUS ENTER AT THE END).
第一部分是允许的最大字符数（加上最后的ENTER键，即char 0Dh），这就是为什么如果我们想要4个字符，我们必须定义5.第二部分是用户输入的长度（在用户按下ENTER后定义）。最后一部分是字符串本身，用户输入的字符（最后是ENTER键，这是char 0Dh）。