Assembly x86 Date to Number - 将字符串分成更小的部分
Assembly x86 Date to Number - Breaking a string into smaller sections
我实际上希望在某个问题上指明正确的方向。
我希望将 x86 程序集中的日期从格式 "DD-MMM-YYYY" 转换为唯一数字，以便稍后可以对其进行冒泡排序并最终转换回来。
所以，当我有一个字符串输入时，即：
。数据
inDate dw "08-SEP-1993"
我想把它分成
day = "08"
month = "SEP"
year = "1993"
以便我可以进一步处理它（我会将 SEP 转换为“7”等）
所以我的问题是什么是分解日期（代码方面）的简单、有效的方法？我知道我需要转换日期格式以允许排序，但我是 Assembly 的新手所以我不确定如何分解字符串以便我可以转换它。
此外，作为第二个问题，您如何将数字从字符串转换为实际数值？
谢谢！
注意：我想应该注意我使用的是 masm32
接下来用EMU8086（16位）做了个小程序，把键盘上的数字抓取成字符串，转成数字比较，最后把数字转成字符串显示。请注意，数字是用 0AH 捕获的，这需要一个 3 级变量 "str"。您需要的转换过程在代码的底部（string2number 和 number2string）。
.model small
.stack 100h
.data
counter dw ?
msj1 db 'Enter a number: $'
msj2 db 'The highest number is: $'
break db 13,10,'$'
str db 6 ;MAX NUMBER OF CHARACTERS ALLOWED (4).
db ? ;NUMBER OF CHARACTERS ENTERED BY USER.
db 6 dup (?) ;CHARACTERS ENTERED BY USER.
highest dw 0
buffer db 6 dup(?)
.code
;INITIALIZE DATA SEGMENT.
mov ax, @data
mov ds, ax
;-----------------------------------------
;CAPTURE 5 NUMBERS AND DETERMINE THE HIGHEST.
mov counter, 5 ;HOW MANY NUMBERS TO CAPTURE.
enter_numbers:
;DISPLAY MESSAGE.
mov dx, offset msj1
call printf
;CAPTURE NUMBER AS STRING.
mov dx, offset str
call scanf
;DISPLAY LINE BREAK.
mov dx, offset break
call printf
;CONVERT CAPTURED NUMBER FROM STRING TO NUMERIC.
mov si, offset str ;PARAMETER (STRING TO CONVERT).
call string2number ;NUMBER RETURNS IN BX.
;CHECK IF CAPTURED NUMBER IS THE HIGHEST.
cmp highest, bx
jae ignore ;IF (HIGHEST >= BX) IGNORE NUMBER.
;IF NO JUMP TO "IGNORE", CURRENT NUMBER IS HIGHER THAN "HIGHEST".
mov highest, bx ;CURRENT NUMBER IS THE HIGHEST.
ignore:
;CHECK IF WE HAVE CAPTURED 5 NUMBERS ALREADY.
dec counter
jnz enter_numbers
;-----------------------------------------
;DISPLAY HIGHEST NUMBER.
;FIRST, FILL BUFFER WITH '$' (NECESSARY TO DISPLAY).
mov si, offset buffer
call dollars
;SECOND, CONVERT HIGHEST NUMBER TO STRING.
mov ax, highest
mov si, offset buffer
call number2string
;THIRD, DISPLAY STRING.
mov dx, offset msj2
call printf
mov dx, offset buffer
call printf
;FINISH PROGRAM.
mov ax, 4c00h
int 21h
;-----------------------------------------
;PARAMETER : DX POINTING TO '$' FINISHED STRING.
proc printf
mov ah, 9
int 21h
ret
endp
;-----------------------------------------
;PARAMETER : DX POINTING TO BUFFER TO STORE STRING.
proc scanf
mov ah, 0Ah
int 21h
ret
endp
;------------------------------------------
;CONVERT STRING TO NUMBER.
;PARAMETER : SI POINTING TO CAPTURED STRING.
;RETURN : NUMBER IN BX.
proc string2number
;MAKE SI TO POINT TO THE LEAST SIGNIFICANT DIGIT.
inc si ;POINTS TO THE NUMBER OF CHARACTERS ENTERED.
mov cl, [ si ] ;NUMBER OF CHARACTERS ENTERED.
mov ch, 0 ;CLEAR CH, NOW CX==CL.
add si, cx ;NOW SI POINTS TO LEAST SIGNIFICANT DIGIT.
;CONVERT STRING.
mov bx, 0
mov bp, 1 ;MULTIPLE OF 10 TO MULTIPLY EVERY DIGIT.
repeat:
;CONVERT CHARACTER.
mov al, [ si ] ;CHARACTER TO PROCESS.
sub al, 48 ;CONVERT ASCII CHARACTER TO DIGIT.
mov ah, 0 ;CLEAR AH, NOW AX==AL.
mul bp ;AX*BP = DX:AX.
add bx, ax ;ADD RESULT TO BX.
;INCREASE MULTIPLE OF 10 (1, 10, 100...).
mov ax, bp
mov bp, 10
mul bp ;AX*10 = DX:AX.
mov bp, ax ;NEW MULTIPLE OF 10.
;CHECK IF WE HAVE FINISHED.
dec si ;NEXT DIGIT TO PROCESS.
loop repeat ;COUNTER CX-1, IF NOT ZERO, REPEAT.
ret
endp
;------------------------------------------
;FILLS VARIABLE WITH '$'.
;USED BEFORE CONVERT NUMBERS TO STRING, BECAUSE
;THE STRING WILL BE DISPLAYED.
;PARAMETER : SI = POINTING TO STRING TO FILL.
proc dollars
mov cx, 6
six_dollars:
mov bl, '$'
mov [ si ], bl
inc si
loop six_dollars
ret
endp
;------------------------------------------
;CONVERT A NUMBER IN STRING.
;ALGORITHM : EXTRACT DIGITS ONE BY ONE, STORE
;THEM IN STACK, THEN EXTRACT THEM IN REVERSE
;ORDER TO CONSTRUCT STRING (STR).
;PARAMETERS : AX = NUMBER TO CONVERT.
; SI = POINTING WHERE TO STORE STRING.
proc number2string
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.
cycle2:
pop dx
add dl, 48 ;CONVERT DIGIT TO CHARACTER.
mov [ si ], dl
inc si
loop cycle2
ret
endp
现在是 32 位版本。接下来是一个小程序，分配给 EAX 一个大数字，将其转换为字符串并将其转换回数字，这里是：
.model small
.586
.stack 100h
.data
msj1 db 13,10,'Original EAX = $'
msj2 db 13,10,'Flipped EAX = $'
msj3 db 13,10,'New EAX = $'
buf db 11
db ?
db 11 dup (?)
.code
start:
;INITIALIZE DATA SEGMENT.
mov ax, @data
mov ds, ax
;CONVERT EAX TO STRING TO DISPLAY IT.
call dollars ;NECESSARY TO DISPLAY.
mov eax, 1234567890
call number2string ;PARAMETER:AX. RETURN:VARIABLE BUF.
;DISPLAY 'ORIGINAL EAX'.
mov ah, 9
mov dx, offset msj1
int 21h
;DISPLAY BUF (EAX CONVERTED TO STRING).
mov ah, 9
mov dx, offset buf
int 21h
;FLIP EAX.
call dollars ;NECESSARY TO DISPLAY.
mov eax, 1234567890
call flip_eax ;PARAMETER:AX. RETURN:VARIABLE BUF.
;DISPLAY 'FLIPPED EAX'.
mov ah, 9
mov dx, offset msj2
int 21h
;DISPLAY BUF (EAX FLIPPED CONVERTED TO STRING).
mov ah, 9
mov dx, offset buf
int 21h
;CONVERT STRING TO NUMBER (FLIPPED EAX TO EAX).
mov si, offset buf ;STRING TO REVERSE.
call string2number ;RETURN IN EBX.
mov eax, ebx ;THIS IS THE NEW EAX FLIPPED.
;CONVERT EAX TO STRING TO DISPLAY IT.
call dollars ;NECESSARY TO DISPLAY.
call number2string ;PARAMETER:EAX. RETURN:VARIABLE BUF.
;DISPLAY 'NEW EAX'.
mov ah, 9
mov dx, offset msj3
int 21h
;DISPLAY BUF (EAX CONVERTED TO STRING).
mov ah, 9
mov dx, offset buf
int 21h
;WAIT UNTIL USER PRESS ANY KEY.
mov ah, 7
int 21h
;FINISH PROGRAM.
mov ax, 4c00h
int 21h
;------------------------------------------
flip_eax proc
mov si, offset buf ;DIGITS WILL BE STORED IN BUF.
mov bx, 10 ;DIGITS ARE EXTRACTED DIVIDING BY 10.
mov cx, 0 ;COUNTER FOR EXTRACTED DIGITS.
extracting:
;EXTRACT ONE DIGIT.
mov edx, 0 ;NECESSARY TO DIVIDE BY EBX.
div ebx ;EDX:EAX / 10 = EAX:QUOTIENT EDX:REMAINDER.
;INSERT DIGIT IN STRING.
add dl, 48 ;CONVERT DIGIT TO CHARACTER.
mov [ si ], dl
inc si
;NEXT DIGIT.
cmp eax, 0 ;IF NUMBER IS
jne extracting ;NOT ZERO, REPEAT.
ret
flip_eax endp
;------------------------------------------
;CONVERT STRING TO NUMBER IN EBX.
;SI MUST ENTER POINTING TO THE STRING.
string2number proc
;COUNT DIGITS IN STRING.
mov cx, 0
find_dollar:
inc cx ;DIGIT COUNTER.
inc si ;NEXT CHARACTER.
mov bl, [ si ]
cmp bl, '$'
jne find_dollar ;IF BL != '$' JUMP.
dec si ;BECAUSE IT WAS OVER '$', NOT OVER THE LAST DIGIT.
;CONVERT STRING.
mov ebx, 0
mov ebp, 1 ;MULTIPLE OF 10 TO MULTIPLY EVERY DIGIT.
repeat:
;CONVERT CHARACTER.
mov eax, 0 ;NOW EAX==AL.
mov al, [ si ] ;CHARACTER TO PROCESS.
sub al, 48 ;CONVERT ASCII CHARACTER TO DIGIT.
mul ebp ;EAX*EBP = EDX:EAX.
add ebx, eax ;ADD RESULT TO BX.
;INCREASE MULTIPLE OF 10 (1, 10, 100...).
mov eax, ebp
mov ebp, 10
mul ebp ;AX*10 = EDX:EAX.
mov ebp, eax ;NEW MULTIPLE OF 10.
;CHECK IF WE HAVE FINISHED.
dec si ;NEXT DIGIT TO PROCESS.
loop repeat ;CX-1, IF NOT ZERO, REPEAT.
ret
string2number endp
;------------------------------------------
;FILLS VARIABLE STR WITH '$'.
;USED BEFORE CONVERT NUMBERS TO STRING, BECAUSE
;THE STRING WILL BE DISPLAYED.
dollars proc
mov si, offset buf
mov cx, 11
six_dollars:
mov bl, '$'
mov [ si ], bl
inc si
loop six_dollars
ret
dollars endp
;------------------------------------------
;NUMBER TO CONVERT MUST ENTER IN EAX.
;ALGORITHM : EXTRACT DIGITS ONE BY ONE, STORE
;THEM IN STACK, THEN EXTRACT THEM IN REVERSE
;ORDER TO CONSTRUCT STRING (BUF).
number2string proc
mov ebx, 10 ;DIGITS ARE EXTRACTED DIVIDING BY 10.
mov cx, 0 ;COUNTER FOR EXTRACTED DIGITS.
cycle1:
mov edx, 0 ;NECESSARY TO DIVIDE BY EBX.
div ebx ;EDX:EAX / 10 = EAX:QUOTIENT EDX:REMAINDER.
push dx ;PRESERVE DIGIT EXTRACTED (DL) FOR LATER.
inc cx ;INCREASE COUNTER FOR EVERY DIGIT EXTRACTED.
cmp eax, 0 ;IF NUMBER IS
jne cycle1 ;NOT ZERO, LOOP.
;NOW RETRIEVE PUSHED DIGITS.
mov si, offset buf
cycle2:
pop dx
add dl, 48 ;CONVERT DIGIT TO CHARACTER.
mov [ si ], dl
inc si
loop cycle2
ret
number2string endp
end start
我实际上希望在某个问题上指明正确的方向。
我希望将 x86 程序集中的日期从格式 "DD-MMM-YYYY" 转换为唯一数字，以便稍后可以对其进行冒泡排序并最终转换回来。
所以，当我有一个字符串输入时，即： 。数据 inDate dw "08-SEP-1993"
我想把它分成
day = "08"
month = "SEP"
year = "1993"
以便我可以进一步处理它（我会将 SEP 转换为“7”等）
所以我的问题是什么是分解日期（代码方面）的简单、有效的方法？我知道我需要转换日期格式以允许排序，但我是 Assembly 的新手所以我不确定如何分解字符串以便我可以转换它。
此外，作为第二个问题，您如何将数字从字符串转换为实际数值？
谢谢！
注意：我想应该注意我使用的是 masm32
接下来用EMU8086（16位）做了个小程序，把键盘上的数字抓取成字符串，转成数字比较，最后把数字转成字符串显示。请注意，数字是用 0AH 捕获的，这需要一个 3 级变量 "str"。您需要的转换过程在代码的底部（string2number 和 number2string）。
.model small
.stack 100h
.data
counter dw ?
msj1 db 'Enter a number: $'
msj2 db 'The highest number is: $'
break db 13,10,'$'
str db 6 ;MAX NUMBER OF CHARACTERS ALLOWED (4).
db ? ;NUMBER OF CHARACTERS ENTERED BY USER.
db 6 dup (?) ;CHARACTERS ENTERED BY USER.
highest dw 0
buffer db 6 dup(?)
.code
;INITIALIZE DATA SEGMENT.
mov ax, @data
mov ds, ax
;-----------------------------------------
;CAPTURE 5 NUMBERS AND DETERMINE THE HIGHEST.
mov counter, 5 ;HOW MANY NUMBERS TO CAPTURE.
enter_numbers:
;DISPLAY MESSAGE.
mov dx, offset msj1
call printf
;CAPTURE NUMBER AS STRING.
mov dx, offset str
call scanf
;DISPLAY LINE BREAK.
mov dx, offset break
call printf
;CONVERT CAPTURED NUMBER FROM STRING TO NUMERIC.
mov si, offset str ;PARAMETER (STRING TO CONVERT).
call string2number ;NUMBER RETURNS IN BX.
;CHECK IF CAPTURED NUMBER IS THE HIGHEST.
cmp highest, bx
jae ignore ;IF (HIGHEST >= BX) IGNORE NUMBER.
;IF NO JUMP TO "IGNORE", CURRENT NUMBER IS HIGHER THAN "HIGHEST".
mov highest, bx ;CURRENT NUMBER IS THE HIGHEST.
ignore:
;CHECK IF WE HAVE CAPTURED 5 NUMBERS ALREADY.
dec counter
jnz enter_numbers
;-----------------------------------------
;DISPLAY HIGHEST NUMBER.
;FIRST, FILL BUFFER WITH '$' (NECESSARY TO DISPLAY).
mov si, offset buffer
call dollars
;SECOND, CONVERT HIGHEST NUMBER TO STRING.
mov ax, highest
mov si, offset buffer
call number2string
;THIRD, DISPLAY STRING.
mov dx, offset msj2
call printf
mov dx, offset buffer
call printf
;FINISH PROGRAM.
mov ax, 4c00h
int 21h
;-----------------------------------------
;PARAMETER : DX POINTING TO '$' FINISHED STRING.
proc printf
mov ah, 9
int 21h
ret
endp
;-----------------------------------------
;PARAMETER : DX POINTING TO BUFFER TO STORE STRING.
proc scanf
mov ah, 0Ah
int 21h
ret
endp
;------------------------------------------
;CONVERT STRING TO NUMBER.
;PARAMETER : SI POINTING TO CAPTURED STRING.
;RETURN : NUMBER IN BX.
proc string2number
;MAKE SI TO POINT TO THE LEAST SIGNIFICANT DIGIT.
inc si ;POINTS TO THE NUMBER OF CHARACTERS ENTERED.
mov cl, [ si ] ;NUMBER OF CHARACTERS ENTERED.
mov ch, 0 ;CLEAR CH, NOW CX==CL.
add si, cx ;NOW SI POINTS TO LEAST SIGNIFICANT DIGIT.
;CONVERT STRING.
mov bx, 0
mov bp, 1 ;MULTIPLE OF 10 TO MULTIPLY EVERY DIGIT.
repeat:
;CONVERT CHARACTER.
mov al, [ si ] ;CHARACTER TO PROCESS.
sub al, 48 ;CONVERT ASCII CHARACTER TO DIGIT.
mov ah, 0 ;CLEAR AH, NOW AX==AL.
mul bp ;AX*BP = DX:AX.
add bx, ax ;ADD RESULT TO BX.
;INCREASE MULTIPLE OF 10 (1, 10, 100...).
mov ax, bp
mov bp, 10
mul bp ;AX*10 = DX:AX.
mov bp, ax ;NEW MULTIPLE OF 10.
;CHECK IF WE HAVE FINISHED.
dec si ;NEXT DIGIT TO PROCESS.
loop repeat ;COUNTER CX-1, IF NOT ZERO, REPEAT.
ret
endp
;------------------------------------------
;FILLS VARIABLE WITH '$'.
;USED BEFORE CONVERT NUMBERS TO STRING, BECAUSE
;THE STRING WILL BE DISPLAYED.
;PARAMETER : SI = POINTING TO STRING TO FILL.
proc dollars
mov cx, 6
six_dollars:
mov bl, '$'
mov [ si ], bl
inc si
loop six_dollars
ret
endp
;------------------------------------------
;CONVERT A NUMBER IN STRING.
;ALGORITHM : EXTRACT DIGITS ONE BY ONE, STORE
;THEM IN STACK, THEN EXTRACT THEM IN REVERSE
;ORDER TO CONSTRUCT STRING (STR).
;PARAMETERS : AX = NUMBER TO CONVERT.
; SI = POINTING WHERE TO STORE STRING.
proc number2string
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.
cycle2:
pop dx
add dl, 48 ;CONVERT DIGIT TO CHARACTER.
mov [ si ], dl
inc si
loop cycle2
ret
endp
现在是 32 位版本。接下来是一个小程序，分配给 EAX 一个大数字，将其转换为字符串并将其转换回数字，这里是：
.model small
.586
.stack 100h
.data
msj1 db 13,10,'Original EAX = $'
msj2 db 13,10,'Flipped EAX = $'
msj3 db 13,10,'New EAX = $'
buf db 11
db ?
db 11 dup (?)
.code
start:
;INITIALIZE DATA SEGMENT.
mov ax, @data
mov ds, ax
;CONVERT EAX TO STRING TO DISPLAY IT.
call dollars ;NECESSARY TO DISPLAY.
mov eax, 1234567890
call number2string ;PARAMETER:AX. RETURN:VARIABLE BUF.
;DISPLAY 'ORIGINAL EAX'.
mov ah, 9
mov dx, offset msj1
int 21h
;DISPLAY BUF (EAX CONVERTED TO STRING).
mov ah, 9
mov dx, offset buf
int 21h
;FLIP EAX.
call dollars ;NECESSARY TO DISPLAY.
mov eax, 1234567890
call flip_eax ;PARAMETER:AX. RETURN:VARIABLE BUF.
;DISPLAY 'FLIPPED EAX'.
mov ah, 9
mov dx, offset msj2
int 21h
;DISPLAY BUF (EAX FLIPPED CONVERTED TO STRING).
mov ah, 9
mov dx, offset buf
int 21h
;CONVERT STRING TO NUMBER (FLIPPED EAX TO EAX).
mov si, offset buf ;STRING TO REVERSE.
call string2number ;RETURN IN EBX.
mov eax, ebx ;THIS IS THE NEW EAX FLIPPED.
;CONVERT EAX TO STRING TO DISPLAY IT.
call dollars ;NECESSARY TO DISPLAY.
call number2string ;PARAMETER:EAX. RETURN:VARIABLE BUF.
;DISPLAY 'NEW EAX'.
mov ah, 9
mov dx, offset msj3
int 21h
;DISPLAY BUF (EAX CONVERTED TO STRING).
mov ah, 9
mov dx, offset buf
int 21h
;WAIT UNTIL USER PRESS ANY KEY.
mov ah, 7
int 21h
;FINISH PROGRAM.
mov ax, 4c00h
int 21h
;------------------------------------------
flip_eax proc
mov si, offset buf ;DIGITS WILL BE STORED IN BUF.
mov bx, 10 ;DIGITS ARE EXTRACTED DIVIDING BY 10.
mov cx, 0 ;COUNTER FOR EXTRACTED DIGITS.
extracting:
;EXTRACT ONE DIGIT.
mov edx, 0 ;NECESSARY TO DIVIDE BY EBX.
div ebx ;EDX:EAX / 10 = EAX:QUOTIENT EDX:REMAINDER.
;INSERT DIGIT IN STRING.
add dl, 48 ;CONVERT DIGIT TO CHARACTER.
mov [ si ], dl
inc si
;NEXT DIGIT.
cmp eax, 0 ;IF NUMBER IS
jne extracting ;NOT ZERO, REPEAT.
ret
flip_eax endp
;------------------------------------------
;CONVERT STRING TO NUMBER IN EBX.
;SI MUST ENTER POINTING TO THE STRING.
string2number proc
;COUNT DIGITS IN STRING.
mov cx, 0
find_dollar:
inc cx ;DIGIT COUNTER.
inc si ;NEXT CHARACTER.
mov bl, [ si ]
cmp bl, '$'
jne find_dollar ;IF BL != '$' JUMP.
dec si ;BECAUSE IT WAS OVER '$', NOT OVER THE LAST DIGIT.
;CONVERT STRING.
mov ebx, 0
mov ebp, 1 ;MULTIPLE OF 10 TO MULTIPLY EVERY DIGIT.
repeat:
;CONVERT CHARACTER.
mov eax, 0 ;NOW EAX==AL.
mov al, [ si ] ;CHARACTER TO PROCESS.
sub al, 48 ;CONVERT ASCII CHARACTER TO DIGIT.
mul ebp ;EAX*EBP = EDX:EAX.
add ebx, eax ;ADD RESULT TO BX.
;INCREASE MULTIPLE OF 10 (1, 10, 100...).
mov eax, ebp
mov ebp, 10
mul ebp ;AX*10 = EDX:EAX.
mov ebp, eax ;NEW MULTIPLE OF 10.
;CHECK IF WE HAVE FINISHED.
dec si ;NEXT DIGIT TO PROCESS.
loop repeat ;CX-1, IF NOT ZERO, REPEAT.
ret
string2number endp
;------------------------------------------
;FILLS VARIABLE STR WITH '$'.
;USED BEFORE CONVERT NUMBERS TO STRING, BECAUSE
;THE STRING WILL BE DISPLAYED.
dollars proc
mov si, offset buf
mov cx, 11
six_dollars:
mov bl, '$'
mov [ si ], bl
inc si
loop six_dollars
ret
dollars endp
;------------------------------------------
;NUMBER TO CONVERT MUST ENTER IN EAX.
;ALGORITHM : EXTRACT DIGITS ONE BY ONE, STORE
;THEM IN STACK, THEN EXTRACT THEM IN REVERSE
;ORDER TO CONSTRUCT STRING (BUF).
number2string proc
mov ebx, 10 ;DIGITS ARE EXTRACTED DIVIDING BY 10.
mov cx, 0 ;COUNTER FOR EXTRACTED DIGITS.
cycle1:
mov edx, 0 ;NECESSARY TO DIVIDE BY EBX.
div ebx ;EDX:EAX / 10 = EAX:QUOTIENT EDX:REMAINDER.
push dx ;PRESERVE DIGIT EXTRACTED (DL) FOR LATER.
inc cx ;INCREASE COUNTER FOR EVERY DIGIT EXTRACTED.
cmp eax, 0 ;IF NUMBER IS
jne cycle1 ;NOT ZERO, LOOP.
;NOW RETRIEVE PUSHED DIGITS.
mov si, offset buf
cycle2:
pop dx
add dl, 48 ;CONVERT DIGIT TO CHARACTER.
mov [ si ], dl
inc si
loop cycle2
ret
number2string endp
end start