在汇编中将 Char 转换为 ASCII 码（十进制）
Convert Char to ASCII Code (Decimal) in Assembly
我是 Assembly 的新手 - 不到一周的经验。我试图从提示中接收一个字符，然后以十进制显示 ASCII 代码。我已经为此工作了几个小时，并决定来这里。
当我在提示符中输入 'a' 时，我的模拟器显示我的 char 变量正在存储 61 个十六进制数，这是正确的。我似乎无法将其打印到屏幕上（十六进制或十进制）。
编辑：将代码更改为循环并除以 10。我有正确的字符打印出来，但相反。数字 97 打印出 79
到目前为止，这是我得到的：
; **************** MACROS *********************
; START PROGRAM
START MACRO
MOV AX, DATA ; Data Segment to AX Register
MOV DS, AX ; HAVE DS Point to Data Segment
ENDM
; END PROGRAM
END MACRO
MOV AH, 4CH ; END Program
INT 21H ; Call DOS Service
ENDM
; PRINT STRING TO OUTPUT
PSTRING MACRO STR
MOV AH,09H
LEA DX,STR
INT 21H
ENDM
; Creates a New Line
NEWLINE MACRO
MOV DX, 0AH ; Input of string to DX
MOV AH, 02H ; Write Char to standard output
INT 21H
MOV DL, 0DH ;
MOV AH, 02H ; Carriage Return
INT 21H ;
ENDM
; Get CHAR Input
GETINPUT MACRO INPUT
MOV AH, 01H ; Receive input
INT 21H ; Call DOS Service
MOV INPUT, AL ; Store into Input Variable
ENDM
; ********** END MACROS *******************
.MODEL
.DATA
MSG1 db 'Choose A Char to Convert To ASCII: $'
CHAR db ? ; Store Input Char
REM db ? ; Remainder 8-bit storate
QUOT db ? ; Quotient 8-bit storage
.CODE
MAIN PROC
START
PSTRING MSG1
GETINPUT CHAR
NEWLINE
MOV AX, 0 ; Clear AX Register
MOV AL, CHAR ; Move input to AL
MOV DX, 0 ; Clear DX Register
MOV DL, 10 ; Add 10 to DL Register
DIV DL ; Divide By Number of Students For Average
; ********************** ;
; QUOTIENT STORED IN AL ;
; REMAINDER STORED IN AH ;
; ********************** ;
myLoop:
MOV REM, AH ; Move Remainder into REM Variable
MOV QUOT, AL ; Move Quotient into QUOT Variable
MOV DL, REM ; Move Data we want printed into DL
ADD DL, '0' ; Make into Ascii Char
MOV AH,02H ; Output Char Service
INT 21H ; Call DOS Service
MOV AL, QUOT ; Place Quotient Into AL
MOV AH, 0 ; AH was altered, Zero it out
MOV DL, 10 ; Set Divisor to 10
DIV DL ; Divide AX by 10
CMP AH, 0 ; If No Quotient Remains we can exit
JNZ myLoop ; Jump if NOT zero
MOV AH, 4CH ; End Program
INT 21H
END
MAIN ENDP
END MAIN
好的，我解决了这个问题。这是最终代码。请记住我只是一个初学者 - 所以如果这段代码不好，请不要给我一个新的。虽然它有效！
; **************** MACROS *********************
; START PROGRAM
START MACRO
MOV AX, DATA ; Data Segment to AX Register
MOV DS, AX ; HAVE DS Point to Data Segment
ENDM
; END PROGRAM
END MACRO
MOV AH, 4CH ; END Program
INT 21H ; Call DOS Service
ENDM
; PRINT STRING TO OUTPUT
PSTRING MACRO STR
MOV AH,09H
LEA DX,STR
INT 21H
ENDM
; Creates a New Line
NEWLINE MACRO
MOV DX, 0AH ; Input of string to DX
MOV AH, 02H ; Write Char to standard output
INT 21H
MOV DL, 0DH ;
MOV AH, 02H ; Carriage Return
INT 21H ;
ENDM
; Get CHAR Input
GETINPUT MACRO INPUT
MOV AH, 01H ; Receive input
INT 21H ; Call DOS Service
MOV INPUT, AL ; Store into Input Variable
ENDM
; ********** END MACROS *******************
.MODEL
.STACK 100H
.DATA
MSG1 db 'Choose A Char to Convert To ASCII: $'
CHAR db ? ; Store Input Char
REM db ? ; Remainder 8-bit storate
QUOT db ? ; Quotient 8-bit storage
COUNT db 0 ; Counts the stacks
.CODE
MAIN PROC
START
PSTRING MSG1
GETINPUT CHAR
NEWLINE
MOV AX, 0 ; Clear AX Register
MOV AL, CHAR ; Move input to AL
MOV DX, 0 ; Clear DX Register
; ********************** ;
; QUOTIENT STORED IN AL ;
; REMAINDER STORED IN AH ;
; ********************** ;
myLoop:
MOV DL, 10 ; Set Divisor to 10
DIV DL ; Divide AX by 10
MOV REM, AH ; Move Remainder into REM Variable
MOV QUOT, AL ; Move Quotient into QUOT Variable
MOV AX, 0 ; Clear AX
MOV AL, REM ; Move REM to AL
PUSH AX ; Push AX to Stack
INC COUNT ; Increase Count by 1
MOV AL, QUOT ; Place Quotient Into AL
MOV AH, 0 ; AH was altered, Zero it out
CMP AL, 0 ; If No Quotient Remains we can exit
JNZ myLoop ; Jump if NOT zero
myLoop2:
POP DX ; Pop from the stack into DX
ADD DX, '0' ; To Ascii Char
MOV AH, 02H ; Print Char Command
INT 21H ; Call to DOS System
DEC COUNT ; Decrement COUNT
CMP COUNT, 0 ; Compare COUNT to 0
JNZ myLoop2
END
MAIN ENDP
END MAIN
我是 Assembly 的新手 - 不到一周的经验。我试图从提示中接收一个字符，然后以十进制显示 ASCII 代码。我已经为此工作了几个小时，并决定来这里。
当我在提示符中输入 'a' 时，我的模拟器显示我的 char 变量正在存储 61 个十六进制数，这是正确的。我似乎无法将其打印到屏幕上（十六进制或十进制）。
编辑：将代码更改为循环并除以 10。我有正确的字符打印出来，但相反。数字 97 打印出 79
到目前为止，这是我得到的：
; **************** MACROS *********************
; START PROGRAM
START MACRO
MOV AX, DATA ; Data Segment to AX Register
MOV DS, AX ; HAVE DS Point to Data Segment
ENDM
; END PROGRAM
END MACRO
MOV AH, 4CH ; END Program
INT 21H ; Call DOS Service
ENDM
; PRINT STRING TO OUTPUT
PSTRING MACRO STR
MOV AH,09H
LEA DX,STR
INT 21H
ENDM
; Creates a New Line
NEWLINE MACRO
MOV DX, 0AH ; Input of string to DX
MOV AH, 02H ; Write Char to standard output
INT 21H
MOV DL, 0DH ;
MOV AH, 02H ; Carriage Return
INT 21H ;
ENDM
; Get CHAR Input
GETINPUT MACRO INPUT
MOV AH, 01H ; Receive input
INT 21H ; Call DOS Service
MOV INPUT, AL ; Store into Input Variable
ENDM
; ********** END MACROS *******************
.MODEL
.DATA
MSG1 db 'Choose A Char to Convert To ASCII: $'
CHAR db ? ; Store Input Char
REM db ? ; Remainder 8-bit storate
QUOT db ? ; Quotient 8-bit storage
.CODE
MAIN PROC
START
PSTRING MSG1
GETINPUT CHAR
NEWLINE
MOV AX, 0 ; Clear AX Register
MOV AL, CHAR ; Move input to AL
MOV DX, 0 ; Clear DX Register
MOV DL, 10 ; Add 10 to DL Register
DIV DL ; Divide By Number of Students For Average
; ********************** ;
; QUOTIENT STORED IN AL ;
; REMAINDER STORED IN AH ;
; ********************** ;
myLoop:
MOV REM, AH ; Move Remainder into REM Variable
MOV QUOT, AL ; Move Quotient into QUOT Variable
MOV DL, REM ; Move Data we want printed into DL
ADD DL, '0' ; Make into Ascii Char
MOV AH,02H ; Output Char Service
INT 21H ; Call DOS Service
MOV AL, QUOT ; Place Quotient Into AL
MOV AH, 0 ; AH was altered, Zero it out
MOV DL, 10 ; Set Divisor to 10
DIV DL ; Divide AX by 10
CMP AH, 0 ; If No Quotient Remains we can exit
JNZ myLoop ; Jump if NOT zero
MOV AH, 4CH ; End Program
INT 21H
END
MAIN ENDP
END MAIN
好的，我解决了这个问题。这是最终代码。请记住我只是一个初学者 - 所以如果这段代码不好，请不要给我一个新的。虽然它有效！
; **************** MACROS *********************
; START PROGRAM
START MACRO
MOV AX, DATA ; Data Segment to AX Register
MOV DS, AX ; HAVE DS Point to Data Segment
ENDM
; END PROGRAM
END MACRO
MOV AH, 4CH ; END Program
INT 21H ; Call DOS Service
ENDM
; PRINT STRING TO OUTPUT
PSTRING MACRO STR
MOV AH,09H
LEA DX,STR
INT 21H
ENDM
; Creates a New Line
NEWLINE MACRO
MOV DX, 0AH ; Input of string to DX
MOV AH, 02H ; Write Char to standard output
INT 21H
MOV DL, 0DH ;
MOV AH, 02H ; Carriage Return
INT 21H ;
ENDM
; Get CHAR Input
GETINPUT MACRO INPUT
MOV AH, 01H ; Receive input
INT 21H ; Call DOS Service
MOV INPUT, AL ; Store into Input Variable
ENDM
; ********** END MACROS *******************
.MODEL
.STACK 100H
.DATA
MSG1 db 'Choose A Char to Convert To ASCII: $'
CHAR db ? ; Store Input Char
REM db ? ; Remainder 8-bit storate
QUOT db ? ; Quotient 8-bit storage
COUNT db 0 ; Counts the stacks
.CODE
MAIN PROC
START
PSTRING MSG1
GETINPUT CHAR
NEWLINE
MOV AX, 0 ; Clear AX Register
MOV AL, CHAR ; Move input to AL
MOV DX, 0 ; Clear DX Register
; ********************** ;
; QUOTIENT STORED IN AL ;
; REMAINDER STORED IN AH ;
; ********************** ;
myLoop:
MOV DL, 10 ; Set Divisor to 10
DIV DL ; Divide AX by 10
MOV REM, AH ; Move Remainder into REM Variable
MOV QUOT, AL ; Move Quotient into QUOT Variable
MOV AX, 0 ; Clear AX
MOV AL, REM ; Move REM to AL
PUSH AX ; Push AX to Stack
INC COUNT ; Increase Count by 1
MOV AL, QUOT ; Place Quotient Into AL
MOV AH, 0 ; AH was altered, Zero it out
CMP AL, 0 ; If No Quotient Remains we can exit
JNZ myLoop ; Jump if NOT zero
myLoop2:
POP DX ; Pop from the stack into DX
ADD DX, '0' ; To Ascii Char
MOV AH, 02H ; Print Char Command
INT 21H ; Call to DOS System
DEC COUNT ; Decrement COUNT
CMP COUNT, 0 ; Compare COUNT to 0
JNZ myLoop2
END
MAIN ENDP
END MAIN