问题
Hi i m using dosbox and masm compilor. I want to prompts the user to enter a character, and prints the ASCII code of the character in hex and in binary on the next line. Repeat this process untill the user types a carriage return. The following code work fine to display the charcters in hexadecimal until a carriage return how can i modify this code to display the binary of charcter as well.?
.MODEL SMALL
.STACK 100H
.DATA
PROMPT_1 DB 0DH,0AH,'Enter the character : $'
PROMPT_2 DB 0DH,0AH,'The ASCII code of the given number in HEX form is : $'
.CODE
MAIN PROC
MOV AX, @DATA ; initialize DS
MOV DS, AX
@START: ; jump label
LEA DX, PROMPT_1 ; load and display the string PROMPT_1
MOV AH, 9
INT 21H
MOV AH, 1 ; read a character
INT 21H
MOV BL, AL ; move AL to BL
CMP BL, 0DH ; compare BL with CR
JE @END ; jump to label @END if BL=CR
LEA DX, PROMPT_2 ; load and display the string PROMPT_2
MOV AH, 9
INT 21H
XOR DX, DX ; clear DX
MOV CX, 4 ; move 4 to CX
@LOOP_1: ; loop label
SHL BL, 1 ; shift BL towards left by 1 position
RCL DL, 1 ; rotate DL towards left by 1 position
; through carry
LOOP @LOOP_1 ; jump to label @LOOP_1 if CX!=0
MOV CX, 4 ; move 4 to CX
@LOOP_2: ; loop label
SHL BL, 1 ; shift BL towards left by 1 position
RCL DH, 1 ; rotate DH towards left by 1 position
; through carry
LOOP @LOOP_2 ; jump to label @LOOP_2 if CX!=0
MOV BX, DX ; move DX to BX
MOV CX, 2 ; initialize loop counter
@LOOP_3: ; loop label
CMP CX, 1 ; compare CX wiht 1
JE @SECOND_DIGIT ; jump to label @SECOND_DIGIT if CX=1
MOV DL, BL ; move BL to DL
JMP @NEXT ; jump to label @NEXT
@SECOND_DIGIT: ; jump label
MOV DL, BH ; move BH to DL
@NEXT: ; jump label
MOV AH, 2 ; set output function
CMP DL, 9 ; compare DL with 9
JBE @NUMERIC_DIGIT ; jump to label @NUMERIC_DIGIT if DL<=9
SUB DL, 9 ; convert it to number i.e. 1,2,3,4,5,6
OR DL, 40H ; convert number to letter i.e. A,B...F
JMP @DISPLAY ; jump to label @DISPLAY
@NUMERIC_DIGIT: ; jump label
OR DL, 30H ; convert decimal to ascii code
@DISPLAY: ; jump label
INT 21H ; print the character
LOOP @LOOP_3 ; jump to label @LOOP_3 if CX!=0
JMP @START ; jump to label @START
@END: ; jump label
MOV AH, 4CH ; return control to DOS
INT 21H
MAIN ENDP
END MAIN
回答1:
When you have the value in register, it is stored in CPU in bits (0/1 encoded as low/high current voltage), so it's actually "formatted" in binary!
You just need to output eight characters '0'/'1' per bit, starting from most significant one.
At the moment where you have the character in AL, the code to output the binary form may look like this:
cx = 8 ; 8 bits to output
bin_loop:
rcl al,1 ; move most significant bit into CF
setc bl ; bl = 0 or 1 by CF (80386 instruction)
add bl,'0' ; turn that 0/1 into '0'/'1' ASCII char
call display_bl ; must preserve al and cx
loop bin_loop
Judging by your usage of cx for loop you are in 16b real mode. So if you also can't use 80386 instructions (setc) (when targetting 8086/80186/80286 CPU, like emu8086 emulator), then this can be achieved in other way (like two instructions for example, instead of one).
From your usage of CF in those loops with shl/rcl I'm sure you will figure something out, it's very similar.
回答2:
Ok, Sarah, I made the changes to your code, now it displays binary, I added two variables, all the changes are pointed by little ◄■ arrows :
.MODEL SMALL
.STACK 100H
.DATA
PROMPT_1 DB 0DH,0AH,'Enter the character : $'
PROMPT_2 DB 0DH,0AH,'The ASCII code of the given number in HEX form is : $'
PROMPT_3 DB 0DH,0AH,'The ASCII code of the given number in BIN form is : $'
MY_CHAR DB ? ; ◄■ char entered by user.
BINARY DB 9 DUP('$') ; ◄■ zeroes and ones.
.CODE
MAIN PROC
MOV AX, @DATA ; initialize DS
MOV DS, AX
@START: ; jump label
LEA DX, PROMPT_1 ; load and display the string PROMPT_1
MOV AH, 9
INT 21H
MOV AH, 1 ; read a character
INT 21H
MOV MY_CHAR, AL ; ◄■ save char to use in binary conversion.
MOV BL, AL ; move AL to BL
CMP BL, 0DH ; compare BL with CR
JE @END ; jump to label @END if BL=CR
LEA DX, PROMPT_2 ; load and display the string PROMPT_2
MOV AH, 9
INT 21H
XOR DX, DX ; clear DX
MOV CX, 4 ; move 4 to CX
@LOOP_1: ; loop label
SHL BL, 1 ; shift BL towards left by 1 position
RCL DL, 1 ; rotate DL towards left by 1 position
; through carry
LOOP @LOOP_1 ; jump to label @LOOP_1 if CX!=0
MOV CX, 4 ; move 4 to CX
@LOOP_2: ; loop label
SHL BL, 1 ; shift BL towards left by 1 position
RCL DH, 1 ; rotate DH towards left by 1 position
; through carry
LOOP @LOOP_2 ; jump to label @LOOP_2 if CX!=0
MOV BX, DX ; move DX to BX
MOV CX, 2 ; initialize loop counter
@LOOP_3: ; loop label
CMP CX, 1 ; compare CX wiht 1
JE @SECOND_DIGIT ; jump to label @SECOND_DIGIT if CX=1
MOV DL, BL ; move BL to DL
JMP @NEXT ; jump to label @NEXT
@SECOND_DIGIT: ; jump label
MOV DL, BH ; move BH to DL
@NEXT: ; jump label
MOV AH, 2 ; set output function
CMP DL, 9 ; compare DL with 9
JBE @NUMERIC_DIGIT ; jump to label @NUMERIC_DIGIT if DL<=9
SUB DL, 9 ; convert it to number i.e. 1,2,3,4,5,6
OR DL, 40H ; convert number to letter i.e. A,B...F
JMP @DISPLAY ; jump to label @DISPLAY
@NUMERIC_DIGIT: ; jump label
OR DL, 30H ; convert decimal to ascii code
@DISPLAY: ; jump label
INT 21H ; print the character
LOOP @LOOP_3 ; jump to label @LOOP_3 if CX!=0
;▼ FROM CHAR TO BINARY ▼
LEA SI, BINARY+7 ; ◄■ point to string in data segment.
MOV CX, 8 ; ◄■ maximum number of binary digits.
@BIN_CONVERSION:
SHR MY_CHAR,1 ; ◄■ get rightmost bit.
JC @BIT1
MOV [BYTE PTR SI], '0'
JMP @BIN_SKIP
@BIT1:
MOV [BYTE PTR SI], '1'
@BIN_SKIP:
DEC SI
LOOP @BIN_CONVERSION
LEA DX, PROMPT_3 ; ◄■ display message.
MOV AH, 9
INT 21H
LEA DX, BINARY ; ◄■ display binary.
MOV AH, 9
INT 21H
JMP @START ; jump to label @START
@END: ; jump label
MOV AH, 4CH ; return control to DOS
INT 21H
MAIN ENDP
END MAIN
来源：https://stackoverflow.com/questions/40769766/convert-character-to-binary-assembly-language