好吧,这是我的ASM代码的一部分:
MOV AH, 0 ;reset AH before division
MOV AL,[myNum] ;move the inputed number to AL
DIV two ;div 123 / 2;
CMP AH,0
JNE inputIsPrime
;If 123 % 2 = 0, output 123 / 2.
DIV ten
MOV DH,AH
SUB AL,'0'
MOV AH,2
MOV DL, AL
INT 21h
MOV divisionCalc,DH
MOV AH,2
MOV DL,DH
INT 21h
JMP endProg
我要达到的目的是,在键盘上输入“ 60”并按Enter键后,输出错误。我想得到数字“ 30”,因为60/2 =30。如果输入42,我想得到21作为输出。
为什么我的代码失败的任何想法?
这是完整的代码:
.MODEL SMALL
.STACK 100h
.DATA
DisplayString DB 'Enter number up to 120:', 13,10,'$'
isPrimeNum DB 'is prime', 13,10,'$'
goodMSG DB 'good', 13,10,'$'
badMSG DB 'bad', 13,10,'$'
divisionCalc DB ?
myNum DB ?
two DB 2
three DB 3
five DB 5
seven DB 7
ten DB 10
eleven DB 11
.CODE
Begin:
MOV AX,@DATA
MOV DS,AX
MOV AH,9
MOV DX,OFFSET DisplayString
INT 21h
MOV BL,0 ; Initialize BL to zero!
; //READ 3 DIGITS // ;
;read first digit for e.g. '1'
MOV ah,1h
INT 21h ;read into AL
SUB AL,30h ; Convert the digit from ASCII to DECIMAL
MOV myNum,AL
MOV AH,1
INT 21h
CMP AL,13 ;is it Enter?
JE endInput
SUB AL,'0' ;Not enter, let's save this new char
MOV CL, AL ; we save the 2nd char to CL
MOV AL, myNum ; lets move our first char to AL
MUL Ten ; multiply by 10 the first char
MOV myNum,AL ;move that to myNum
ADD myNum,CL ; add to AL the 2nd char.
MOV AH,1
INT 21h
CMP AL,13 ; is it enter?
JE endInput
SUB AL,'0' ;Not enter, let's save this new char
MOV CL, AL ; we save the 2nd char to CL
MOV AL, myNum ; lets move our first char to AL
MUL Ten ; multiply by 10 the first char
MOV myNum,AL ;move that to myNum
ADD myNum,CL ; add to AL the 2nd char.
mov AH,1 ; if the number is 3 chars then this will be the enter now.
int 21h
; // FINISH READING 3 DIGITS // ;
endInput:
; AL = AX / two
; AH = AX % two
MOV AH, 0 ;reset AH before division
MOV AL,myNum ;move the inputed number to AL
DIV two ;div 123 / 2;
CMP AH,0
JNE inputIsPrime
;If 123 % 2 = 0, output 123 / 2.
DIV ten
MOV DH,AH
SUB AL,'0'
MOV AH,2
MOV DL, AL
INT 21h
MOV divisionCalc,DH
MOV AH,2
MOV DL,DH
INT 21h
JMP endProg
inputIsPrime:
MOV AH,9
MOV DX,OFFSET isPrimeNum
INT 21h
endProg:
MOV AH,4Ch
INT 21h
END Begin
除以2的最简单方法是右移一位的值,
该位移将存储在进位标志中
所以
shr AX,1
jnc isDividable ; no carry set = there was no division "rest"
将检查AX是否可被2除
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我来说两句