Im trying to give a one digit number, and know if the parity is odd or even, for example, give 9 and print that is an odd number.
This is what I have:
assume cs:cseg,ds:dseg,ss:sseg
cseg segment
start:
mov ax, dseg
mov ds, ax
mov ah, 01h ; Here, im adding a number
int 21h
jp even
jnp odd
even:
mov ah,09
lea dx,par
int 21h
jmp exit
odd:
mov ah,09
lea dx,odd1
int 21h
jmp salir
salir:
mov ax,4C00h
int 21h
cseg ends
dseg segment byte
even Db 'Even number$'
odd11 Db 'Odd number$'
dseg ends
sseg segment stack
db 100h dup(?)
sseg ends
end start
Thanks! And sorry for my bad english.
To test if a number is odd or even, you check bit 0 of this number. if bit0 is set, then the number is odd. nothing else. Don't confuse PF (parity flag) and JP/JNP conditions. Parity Flag shows if the number of bits set in the least significant byte is even or odd.
From documentation:
PF (bit 2) Parity flag — Set if the least-significant byte of the result contains an even number of 1 bits; cleared otherwise.
Number 1 - odd number, 00000001b PF is cleared because there's only one bit set
Number 2 - even number, 00000010b, but PF is cleared again! because there's only one bit set
Number 3 - odd number, 00000011b, but PF set! because there're two bits set
While Alexander Zhak's answer pointed out the flaws in your code checking the parity flag, the fastest way to check the parity of the number in al in a non-desctructive way is:
test al,1
jz even
odd: ...
even: ...
To test if a number is odd or even, just shift it right once and check the carry flag:
mov al, 09
shr al ; shifts all bits right 1, lowest bit to carry flag
jc _odd ; carry set - there was a bit in lowest position
_even:
; Do something with even numbers and exit
_odd:
; Do something with odd numbers and exit
Understanding the carry flag is very useful for a whole variety of coding tricks. For instance, "Parity" is counting up the total number of bits in the number and we can use a similar trick for that:
mov ah, 09
xor al, al ; al = 0
_loop:
shr ah ; lower bit into carry flag
adc al, 0 ; Add 0 + carry flag to al
and ah, ah ; sets Z flag if ah is zero
bne _loop
; al now contains the total number of bits contained in ah
I found the solution!
Just add "and al,00000001" above of jp even
assume cs:cseg,ds:dseg,ss:sseg
cseg segment
start:
mov ax, dseg
mov ds, ax
mov ah, 01h ; Here, im adding a number
int 21h
and al,00000001
jp even
jnp odd
even:
mov ah,09
lea dx,par
int 21h
jmp exit
odd:
mov ah,09
lea dx,odd1
int 21h
jmp salir
salir:
mov ax,4C00h
int 21h
cseg ends
dseg segment byte
even Db 'Even number$'
odd11 Db 'Odd number$'
dseg ends
sseg segment stack
db 100h dup(?)
sseg ends
end start
.MODEL SMALL
.STACK 100H
.DATA
PROMPT_1 DB 'Enter a number: $'
PROMPT_2 DB 0DH,0AH,'The given number in binary is: $'
PROMPT_3 DB 0DH,0AH,'$'
MSG1 DB 10,13,’Number is Odd $’
MSG2 DB 10,13,’Number is Even $’
.CODE
MAIN PROC
MOV AX, @DATA
MOV DS, AX
LEA DX, PROMPT_1
MOV AH, 9
INT 21H
XOR BL, BL
MOV CX, 8
MOV AH, 1
@INPUT:
INT 21H
CMP AL, 0DH
JE @END
AND AL, 0FH
SHL BL, 4
OR BL, AL
LOOP @INPUT
@END:
MOV AL, BL
MOV CX, 8
@LOOP:
SHR AL, 1
RCL BL, 0
LOOP @LOOP
LEA DX, PROMPT_2
MOV AH, 9
INT 21H
MOV CX, 8
MOV AH, 2
@OUTPUT:
SHL BL, 1
JNC @ZERO
MOV DL, 31H
JMP @DISPLAY
@ZERO:
MOV DL, 30H
@DISPLAY:
INT 21H
LOOP @OUTPUT
MOV AH,01H
INT 21H
mov ah,01h
SAR AL,01h
JC ODD
LEA SI,MSG1
CALL PRINT
JMP TERMINATE
ODD:
LEA SI,MSG2
CALL PRINT
TERMINATE:
MOV AH,4CH
INT 21H
PRINT PROC
MOV DX,SI
MOV AH,09H
INT 21H
MOV AH, 4CH
INT 21H
END MAIN
来源：https://stackoverflow.com/questions/29292455/parity-of-a-number-assembly-8086