我第一次来这里。我是一名大学生。我使用汇编语言创建了一个简单的程序。我想知道我是否可以使用循环方法来运行它，就像它在我发布的程序下面所做的一样。我也渴望找到可以通过 MSN Messenger 交谈的人，这样我就可以马上问你问题。（如果可能的话）好的，谢谢
.MODEL small
.STACK 400h
.data
prompt db 10,13,'Please enter a 3 digit number, example 100:',10,13,'$' ;10,13 cause to go to next line
first_digit db 0d
second_digit db 0d
third_digit db 0d
Not_prime db 10,13,'This number is not prime!',10,13,'$'
prime db 10,13,'This number is prime!',10,13,'$'
question db 10,13,'Do you want to contine Y/N $'
counter dw 0d
number dw 0d
half dw ?
.code
Start:
mov ax, @data ;establish access to the data segment
mov ds, ax
mov number, 0d
LetsRoll:
mov dx, offset prompt ; print the string (please enter a 3 digit...)
mov ah, 9h
int 21h ;execute
;read FIRST DIGIT
mov ah, 1d ;bios code for read a keystroke
int 21h ;call bios, it is understood that the ascii code will be returned in al
mov first_digit, al ;may as well save a copy
sub al, 30h ;Convert code to an actual integer
cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and
;extends it to ax, doubling its size from 8 bits to 16 bits
;The first digit now occupies all of ax as an integer
mov cx, 100d ;This is so we can calculate 100*1st digit +10*2nd digit + 3rd digit
mul cx ;start to accumulate the 3 digit number in the variable imul cx
;it is understood that the other operand is ax
;AND that the result will use both dx::ax
;but we understand that dx will contain only leading zeros
add number, ax ;save
;variable <number> now contains 1st digit * 10
;----------------------------------------------------------------------
;read SECOND DIGIT, multiply by 10 and add in
mov ah, 1d ;bios code for read a keystroke
int 21h ;call bios, it is understood that the ascii code will be returned in al
mov second_digit, al ;may as well save a copy
sub al, 30h ;Convert code to an actual integer
cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and
;extends it to ax, boubling its size from 8 bits to 16 bits
;The first digit now occupies all of ax as an integer
mov cx, 10d ;continue to accumulate the 3 digit number in the variable
mul cx ;it is understood that the other operand is ax, containing first digit
;AND that the result will use both dx::ax
;but we understand that dx will contain only leading zeros. Ignore them
add number, ax ;save -- nearly finished
;variable <number> now contains 1st digit * 100 + second digit * 10
;----------------------------------------------------------------------
;read THIRD DIGIT, add it in (no multiplication this time)
mov ah, 1d ;bios code for read a keystroke
int 21h ;call bios, it is understood that the ascii code will be returned in al
mov third_digit, al ;may as well save a copy
sub al, 30h ;Convert code to an actual integer
cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and
;extends it to ax, boubling its size from 8 bits to 16 bits
;The first digit now occupies all of ax as an integer
add number, ax ;Both my variable number and ax are 16 bits, so equal size
mov ax, number ;copy contents of number to ax
mov cx, 2h
div cx ;Divide by cx
mov half, ax ;copy the contents of ax to half
mov cx, 2h;
mov ax, number; ;copy numbers to ax
xor dx, dx ;flush dx
jmp prime_check ;jump to prime check
print_question:
mov dx, offset question ;print string (do you want to continue Y/N?)
mov ah, 9h
int 21h ;execute
mov ah, 1h
int 21h ;execute
cmp al, 4eh ;compare
je Exit ;jump to exit
cmp al, 6eh ;compare
je Exit ;jump to exit
cmp al, 59h ;compare
je Start ;jump to start
cmp al, 79h ;compare
je Start ;jump to start
prime_check:
div cx; ;Divide by cx
cmp dx, 0h ;reset the value of dx
je print_not_prime ;jump to not prime
xor dx, dx; ;flush dx
mov ax, number ;copy the contents of number to ax
cmp cx, half ;compare half with cx
je print_prime ;jump to print prime section
inc cx; ;increment cx by one
jmp prime_check ;repeat the prime check
print_prime:
mov dx, offset prime ;print string (this number is prime!)
mov ah, 9h
int 21h ;execute
jmp print_question ;jumps to question (do you want to continue Y/N?) this is for repeat
print_not_prime:
mov dx, offset Not_prime ;print string (this number is not prime!)
mov ah, 9h
int 21h ;execute
jmp print_question ;jumps to question (do you want to continue Y/N?) this is for repeat
Exit:
mov ah, 4ch
int 21h ;execute exit
END Start