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;Euclids Algorithm to find the highest common factor of 2 positive integers


.MODEL SMALL
.STACK 200h
.DATA
WELCOME DB 'Euclids Algorithm for HCF', 10d, 13d, 'for highest common factor', 10d, 13d, '$'

PROMPT1 DB 'Please type the first integer $'
PROMPT2 DB 10d,13d, 'Please type second integer$'
PROMPT3 DB 10d,13d, 'HCF is $'

.CODE
MOV BX, @DATA ; this two state move sets the
MOV DS,BX ; DS Register
MOV AH, 9d ; Bios for print a string
MOV DX,(offset)     WELCOME ; code 9 expects the offset
INT 21h	; to be in DX - prints string at DS::DX
MOV AH,9d
MOV DX, (offset) PROMPT1
INT 21h ; after this , headings are complete
MOV AH, 1d ; get ready to read first integer
INT 21h  ; invoke bios.  Result left in AC. ASCII Code!
SUB AL, 30h  ; sets AC to ACSII minus 30 hex
MOV CL,AL ; save first integer in CL
MOV AH,9d ; get ready to print second prompt
MOV DX, (offset) PROMPT2
INT21h   ; after this read second integer's code
MOV AH,1d
INT21h
SUB AL,30h ; convert ASCII toreal value
REPEAT:
CMP AL,CL
JE PRINT_AND_EXIT
; if they are not equal
JG A_GREATER
; if here, b or CL is greater than AL
SUB CL,AL ; sets CL  to CL-AL
JMP REPEAT
A_GREATER:
SUB AL,CL ; set AL = AL-CL
JMP REPEAT
PRINT_AND_EXIT:
MOV AH,0d
MOV DX,(offseT) PROMPT3
INT 21h
MOV DL,AL
ADD DL,30h
MOV AH,2d
INT 21h
; need to quit
MOV AH,4ch
INT 21h
END

.model small
.stack 200h
.data
	welcome	db	'Running Euclids algorithm to find the HCF', 0Ah, 0Dh, 'of two small positive integers.', 0Ah, 0Dh, '$'
	prompt2		db	0Ah, 0Dh, 'Please type the first integer $'
	prompt3		db	0Ah, 0Dh, 'Please type the second integer $'
	prompt4		db	0Ah, 0Dh, 'The HCF is $'
	goodbye	db	0AH, 0Dh, 'quitting now$'
.code
	;first, set the pointer to the start of the data segment to provide access to the strings
	mov	dx, @data
	mov	ds, dx

	;print opening prompt
	mov	dx, offset welcome	;set the offset to the 'welcome' string
	mov	ah, 9d		;BIOS code for print a dollar terminated string
	int 21h			;invoke BIOS -- prints the string at DS::DX, always

	;begin algorithm by reading in two 1-digit integers
	;print the prompt first, using BIOS code '9d'
	mov	dx, offset prompt2
	mov	ah, 9d
	int 	21h

	;now get the keystroke
	mov	ah, 1d		;BIOS code for read a keystroke
	int	21h		;invoke BIOS, ASCII code will be in register 'al' after this command
	;save the keystroke in register 'cl'
	;after we get the integer value from the ASCII code
	;by subtracting hex 30 from the ASCII code value
	sub	al, 30h
	mov	cl,al

	;now read the second integer as above
	;first print the prompt using BIOS code '1d'
	mov	dx, offset prompt3	;set the offset for DS::DX
	mov	ah, 9d		;set BIOS code
	int	21h		;invoke BIOS

	mov	ah, 1d		;BIOS code for read a keystroke
	int	21h		;invoke BIOS, ASCII code will be in register 'al' after this command
	sub	ah, 30h
	; we now have two integers, actual values not ASCII codes, one in 'cl' and the other in 'al'


repeat:	
	cmp	cl, al		;compare the two integers
	jz	print_and_exit	;print the result and return control to operating system
	;if here, the two numbers were **NOT** the same
	;note that the flags are still set from the compare statement, but it's probably wise to do another
	;one in case we change or edit the file. 
	cmp	cl, al
	jg	greater	;test to see if cl > al, and if it is, jump to label 'greater'
	;OK, so if we get to here, cl **must** be less than (<) al
	sub	al, cl 	;diminish 'al' by 'cl'
	jmp	repeat	;go round the loop until the variables become equal

greater:
	;if we get here, it was shown that cl > al, so diminish 'cl'
	sub	cl, al
	jmp	repeat	;go round the loop until the variables become equal


print_and_exit:
	;there's only one way to get here and that's when the integers are equal
	;so print and then we're done.

	;first the prompt
	mov	dx, offset prompt4
	mov	ah, 9d	;BIOS code for print a string
	int	21h

	;last, the actual result
	mov	ah, 2d	;BIOS code for print to screen
	;character's ASCII code must be in 'dl' for this to work
	mov	dl, cl	;starting to print 'cl's value
	add	dl, 30h	;convert value to ASCII code
	int	21h

	mov dx, offset goodbye
	mov ah, 9d
	int 21h

	mov	ah, 4ch	;BIOS code for exit/quit
	int 	21h

end

.model small
.stack 200h
.data
	welcome	db	'Running Euclids algorithm to find the HCF', 0Ah, 0Dh, 'of two small positive integers.', 0Ah, 0Dh, '$'
	nameoutput	db	'Rose, Matthew 0471844',0Ah, 0Dh, '$'
	prompt2		db	0Ah, 0Dh, 'Please type the first integer $'
	prompt3		db	0Ah, 0Dh, 'Please type the second integer $'
	prompt4		db	0Ah, 0Dh, 'The HCF is $'
	goodbye	db	0AH, 0Dh, 'quitting now$'
.code
	;first, set the pointer to the start of the data segment to provide access to the strings
	mov	dx, @data
	mov	ds, dx
	;print name and ID
	mov dx, offset nameoutput  ;sets the offset to the nameoutput string
	mov ah,9d  		;code to print a dollar terminated string
	int 21h 		;prints the string at DS::DX
	;print opening prompt
	mov	dx, offset welcome	;set the offset to the 'welcome' string
	mov	ah, 9d		;BIOS code for print a dollar terminated string
	int 21h			;invoke BIOS -- prints the string at DS::DX, always

	;begin algorithm by reading in integers
	;print the prompt first, using BIOS code '9d'
	mov	dx, offset prompt2
	mov	ah, 9d
	int 	21h
	MOV AL, 0d
	MUL CX
	MUL BX
enter_loop1:
	;now get the keystroke
	mov	ah, 1d		;BIOS code for read a keystroke
	int	21h		;invoke BIOS, ASCII code will be in register 'al' after this command
	cmp AL,13d
	JZ next_integer
	;save the keystroke in register 'cl'
	;after we get the integer value from the ASCII code
	;by subtracting hex 30 from the ASCII code value
	sub	al, 30h
	add	cl,al
	JMP enter_loop1
	

next_integer:
	;now read the second integer as above
	;first print the prompt using BIOS code '1d'
	mov	dx, offset prompt3	;set the offset for DS::DX
	mov	ah, 9d		;set BIOS code
	int	21h		;invoke BIOS
	
	mov	ah, 1d		;BIOS code for read a keystroke
	int	21h		;invoke BIOS, ASCII code will be in register 'al' after this command
	cmp     al,13d
	JZ repeat
	sub	ah, 30h
	add	bl,al
	JMP next_integer
	; we now have two integers, actual values not ASCII codes, one in 'cl' and the other in 'al'


repeat:	
	cmp	cl, bl		;compare the two integers
	jz	print_and_exit	;print the result and return control to operating system
	;if here, the two numbers were **NOT** the same
	;note that the flags are still set from the compare statement, but it's probably wise to do another
	;one in case we change or edit the file. 
	cmp	cl, bl
	jg	greater	;test to see if cl > bl, and if it is, jump to label 'greater'
	;OK, so if we get to here, cl **must** be less than (<) bl
	sub	bl, cl 	;diminish 'bl' by 'cl'
	jmp	repeat	;go round the loop until the variables become equal

greater:
	;if we get here, it was shown that cl > bl, so diminish 'cl'
	sub	cl, bl
	jmp	repeat	;go round the loop until the variables become equal


print_and_exit:
	;there's only one way to get here and that's when the integers are equal
	;so print and then we're done.

	;first the prompt
	mov	dx, offset prompt4
	mov	ah, 9d	;BIOS code for print a string
	int	21h

	;last, the actual result
	mov	ah, 2d	;BIOS code for print to screen
	;character's ASCII code must be in 'dl' for this to work
	mov	dl, cl	;starting to print 'cl's value
	add	dl, 30h	;convert value to ASCII code
	int	21h

	mov dx, offset goodbye
	mov ah, 9d
	int 21h

	mov	ah, 4ch	;BIOS code for exit/quit
	int 	21h

end

next_integer:
	;now read the second integer as above
	;first print the prompt using BIOS code '1d'
	mov	dx, offset prompt3	;set the offset for DS::DX
	mov	ah, 9d		;set BIOS code
	int	21h		;invoke BIOS

	mov	ah, 1d		;BIOS code for read a keystroke
	int	21h		;invoke BIOS, ASCII code will be in register 'al' after this command
	cmp     al,13d
	JZ repeat
	sub	ah, 30h
	add	bl,al
	JMP next_integer

	;print the prompt first, using BIOS code '9d'
	mov	dx, offset prompt2
	mov	ah, 9d
	int 	21h
	MOV AL, 0d
	MUL CX
	MUL BX
enter_loop1:
	;now get the keystroke
	mov	ah, 1d		;BIOS code for read a keystroke
	int	21h		;invoke BIOS, ASCII code will be in register 'al' after this command
	cmp AL,13d
	JZ next_integer
	;save the keystroke in register 'cl'
	;after we get the integer value from the ASCII code
	;by subtracting hex 30 from the ASCII code value
	sub	al, 30h
	add	cl,al
	JMP enter_loop1

[org 0x0100]

dos	equ	0x021  	; interrupt for DOS function selector
getchar	equ	0x02	; DOS routine code "read character"
putchar	equ	0x02	; DOS routine code "print character"
puts	equ	0x09	; DOS routine code "print string"
quit	equ	0x4c	; DOS routine code "return to DOS"

CR	equ	0x0A	; ASCII carriage return
LF	equ	0x0D	; ASCII line feed
off_zed	equ	0x30	; ASCII zero


[section code]

	;print name and ID
	mov 	dx, nameoutput  ;sets the offset to the nameoutput string
	mov 	ah, puts	;code to print a dollar terminated string
	int 	dos  		;prints the string at DS::DX

	;print opening prompt
	mov	dx, welcome	;set the offset to the 'welcome' string
	mov	ah, puts	;DOS code for print a dollar terminated string
	int 	dos		;invoke DOS -- prints the string at DS::DX, always

	;begin algorithm by reading in integers
	;print the prompt first, using DOS code '9d'
	mov	dx, prompt2
	mov	ah, puts
	int 	dos
	mov 	al, 0
	mul 	cx
	mul 	bx

enter_loop1:
	;now get the keystroke
	mov	ah, getchar	;DOS code for read a keystroke
	int	dos		;invoke DOS, ASCII code will be in register 'al' after this command
	cmp 	al, LF
	jz 	next_integer

	;save the keystroke in register 'cl'
	;after we get the integer value from the ASCII code
	;by subtracting hex 30 from the ASCII code value
	sub	al, off_zed
	add	cl, al
	jmp 	enter_loop1


next_integer:
	;now read the second integer as above
	;first print the prompt using DOS code '1d'
	mov	dx, prompt3	;set the offset for DS::DX
	mov	ah, puts	;set dos code
	int	dos		;invoke dos

enter_loop2:	
	mov	ah, getchar	;dos code for read a keystroke
	int	dos		;invoke dos, ASCII code will be in register 'al' after this command
	cmp     al, LF
	jz 	repeat
	sub	ah, off_zed
	add	bl, al
	jmp 	enter_loop2
	; we now have two integers, actual values not ASCII codes, one in 'cl' and the other in 'al'


repeat:	
	cmp	cl, bl		;compare the two integers
	jz	print_and_exit	;print the result and return control to operating system
	;if here, the two numbers were **NOT** the same

	;note that the flags are still set from the compare statement, but it's probably wise to do another
	;one in case we change or edit the file. 
	cmp	cl, bl
	jg	greater		;test to see if cl > bl, and if it is, jump to label 'greater'

	;OK, so if we get to here, cl **must** be less than (<) bl
	sub	bl, cl 		;diminish 'bl' by 'cl'
	jmp	repeat		;go round the loop until the variables become equal

greater:
	;if we get here, it was shown that cl > bl, so diminish 'cl'
	sub	cl, bl
	jmp	repeat	;go round the loop until the variables become equal


print_and_exit:
	;there's only one way to get here and that's when the integers are equal
	;so print and then we're done.

	;first the prompt
	mov	dx, prompt4
	mov	ah, puts	;dos code for print a string
	int	dos

	;last, the actual result
	mov	ah, putchar	;dos code for print to screen

	;character's ASCII code must be in 'dl' for this to work
	mov	dl, cl		;starting to print 'cl's value
	add	dl, off_zed	;convert value to ASCII code
	int	dos

	mov 	dx, goodbye
	mov 	ah, puts
	int 	dos

	mov	ah, quit	;dos code for exit/quit
	int 	dos

[section data]
welcome		db	"Running Euclids algorithm to find the HCF", CR, LF
		db 	"of two small positive integers.", CR, LF, "$"
nameoutput	db	"Rose, Matthew 0471844", CR, LF, "$"
prompt2		db	CR, LF, "Please type the first integer $"
prompt3		db	CR, LF, "Please type the second integer $"
prompt4		db	CR, LF, "The HCF is $"
goodbye		db	CR, LF, "quitting now$"