ECOAR_RISCV_Musab_Alsaifi.asm

/*
Musab Alsaifi
*/
.data
prompt_b:	.string		"Enter the value for Blue value: "
prompt_g:	.string		"Enter the valur for Green value: "
prompt_r: 	.string		"Enter the value for Red value: "
prompt_dx:	.string		"Enter the dx direction: "
prompt_dy:	.string 	"Enter the dy direction: "

#Const.
.eqv    WIDTH 		512
.eqv    HEIGHT 		256
.eqv    PRINT_STR 	4
.eqv    READ_INT 	5
.eqv    EXIT_0      	10
.eqv    ADDRESS 	268697600
.eqv    SLEEP      	32

.text
	#print prompt_r 
	li		a7, PRINT_STR
	la		a0, prompt_r
	ecall
	
	#read and store the value for Red
	li		a7, READ_INT	
	ecall
    	slli    	s0, a0, 16
	
	#print prompt_g 
	li		a7, PRINT_STR
	la		a0, prompt_g
	ecall
	
	#read and store the value for Grean
	li		a7, READ_INT
	ecall
    	slli    	a0, a0, 8
	add 		s0, s0, a0
	
	#print prompt_b
	li		a7, PRINT_STR
	la		a0, prompt_b
	ecall
	
	#read and store the value for Blue
	li		a7, READ_INT
	ecall
	add 		s0, s0, a0

	#print prompt_dx
	li		a7, PRINT_STR
	la		a0, prompt_dx
	ecall
	
	#read and store the value for dx
	li		a7, READ_INT
	ecall
	mv 		s3, a0

	#print prompt_dy
	li		a7, PRINT_STR
	la		a0, prompt_dy
	ecall
	
	#read and store the value for dy
	li		a7, READ_INT
	ecall
	mv 		s4, a0

	li 		s5, ADDRESS
    	#bytes per line
	li		t0, WIDTH	
	slli	t0, t0, 2

    	#number of lines to skip to get to the middle line
	li		t1, HEIGHT
	srli 	t1, t1, 1
    	mv      s6, t1
	li		t2, -1

#number of divisions to go to the middle of the height. So, if height is 256, then half of it is 128 which is 2^7
loop:
	beqz	t1, middle
	srli	t1, t1, 1
	addi	t2, t2, 1
	j		loop

#go to the middle of the height  
middle:
	sll		t3, t0, t2		#t0 * 7, so now we are at the right of middle screen
	li		t1, WIDTH
	slli		t1, t1, 1		#divide width by 2
	sub		t3, t3, t1		#substract half of the width to go to the middle of the screen (0, 0)
	add 		s5, s5, t3		#go to point (0, 0) which is the middle of the screen

    	li      	s7, WIDTH	
    	srli    	s7, s7, 1		#store in s7 half of the width cz we need to use it later

#implementation of bresenham algorithm
	sub		t2, s4, s3 		#e
    	slli    	t2, t2, 1
    	li      	t3, 0       	#x
    	li      	t4, 0       	#y
    	li      	t5, 4
    	li      	a7, SLEEP
    	li      	a0, EXIT_0

neg_x:
	bgtz		s3, neg_y
	neg 		t5, t5
    	neg     	s3, s3

neg_y:
	bgtz		s4, while
	neg 		t0, t0
    	neg     	s4, s4

while:
    	bne     	t4, s6, check_x
    	neg     	t4, t4
    	neg     	t0, t0

check_x:
    	bne     	t3, s7, draw
    	neg     	t3, t3
    	neg     	t5, t5

if:
	bltz 		t2, cont
	sub     	s5, s5, t0
    	addi    	t4, t4, 1
	sub		t2, t2, s3
    	j       	while

cont:
    	addi    	t3, t3, 1
	add	    	s5, s5, t5
    	add     	t2, t2, s4
	j		while

#Drawing pixal
draw:
    	ecall
	sw 		s0, (s5)
	j       	if

exit:
	li	    	a7, EXIT_0
	ecall