doubledabble

.data 
line: .asciiz "\n"
adding: .asciiz "Adding three\n"
.macro print_string($funcArg)
	la $a0, $funcArg
	li $v0, 4
	syscall
.end_macro 

.macro print_bin($toPrint)
	move $a0, $toPrint
	li $v0, 35
	syscall
	print_string(line)
.end_macro

.macro invert($toInvert)
	xori $toInvert, $toInvert, 0xFFFFFFFF
.end_macro 

.macro left_justify($toJustify, $leftMask, $zeros)
	shiftLeft:
		and $t3, $toJustify, $leftMask # Let $t3 = the left most bit
		bnez $t3, end
		sll $toJustify, $toJustify, 1
		addi $zeros, $zeros, 1
		j shiftLeft
	end:
.end_macro

.text
li $t1, -1 # Let $t1 = k
li $s0, 243 # Let $s0 = two's complement int
print_bin($s0)
li $t2, 0x00000001 
sll $t2, $t2, 31 # Let $t2 = left most bit
move $t4, $zero # Let $t4 = result = 0
move $t5, $zero # Let $t5 = number of preceding zeros in $s0
left_justify($s0, $t2, $t5)
subi $t5, $t5, 16
abs $t5, $t5 # Let $t5 = the number of bits to go through

outerloop:
	addi $t1, $t1, 1 # Iterate k
	sll $t4, $t4, 1 # Shift the result over to make room for the incoming bit
	beq $t1, $t5, done # Once we're through the whole register (32 bits) finish
	and $t3, $s0, $t2 # Let $t3 = the left most bit of $s0
	srl $t3, $t3, 31  # Right justify $t3
	or $t4, $t4, $t3  # Set the right most bit of $t4 (result) to $t3
	
	print_bin($s0)
	print_bin($t4)
	sll $s0, $s0, 1 # Move the original number over

	li $t6, -1 # Let $t6 = i
	li $t7, 0x0000000F # Let $t7 = mask
	innerloop: # In the inner loop we have to check if any of the the words are over 4
		addi $t6, $t6, 1 # i++
		beq  $t6, 4, outerloop
		and $t8, $t4, $t7 # Let $t8 = our current word
		bgt $t8, 4, greaterThanFour
		j continue
		greaterThanFour:
			print_string(adding)
			addi $t8, $t8, 3
			invert($t7)
			and $t4, $t4, $t7
			or $t4, $t4, $t8
			invert($t7)
			j outerloop
		continue:
		sll $t7, $t7 4
		j innerloop
done: