MIPS integer implementation of floating point addition - Programmers Heaven

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MIPS integer implementation of floating point addition

So I'm trying to write a program in MIPS assembly code to help me better understand how floating point addition works. I understand how a floating point number is broken up into a 1 bit sign, 8 bit exponent, and 23 bit fraction. I've written a program that gets two inputs from a user, and adds them WITHOUT using any floating point instructions except mtc1 and mfc1 (for input and output). My code has bugs because when I add 1 + 2 I get 2.74804688. I'm still trying to debug the code but can't seem to grasp the problem. If anyone can help, I would greatly appreciate it.

THIS IS MY CODE (excluding the user input...the first floating point value is in $s0, and the second in $s1)

#Integer implementation of floating-point addition
#Initialize variables
add $s0,$t0,$zero #first integer value
add $s1,$t1,$zero #second integer value
add $s2,$zero,$zero #initialize sum variable to 0
add $t3,$zero,$zero #initialize SUM OF SIGNIFICANDS value to 0

#get EXPONENT from values
sll $s5,$s0,1 #getting the exponent value
srl $s5,$s5,24 #$s5 = first value EXPONENT

sll $s6,$s1,1 #getting the exponent value
srl $s6,$s6,24 #$s6 = second value EXPONENT

#get SIGN from values
srl $s3,$s0,31 #$s3 = first value SIGN
srl $s4,$s1,31 #$s4 = second value SIGN

#get FRACTION from values
sll $s7,$s0,9
srl $s7,$s0,9 #$s7 = first value FRACTION
sll $t8,$s1,9
srl $t8,$s1,9 #$t8 = second value FRACTION

#compare the exponents of the two numbers
compareExp: ######################

beq $s5,$s6, addSig
blt $s5,$s6, shift1 #if first < second, go to shift1
blt $s6,$s5, shift2 #if second < first, go to shift2
j compareExp

shift1: #shift the smaller number to the right
srl $s7,$s7,1 #shift to the right 1
addi $s5,$s5,1
j compareExp

shift2: #shift the smaller number to the right
#srl $s0,$s0,1 #shift to the right 1
#j compareExp
srl $t8,$t8,1 #shift to the right 1
addi $s6,$s6,1
j compareExp

addSig:

add $t3,$s7,$t8 #Add the SIGNIFICANDS

li $v0, 4
la $a0, sum
syscall

li $v0, 1
move $a0, $t3
syscall

j result

result:
li $v0, 4
la $a0, newline
syscall

sll $t4,$s3,31 #SIGN
#FRACTION
sll $t5,$s6,23 #EXPONENT
add $t6,$t4,$t5
add $t6,$t6,$t3

li $v0, 4
la $a0, sum2
syscall

li $v0, 1
move $a0, $t6
syscall

li $v0, 4
la $a0, newline
syscall

li $v0, 4
la $a0, sum2
syscall

li $v0,2
mtc1 $t6,$f12
syscall
jr $31
# END OF PROGRAM

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