Three Operands Addition

The assignment content discusses code for three operands and two operands. It explains how the code executes addition instructions and how it differs from adding three operands. The code also includes assembly language instructions, such as lw, addiu, and beq, and a procedure call to demonstrate function calling.

1. Answer
a. Code for Three Operands
add $t0, $s1, $s2
add $s0, $t0, $s3
b. Code for Two Operands
add $s0, $s1
add $s0, $s2
add $s0, $s3
Here s0 is A , s1 is B , s2 is C and s3 is D
in first line s0 will get whatever is s1 , in second line s0 will be added with s2 and in last line s0
( which have addition of s0+s1) will be added with s3 , so you will get all the added values of s1 ,
s2 and s3 in s0.
Difference between 3-Operand Arithmetic instruction, is 2 operands will take extra time to
calculate, and it is executing add instruction three times, however in 3 operands have to execute
add two times only.
2. Answer
i) $T0 = 1
ii) Sorting of Arrays
3. Answer
.text
.globl main
main:
lw $s1, B # Load the word stored in label B
addiu $s0, $s1, -201 # here S0 is A and s1 is B
.data

4. Answer
if ( q != r)
r = r + 1 ;
else
q = r * 2 ;
.text
.globl main
main:
lw $t0, Q # Load the word stored in label Q
lw $t1, R # Load the word stored in label R
beq $t0, $t1, ELSE
add $t1, $t1,1 #
ELSE:
li $s0, 2
mult $t0, $s0 # r * 2
mflo $t0
.data
Q: .word 4
R: .word 4
5. Answer
Address/Immediate
Instruction Type Opcode RS RT RD SHAMT FUNCT
lw $t0,0($s0) I 0x23 NA
addi $s0,$s0,4 I 0x08 NA
slt $t3,$t0,$t1 R 0x00 0x2A

beq
$t3,$zero,rept I 0x04 NA
j loop0 J 0x02
6. Answer
main:
la $t0, n
lw $t1, 100
li $so, 0
lw $t1,0($s0)
Loop:
bne $t1, $zero, EXIT
add $t1, $s0, $t1
addi $t1, $t1, -1
j Loop
exit:
7. Answer
.text
main: #assume value a is already in $t0, b in $t1
add $a0,$0,$t0 # it's the same function as move the value
add $a1,$0,$t1
jal subr # call procedure
add $t3,$0,$v0 # move the return value from $v0 to where we want
syscall
subr:
addi $sp,$sp,-4 # Moving Stack pointer
sw $t0, 0($sp) # Store previous value
sub $t0,$a0,$a1 # Procedure Body
add $v0,$0,$t0 # Result
lw $t0, 0($sp) # Load previous value
addi $sp,$sp,4 # Moving Stack pointer
jr $ra