Question

Question 7 Convert C code into MIPS code: A[11] = A[10] +Y Assume Y in $sl and base address of A in $s3 T T T Arial 3 (12pt) V T.EE 25

Answer 1

We are to Write Code Corresonding to A[11] = A[10] + Y in MIPS

IT is given that Y is stored in $s1 and base address of array A is $s3

We will follow following steps to write full code

1). We will store address of A[10] and A[11] in registers $t0 and $t1

(i) we will store 10 and 11 in register $t0 and $t1 respectively using addi

(ii) multiply the content of $t0 and $t0 by 4 using shift left operation

(iii) will add value of base address of array to $t0 and $t1 and store result in $t0 and $t1 respectively.

$t0 and $t1 now contains address of A[10] and A[11] respectively.

2).

(i)fetching A[10] in register $t2

(ii)then adding A[10] with value of Y which is stored in $s1.

3).

FINALLY we stored A[10]+Y in memory location A[11] using $t1

Following Image contains full code along with explananation of each line of code

។ who register add immediate addi addi SHI 2 ΟΥ shift lgt multiply content of $to legical SI $to, $2000, lo # $to=o+10=10 $tig $2000, 11 # $t1 = 0 + 11 = 11 $to, &to , 2 #deft shiftby, by 4 $t1, $t1, 2 # multiply content of $ti by 2 $to, $to, $53 # $to = 4A [lo] Centoins address of A[o] $tig $t1, $53 # $t1= 4 ACI] عليه add ada dood 여 [ ada $tz, ol$to) # fetching Alloo in registus $tz, $ta, $51 7 calculting YTAG) 4 storing in $12, o($t) register store [ sw # ACID= A[lo]+Y stering in memory Tuiven base bose address of register $534 y in $5,

addi $t0,$zero,10 # $t0 = 10

addi $t1,$zero,11 # $t1 =11

sll $t0,$t0,2 #left shift by 2 or multiply by 4   $t0 = $t0*4

sll $t1,$t1,2 # $t1=$t*4

add $t0,$t0,$s3 # $t0 = & A[11]

add $t1,$t1,$s3 # $t1 = &A[11]

lw $t2,0($t0) #fetching A[10] in register

add $t2,$t2,$s1 # $t2 = Y+A[10]

sw $t2,0($t1) # A[11] = A[10]+Y