the problem is with solution. I do not understand how tehe method to translate f

Question

the problem is with solution. I do not understand how tehe method to translate from C to MIPS. please I need explanation in details for each character in solution.
thanks

pogle.com/drive/folders/OB E4cGBSLAUbYi1rcEdLZ09sZDA Convert the following to MIPS. Be sure to use the following tableTor your conversions Use only true MIPS instructions; no pseudo instructions are to be used in any of your conversions. Please add comments when needed for clarity. Total points: 6 Table 1: For all C to MIPS Conversions C variable MIPS register replacement $s0 $s1 $s2 $s3 $s4$a0 (a) (3 points) Translate the following C code to MIPS Assembly: j = a[2] ; j += 2; a[10] = j; (b) (3 points) Translate to MIPS: i·a[j] ; a[k] =1; 40 3 to, ao, s LI Page 1 of 2

Explanation / Answer

PLEASE FIND THE ANSWE BELOW:

j = a[2]
j+=2
a[10] =j;

To access a[2]. Need the base address of a that is stored in $a0 as shown in table.
now each integer in mips will be of size 4.
So a[0] will store at base address
a[1] will store at base address + 4
a[2] will store at base address + 8
So by doing 8($a0) can access a[2]

to get some value from address need load word lw instruction
that's why j = a[2] will be lw $s2,8($a0)

Now j+=2 increasing j by 2
so same can be done by adding 2 in $s2
that why j+=2 will be addi $s2,$s2,2

again to access a[10] == base address of a + 40 (4 bit for each integer)
and to store at address will use save word sw instruction.
that why a[10] = j will be sw $s2,40($a0)

So final conversion will be

lw $s2,8($a0)
$s2,$s2,2
sw $s2,40($a0)

B)

i=a[j]

a[k]=i;

first a[j] can be accessed by base address of a + ( j * 4) (to convert address for integer)

so doing shift left(sll) one time will multiply number by 2
so to multiply number by 4 will do sll 2 time and will store the result to temp register $t0
so j can be converted to integer index like below

sll $t0,$s2,2 # do the left shift twice and store result to $t0

now add the address to base address

add $t0,$t0,$a0

so instruction i = a[j] will be
lw $s1,0($t0)

again
a[k] will be like
sll $t0,$s3,2 #store k address to t0
add $t0,$t0,$a0 #add address to base address
sw $s1, 0($t0)

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