Translate the following code into MIPS code. A[i - 6] = A[i + 2] + 10; i = i + 7

Question

Translate the following code into MIPS code.

A[i - 6] = A[i + 2] + 10;

i = i + 7;
A[3] = A[i - 1];

a) Assume A is an array of integers (each integer takes 4 bytes). A's address
is stored at register $10. Also assume that the compiler associates the
variable i to the register $11.
b) Assume A is an array of characters (each character takes one byte). A's address
is stored at register $10. Also assume that the compiler associates the variable i to   the register $11.

PLEASE DO NOT COPY DOWN A DIFFERENT SOLUTION

Explanation / Answer

ANSWER:

li $t0,10 // 10 value decleration

li $1,7 // 7 value decleration

lw $t2, 0($a1) // load A[i+2]

lw $t3, 0($a1) // A[i - 1] code

add $t4,$t1,$t0 //A[i - 6] = A[i+2] +10;

add $t1,7 //i=i+7 code

sw $t3,4($a1) //A[3] = A[i - 1];

EXAMPLE:

for(i=0;i < 98; i++)

{

C[i] = A[i+1] -A[i]*B[i+2]

}

MIPS code with explanaation:

li $s0, 0xA000 # Load Address of A

li $s1, 0xB000 # Load Address of B

li $s2, 0xC000 # Load Address of C

li $t0, 0 #Starting index of i

li $t5,98 #Loop bound

loop;

lw $t1, 0($s1) # Load A[i]

lw $t2, 8($s2) # Load B[i+2]

mul $t3, $t1,$t2 # A[i] * B[i+2]

lw $t1, 4($s1) # Load A[i+1]

add $t2, $t1,$t3 # A[i+1] + A[i]*B[i+2]

sw $t2,4($s3) # C[i] = A[i+1] + A[i]*B[i+2]

addi $s1, 4# Go to A[i+1]

addi $s2, 4# Go to B[i+1]

addi $s3, 4# Go to C[i+1]

addi $t0,1# increment index variable

bne $t0, $t5,loop # Compare with loop Bound

halt:

nop

Example 2:

int A[100], B[100];

for(i=1;i<100;i++)

{

A[i] = A[i-1] + B[i];

}

MIPS CODE:

li $t0, 1 # Starting index of i

li $t5, 100 # Loop bound

loop;

lw $t1, 0($a1) # Load A[i-1]

lw $t2, 4($a2) # Load B[i]

add $t3,$t1,$t2 # A[i-1]+B[i]

sw $t3, 4($a1) # A[i] = A[i-1]+B[i]

addi $a1, 4# Go to i+1

addi $a2, 4 # Go to i+1

addi $t0, $t5,loop #Compare with Loop Bound

halt:

nop

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