Question

2) Write a MIPS assembly language version of the following C code segment:

int A[75], B[75];

for (i = 1; i < 73; i ++) {

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

}

Arrays A, B and C start at memory location A000hex, B000hex and C000hex respectively. Try to reduce the total number of instructions and the number of expensive instructions such as multiplies.

WRITE A SIMPLE CODE WORKING FOR MARS 4.5

Answer 1

Please find the answer below:::

.data
A : .word 300 #array of size 75 integer
B : .word 300 #array of size 75 integer
C : .word 300 #array of size 75 integer
.text

la $s0,A #get base address of A
la $s1,B #get base address of B
la $s2,C #get base address of C

li $t0,1 #load i =1
loop:
bge $t0,73,exit #if not i<73 exit
mul $t0,$t0,4 #get base address of i
add $t1,$t0,$s0 #get base address of A[i]
sub $t2,$t1,4 #get base address of A[i-1]
lw $t2,($t2) #get value of A[i-1]
add $t3,$t1,4 #get base address of A[i+1]
lw $t3,($t3) #get value of A[i+1]
add $t4,$t0,8 #get address of i+2
add $t4,$t4,$s1 #get address of B[i+2]
lw $t4,($t4) #get value of A[i]
lw $t1,($t1) #get value of A[i]
add $t5,$t0,$s2 #get address of C[i]

add $t6,$t1,$t3 #get A[i] + A[i+1]
sub $t7,$t4,$t2 #get (B[i+2] -A[i-1])
mul $t6,$t6,$t7 #get (A[i] + A[i+1])*(B[i+2] -A[i-1])
sw $t6,($t5) #store result to c[i]

add $t0,$t0,1 #i++
j loop
exit: