CODEWITHSUNDEEP
cassemblymips
Lord Bahamut
Lord Bahamut

Reputation: 153

Converting C code to MIPS Assembly Language

Here's the C code:

int A[10];
int sum = 0;
int i = 0;

while (i<10){
sum += A[i++];
sum *= 2;
}

Here's my take at converting into MIPS:

**Reg. Allocation Table:
A = $s1
sum = $s2
i = $s3
10 = $s4**

loop: beq $s3, $s4, endloop 

"here's where i get stuck, inside the while loop."

j: loop
endloop:

I understand that in a basic while loop such as: i = $s1, 5 = $s3

i=0;
while(i != 5)
i=i+1;

addi $s1, $zero, 0 #i=0

loop: beq $s1, $s3, endloop
add   $s1, $s1, 1
j     loop:

endloop:

I'm just having trouble, or having a hard time comprehending a slightly more difficult loop where sum+= A[i++]; and sum*= 2; are introduced.

Any help is greatly appreciated. I'm not looking for a complete solution, so please help me think.

Thank you!

Upvotes: 3

Views: 10222

Answers (1)

Konrad Lindenbach
Konrad Lindenbach

Reputation: 4961

Break it down:

sum += A[i++]; becomes sum = sum + A[i]; i = i + 1; which in MIPS assembly could be expressed:

add $t0 $s3 $s1 #index A at i
lw  $t1 0($t0)  #load A at i

add  $s2 $s2 $t1 #add A[i] to sum
addi $s3 $s3 1   #increment i

sll $s2 $s2 1 #double sum

Upvotes: 4

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