CODEWITHSUNDEEP
cassemblymips
Mona Wade
Mona Wade

Reputation: 192

Translate C to MIPS code (understanding memory access in MIPS)

Assume that the variables f and g are assigned to registers $s0, $s1. Assume that the base address of the arrays A and B are in registers $s6 and $s7 respectively. Below is the MIPS code I have written to translate A[2*(f+g)] = B[B[16 + f/2] ]:

# accessing the correct address for A[2*(f+g)] 
line 1. add $t0, $s0, $s1    # $t0 = f + g
line 2. add $s6, $s6, $t0    # A[0] should update to A[(f+g)/4]
line 3. sll $s6, $s6, 3      # A[(f+g)/4] should update to A[(8*((f+g)/4)]

line 4. srl $s0, $s0, 1      # f = f/2
line 5. addi $s0, $s0, 16    # f = f/2 + 16
line 6. sll $s0, $s0, 2      # f = (f/2 + 16) * 4
line 7. add $s7, $s0, $s7    # B[0] should update to B[f/2 + 16] 

line 8. sll $s7, $s7, 2      # B[f/2 + 16] should update to (B[f/2 + 16]) * 4
line 9. add $t0, $s7, $0     # $t0 = (B[f/2 + 16]) * 4
line 10. sw $s6, $t0($s7)     # should be storing B[(B[f/2 + 16]) * 4] in A[2*(f+g)]

I think that I am messing up with access the memory location inside of array B when you need to access it twice. Can someone help?

Upvotes: 0

Views: 187

Answers (1)

Mona Wade
Mona Wade

Reputation: 192

I believe the correct solution is the following:

add $t0, $s0, $s1
sll $t0, $t0, 3
add $s6, $t0, $s6

srl $t1, $s0, 1
addi $t1, $t1, 16
sll $t1, $t1, 2
add $t2, $s7, $t1
lw $t3, 0($t2)
add t4, $s7, $t3
lw $t5, 0($t4)
sw $t5, 0($s6)

Upvotes: 1

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