Question

g[8]=4[A[3j] + B[i+6j]

I and j are variables. When doing MIPS instructions for above statement what needs to be done to 4[A[3j]? Choose your own temporary and store registers. Explain show step by step

Answer 1

g[8] = 4[A[3j] + B[i+6j]

1. In order to expand the above statement we need to understand the Instruction Set Architecture (ISA) & Memory allocation

Example ( instruction format for load instruction )

lw $to, 4($t1)

// here $to is the destination address, $t1 inside the bracket is the base address and 4 is offset . Lets assume the base //address was 4n then the 4($t1) will correspond to 4n+4.

Please look at the table below for memory allocation  

Memory allocation

4n

4n+1

4n+2

4n+3

4n+4

4n+5

4n+6

4n+7

4n+8

4n+9

4n+10

4n+11

4n+12

4n+13

4n+14

4n+15

|---------- A[0]-----------|--------------A[1]------------|--------------A[2]--------------|-------------A[3]---------------------|

2. Now that we have understood the ISA basics we can proceed to the question g[8] = 4[A[3j] + B[i+6j]

here i & j are variables so if we consider 4[A[3j] it will be expanded as A[3*j] so A[3] will be the base address with some offset to be added

so if we consider i = 1 , j = 1 , value of A will become A[3] with memory constant 12, This output willl be further multiplied with 4

B will become B[1+6] which is B[7] , Memory constant to be added will become 28

Assembly:

lw $t0, 12($s1) // A[3] is brought in to $t0 with constant offset 4 added

ori $t1, $zero, 4 // store immediate value for 4 in 4[A[3j]]

mult $t0, $t0, $t1 // multiply the value by 4

lw $t2, 28($s2) // B[7] is brought into $t1

add $t0, $to, $t2 // added and stored in $t0

sw $t0, 32($s4) // $t0 is stored into g[8]