Can someone help me with this MIPS problem, I\'ve asked for help 2 times for thi

Question

Can someone help me with this MIPS problem, I've asked for help 2 times for this problem.

Write a MIPS code that finds the minimum, maximum, sum and average of an array A. In your code, you need to load array A base address in the $s1 register. Once your code completes its execution, minimum, maximum, sum and average will be stored in $s2, $s3, $s4 and $s5 registers, respectively.

Hint:

You can declare arrays with initialization list with the following MIPS declaration.

.data

A: .word 10, 160, 20, 70, 60, 140,150, 80, 90,100, 110, 30, 40, 120,130, 50

Asize: .word 16

ShiftAmount: .word 4

You can assume that the array size is power of 2. You can use shift operation for division calculation. You can use the ShiftAmount variable to specify the total shift.

Also I need to compute the median of the array.

Explanation / Answer

MIPS CODE

.data

A: .word 10, 160, 20, 70, 60, 140,150, 80, 90,100, 110, 30, 40, 120,130, 50

Asize: .word 16

ShiftAmount: .word 4

min: .word      

max; .word     

sum: .word      0

avg: .word       0

.code

la         $t0, A                          ; base address of array           

la         min, 0($t0)                  ; initialize min with first element

la         max, 0($t0)                  ; initialize max with first element       

li          $t1, 0                           ; initiate counter i to access elements as A[i]

LOOP:

sll         $t4,$t1,2                      ; left shift by 2 bits will multiply i by 4 which gives offset of element to be accessed since ith element is stored at 4*i offset from base address

add      $t2, $t0,$t4                 ; add this offset to base address

lw        $t3, 0($t2)                   ; t3 contains A[i]

cmp     $t3, min

blt        min, 0($t3)                  ; if A[i] < min then min = A[i]

cmp     $t3, max

bgt       max, 0($t3)                  ; if A[i] > max then max = A[i]

add      sum, 0($t3)

addi     $t1, #1                         ; increment the counter by 1

cmp     $t1, #16                      

glt        LOOP                         ; if counter <16 then loop again otherwise exit the loop

la         $t1, sum                      ; load sum into t1

sra        $t4,$t1,4                      ; right shift sum by 4 bits to divide it by 16(24) [avg = sum/16]

lw        avg, 0($t4)

EXIT:

_exit

min, max,sum & avg contains corresponding minimum, maximum, sum & average of the array.

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