The largest element

Eight words sit in memory and the program walks them once, keeping the largest one it has seen so far in d0 and the position it was found at in d3. One of the numbers is negative, which is what makes the choice of condition matter.

Adding up an array needs nothing carried between passes. Finding the largest does: every pass has to compare its element against the best one so far, which is the shape of every "find the best" program there is.

count equ 8

    lea numbers, a0     ; a0 points at the first element
    move.w (a0)+, d0    ; best = numbers[0]
    clr.w d3            ; where = 0
    clr.w d4            ; i = 0
    move.w #count-2, d1 ; seven elements left, and dbra counts one more
loop:
    addq.w #1, d4       ; i++
    move.w (a0)+, d2    ; n = *a0++
    cmp.w d0, d2        ; n - best
    ble not_bigger      ; if(n <= best) keep the one we have
    move.w d2, d0       ; best = n
    move.w d4, d3       ; where = i
not_bigger:
    dbra d1, loop

    org $2000
numbers: dc.w 12, -4, 37, 8, 99, 41, 2, 60

The first element is read before the loop, into d0 and with (a0)+ stepping past it, so the loop itself has only seven elements left and starts with an answer that is already right for the part of the array it has seen. Starting d0 at 0 instead would be a different program, one that answers 0 for an array of negative numbers.

d0 comes out at 00000063, which is 99, and d3 at 00000004: 99 is the fifth element and the first one is number 0.

ble is the signed condition, and the -4 in the array is why it has to be. Change ble not_bigger to bls not_bigger, the unsigned one, and run it again: d0 comes out at 0000FFFC and d3 at 1. Read as an unsigned word, FFFC is 65532, so the program decides that -4 is the largest number in the array and nothing else gets a look in. Both versions assemble, both run to the end, and one of them is wrong.