Bubble sort

Eight words in memory, sorted from smallest to largest where they lie. The loop compares each pair of neighbours and swaps them when they are the wrong way round, and it does that as many times as there are elements, so the largest number reaches the end on the first pass and the rest follow.

Every loop up to here read an array once. This one reads it seven times, with an inner loop that walks the array and an outer loop that says how often, and the two counters have to be kept apart.

You need to know: the "Loops and dbra" lecture and the "Fill an array with the numbers from 1 to 10" Example. What is new here is nesting: the inner counter is set inside the outer loop, because it has to start again from the top on every pass.

(a0) and 2(a0) are the pair being compared, the element the pointer is on and the one two bytes after it, which is the next word. Reading both without moving a0 is what makes the swap two plain move.w instructions.

Both the lea numbers, a0 and the move.w d0, d1 belong inside the outer loop. a0 has walked to the end of the array by the time a pass finishes, so it goes back to the start; and d1 is $FFFF by then, which as a counter would run the inner loop 65536 times. Copying d0 into it is also what makes each pass shorter than the one before, since the last element is already in its place after the first pass, the last two after the second, and so on.

Run it with the memory panel on 2000 and the eight words read 1, 4, 8, 15, 16, 23, 42 and 99, in that order. It took 220 instructions to sort eight numbers, and it would take about four times as many to sort sixteen, because both loops grow with the array.

Try changing bge in_order to ble in_order. The same program sorts the other way round, largest first, because the only thing that says which order you wanted is that one condition.