Sum of an array

Six numbers are written into memory by the assembler, and the program adds them up and leaves the total in a. There is one loop, it runs a fixed number of times and there is no condition inside it, so the thing to look at is how the program gets from one number to the next.

An array does not fit in the registers and its elements have no names of their own, so the program keeps the address of the next element in a pair and increments it as it goes.

You need to know: the "Loops and djnz" lecture and the "Arrays, strings and ix" lecture. What is new here is walking memory with a pointer, (hl) reads the byte hl points at and the inc hl under it steps the pointer forward by the size of one element.

(hl) works wherever an 8 bit register works, so add a, (hl) is one instruction and the value never passes through a register of its own. (bc) and (de) cannot do that, they only work with a and only as ld a, (bc) and ld (bc), a, which is why a program that walks one array keeps its pointer in hl.

xor a is the idiom for a = 0. It computes a ^ a, which is zero whatever a held, in one byte where ld a, 0 is two.

The numbers are bytes, one each, so inc hl moves on by one every time round and hl comes out at 9006, one past the last element. a comes out at 6C, which is 108.

Six bytes fit in one byte of total here, and that is the ceiling: the M68K adds words into a long and has room to spare, while this program wraps round as soon as the array adds up to more than 255. A sum that has to be bigger goes in hl, one add hl, de per element, with the byte widened into de first.

Try adding a seventh number to the .db line, say 100. a still comes out at 6C, because count is what the loop counts with and you did not touch it. Change count equ 6 to count equ 7 and it adds the new one too, for D0, which is 208.