Sum of an array

Six numbers are written into memory by the assembler, and the program adds them up and leaves the total in d0. 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 an address register and increments it as it goes.

You need to know: the "Loops and dbra" lecture and the "Arrays, strings and (a0)+" lecture. What is new here is walking memory with a pointer, (a0) reads the memory a0 points at and (a0)+ reads it and then steps a0 forward by the size of the read.

The numbers are words, two bytes each, so add.w (a0)+, d0 leaves a0 two bytes further along every time round. They sit at $1014, right after the code, on the first page the memory panel shows. Step through the loop and you can watch a0 grow by two at every add.w, and when the program stops d0 holds 108.

Try adding a seventh number to the dc.w line. d0 still comes out at 108, 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.