Sum an Array of Bytes

A count sits at 2000H, and that many bytes follow it from 2001H onwards. Add them up.

Adding two numbers needed no loop. Adding n numbers needs three things working together, and getting any one of them wrong gives a plausible-looking answer that is quietly incorrect:

1. A pointer that advances through memory
2. A counter that says how many are left
3. An accumulator holding the running total

The counter is where beginners lose. DCR sets the zero flag when it reaches zero, so the natural shape is "decrement, jump back if not zero". That runs the body n times only if you set the counter up correctly and test it in the right place.

You will build the loop first, then make it accumulate.