A standard 4 step counter using two bits of a byte "knows" the following states:

1. step: XXXX XX00
2. step: XXXX XX01
3. step: XXXX XX10
4. step: XXXX XX11

You can also do it this way:

1. step: X00X XXXX
2. step: X01X XXXX
3. step: X10X XXXX
4. step: X11X XXXX

This is trivial, of course. As the two bits are within the same byte one could call it a "horizontal counter". I don't know whether this term exists. But it shall demonstrate here the difference to another type of counter, namely the "vertical counter". This one uses bits of different bytes, where the bits are at the same position within the byte. In the code example of Hans the two bits sit in the bytes CtrA and CtrB and the counting scheme looks as follows:

1. step: CtrB: XXXX XXX0
         CtrA: XXXX XXX0

2. step: CtrB: XXXX XXX1
         CtrA: XXXX XXX0

3. step: CtrB: XXXX XXX0
         CtrA: XXXX XXX1

4. step: CtrB: XXXX XXX1
         CtrA: XXXX XXX1

You can also do it this way:

1. step: CtrB: XXXX X0XX
         CtrA: XXXX X0XX

2. step: CtrB: XXXX X1XX
         CtrA: XXXX X0XX

3. step: CtrB: XXXX X0XX
         CtrA: XXXX X1XX

4. step: CtrB: XXXX X1XX
         CtrA: XXXX X1XX

or this way:

1. step: CtrB: X0XX XXXX
         CtrA: X0XX XXXX

2. step: CtrB: X1XX XXXX
         CtrA: X0XX XXXX

3. step: CtrB: X0XX XXXX
         CtrA: X1XX XXXX

4. step: CtrB: X1XX XXXX
         CtrA: X1XX XXXX

The interesting question is now: How to increment such a vertical counter?
The answer is simple: Whenever the vertical counter is incremented (by one step) the bit in byte CtrB changes its state, means was it "0" becomes it "1" and was it "1" becomes it "0". This is done by the instruction:

xrl     CtrB,#0ffh

You could also do it with the "CPL" instruction, but then the byte must sit in the accumulator.

The bit in byte CtrA, on the other hand, becomes "1", when the bit in CtrA and the bit in CtrB were of different states. Otherwise it becomes "0". So,

mov 	a,CtrB
xrl     CtrA,a

does the trick!

Hope this helps,

Kai