George,

It does not write into R0 in the loop.

Let's hand-simulate what happens:
R0 gets filled by 7Fh and it starts to write 0 into RAM addressed by R0 and decrement R0. Now let's start our "simulation" at the moment when R0 contains 2:

   mov   @r0,#0  ;moves 0 into address 02, which is R2 in bank 0
   djnz  r0,Loop   ;r0 gets decremented, so it contains 01, 
                   ;   which is nonzero, so it loops back
   mov   @r0,#0    ;moves 0 into address 01, which is R1 in bank 0
   djnz  r0,Loop   ;r0 gets decremented, so it contains 00, 
                   ;   which is zero, so it does not loop back 
                   ;   but it continues - loop exits

So R0 is at the end of the loop 0, but it has not been written to in the loop. If the constant would be different, e.g. mov @r0,#55h, at the end of the loop the whole RAM in the addresses 01-7F would be filled by 55h, but R0 would contain 00.

You can see it in a different way, the loop counter is R0 and it starts with 7Fh, and as it is the pointer at the same time, it fills 7Fh=127 bytes starting at 7Fh downwards, so it does not fill the whole 128 bytes which would be 00h-7Fh, it fills one less -> 01h-7Fh.

Actually, if it would write 0 into R0 inside the loop (through some other mechanism), it would stay in the loop forever, as after djnz R0 would be FFh which is nonzero so it would keep looping.

I understand this kind of "blindness" - I suffer from it too...

Jan Waclawek