Maximum looping of a delay without nesting

[abi072]

MOV R0,#05H
HERE:DJNZ RO,HERE
END
//THE ABOVE LOOP RUNS FOR 5 TIMES.

MOV R0,#0FFH
HERE:DJNZ R0,HERE
END
BUT, I READ IN A BOOK THAT THE ABOVE PROGRAM CAN RUN FOR 256 TIMES. HOW IS THAT POSSIBLE? CAN ANYONE EXPLAIN?

[digsys]

Incorrect - 1st example runs 4 times
Incorrect - 2nd example you load R0 with 0h to get 256 loops
DJNZ means Decrement FIRST and if = 0, DON'T jump. Run it step by step

[abi072]

Thanks for your response,"DIGSYS".
But,
The attached image shows that R5 has to be initialized to FFh for the loop to repeat 256 times

And the example 3.1 of next image says that looping will be for 10 times if initialized to 10.

Please explain that. I'm really confused.

[PA0PBZ]

MOV R0,#05H
HERE:DJNZ RO (4) ,HERE
HERE:DJNZ RO (3) ,HERE
HERE:DJNZ RO (2) ,HERE
HERE:DJNZ RO (1) ,HERE
HERE:DJNZ RO (0) ,HERE
END

So, that is 5 times isn't it? 

The second one repeats 255 times, not 256.

[abi072]

Ya, I agree to you PAOPBZ.

But, the images I have posted tells a different story for my second example

[PA0PBZ]

But, the images I have posted tells a different story for my second example

So the example is wrong.

[abi072]

okay. thanks 

[digsys]

MOV R0,#05H
HERE:DJNZ RO (4) ,HERE
HERE:DJNZ RO (3) ,HERE
HERE:DJNZ RO (2) ,HERE
HERE:DJNZ RO (1) ,HERE
HERE:DJNZ RO (0) ,HERE
END
So, that is 5 times isn't it? 

Except on (0) there is no Jump, it's the exit point. So the routine that would "normally" live in that loop does NOT get executed !!
Which is the whole point of the djnz AND one of it's drawbacks ie You CAN'T use it like a normal index > 00, always have to +1 (or -1)
So to get 256 LOOPS, you load with 00(0?? you say) .. yes because the VERY FIRST instruction is -1cnt ! so the FIRST compare is FFh
And abi ... there are MANY MANY cases of examples being wrong or mis-leading. Trust your own logic

[abi072]

If I modify the example a bit and write it as:

Mov A,#00h
Mov R0,#05h
here: Inc a
DJNZ R0,here
end

Then after executing the above program the value at A=05.

So,

Initially R0=5,A=0

1: A=1. then R0=4. As 4!=0, it goes to here.
2: A=2. R0=3. As 3!=0, it goes to here
3:A=3. R0=2. As 2!=0,ite goes to here
4:A=4.R0=1. As1!=0, it goes to here
5.A=5. R0=0. 0=0. And it terminates.

LOOP means having a common start and end points. RIGHT?? In the 5th case, it doesn't jumps. AGREED. But, The looping takes place 5 times.Because A=5.

Is this logic of mine right?

[digsys]

Well, technically you're correct - yes it does pass that point 5 times ! But you shouldn't think like that -
The purpose of these loops is to run a routine (the actual loop) an exact number of times, in this case 4 !!
There are other types of indexed loops that WILL run it 5 times ! So, it's important to keep that clear in your head.
So when you see a DJNZ Loop, remind to yourself - this will run #val -1 ! It's very easy to miss it and spend hours
wondering why things don't work like they should.

[andersm]

So when you see a DJNZ Loop, remind to yourself - this will run #val -1 !

This is incorrect. The jump will be taken (count-1) times, but the instruction will be reached (count) times. If the DJNZ placed at the bottom of the loop, every time the it's reached counts as an iteration.

Expressed in C, a typical DJNZ loop looks like

Code: [Select]

do {
    // whatever
} while (--count);

[digsys]

This is incorrect. The jump will be taken (count-1) times, but the instruction will be reached (count) times

Ummmm that's what I said -

... yes it does pass that point 5 times .... (the actual loop) .... in this case 4 .... 

[andersm]

You said the loop body will be executed 4 times, which is clearly not the case.