dbra to handle long counter

[Geijer]

Am I thinking right if I do this

loop:
        ...
        dbra    d5,loop
	addq.w  #1,d5
	subq.l  #1,d5
	bge.s   loop

instead of this

loop:
        ...
	subq.l  #1,d5
	bge.s   loop

when I want to make sure the loop logic uses as few cycles as possible on ocs/ecs with only chipmem and d5 is normally $64 but can exceed $ffff

[jotd]

unroll your loops instead if you can.

[meynaf]

If you have a free register - even an address register - you can put $10000 in it and follow your dbra with sub.l rx,d5 / bcc .loop.

[Ernst Blofeld]

I'm not an expert, but I believe your code is correct and isn't inefficient. But, to calculate how many cycles different options will take, I think we need to know the proportion of times that D5 exceeds $ffff, because those become big numbers, and therefore it may be worth doing things a different way. We don't know what your loop is doing, but if you need to do it 65 thousand times every now and then, is there a shortcut?

[Ernst Blofeld]

Originally Posted by meynaf

If you have a free register - even an address register - you can put $10000 in it and follow your dbra with sub.l rx,d5 / bcc .loop.

If the dbra only touches the lower 16 bits, could he do a swap on D5 for the outer loop?

[ross]

If you are low on registers:

    move.l  #$counter-1,dx
.l  ..code..
    dbra    dx,.l
    subi.l  #$10000,dx
    bcc.b   .l

[actually same as meynaf ]

[meynaf]

Originally Posted by Ernst Blofeld

If the dbra only touches the lower 16 bits, could he do a swap on D5 for the outer loop?

Yes, swap d5 on start of loop and a swap d5 + subq.w #1,d5 at the end. But it is probably slower than a single sub.l.

[Geijer]

Originally Posted by jotd

unroll your loops instead if you can.

Unfortunantely already unrolled as much as I want to.

Originally Posted by meynaf

If you have a free register - even an address register - you can put $10000 in it and follow your dbra with sub.l rx,d5 / bcc .loop.

Unfortunantely no free registers at all

Originally Posted by ross

If you are low on registers:

    move.l  #$counter-1,dx
.l  ..code..
    dbra    dx,.l
    subi.l  #$10000,dx
    bcc.b   .l

[actually same as meynaf ]

Good idea, thanks to both of you!

Originally Posted by Ernst Blofeld

If the dbra only touches the lower 16 bits, could he do a swap on D5 for the outer loop?

Need to count that, thanks.

[Thomas Richter]

Originally Posted by Geijer

Am I thinking right if I do this

loop:
        ...
        dbra    d5,loop
    addq.w  #1,d5
    subq.l  #1,d5
    bge.s   loop

instead of this

loop:
        ...
    subq.l  #1,d5
    bge.s   loop

when I want to make sure the loop logic uses as few cycles as possible on ocs/ecs with only chipmem and d5 is normally $64 but can exceed $ffff

First of all, the two constructions are not equivalent. If d5 is negative but of absolute value <= 32768, the two constructions perform different. Second, this looks like a rather pointless micro-optimization. Instead of counting every cycle, think about how you can possibly change the algorithm so the loop is not required in first place, or whether you can do something else in parallel while the blitter is performing the loop, or whether you can partially unroll the loop such that the full counter value does not exceed 2^16.

[meynaf]

It is not necessarily pointless. As the OP said, D5 is usually $64 but sometimes it's bigger.
So he wants to be able to treat that special case without slowing down the normal one.

Thinking about it, perhaps it is possible to duplicate the whole routine, making two versions of it. One will use only single dbra, while the other will have the full longword loop.

[Don_Adan]

I remember one dbra routine for handling longword counter. It looks next if i remember right.

 swap D5 ; swapped counter
loop32
 swap D5 ; to word counter
loop
... routine here
 dbra D5,loop
 swap D5 ; to highword counter
 dbra D5,loop32

[Geijer]

Thank you all for your input.