DDFSTRT/DDFSTOP for Hires reverse calculation.

[a/b]

My first version (no wicked optimizations...): 2x30=60 bytes.

[ross]

Bah... 68 byte

[paraj]

Also 60 bytes, but unlike a/b it's not because I'm saving my good tricks for round 2...
Still have an annoying andi.w #3,d4 in there that I haven't found a space saving way of avoiding.

[meynaf]

Originally Posted by ross

Bah... 68 byte

You're not alone. 66 here

[ross]

Originally Posted by paraj

Still have an annoying andi.w #3,d4 in there that I haven't found a space saving way of avoiding.

Found a way to remove it, but...

[paraj]

Originally Posted by ross

Found a way to remove it, but...

Actually, so did I, 58 bytes

[a/b]

Actually not saving anyhthing, just haven't found any great optimizations ;p.
Now I'm having another look, validation changed a little... it's down to 58 bytes (no, it wasn't #3 ;p).

[ross]

62..

[meynaf]

Ok, now i've definitely missed something important...

[ross]

Ok, 58, but I'm not sure it is right because my brain stopped working

[paraj]

Originally Posted by ross

Ok, 58, but I'm not sure it is right because my brain stopped working

Aren't you testing it?
Attached is what I'm using (harness is in C since I only have so much code golfing time ) Compile with vc +kick13 -O2 -o test.exe main.c fetch.s

[meynaf]

Originally Posted by paraj

Attached is what I'm using (harness is in C since I only have so much code golfing time ) Compile with vc +kick13 -O2 -o test.exe main.c fetch.s

I see your loop for FMODE goes from 0 to 5. But valid FMODE values are 0 to 15 ; bits 14-15 also have a meaning.
Besides, is DDFSTRT/DDFSTOP high byte always guaranteed to be 0 ?

[paraj]

Originally Posted by meynaf

I see your loop for FMODE goes from 0 to 5. But valid FMODE values are 0 to 15 ; bits 14-15 also have a meaning.
Besides, is DDFSTRT/DDFSTOP high byte always guaranteed to be 0 ?

It doesn't run through all possible values to limit runtime to something reasonable while testing. From Ross' initial challenge I would expected DDF* high bytes to be clear at least ("the DDF registers with usual values") but FMODE to have "any value", so it just checks a few FMODE >3.

I.e. in my code I expect FMODE=$0..$FFFF, DDFSTRT=$0.$FF,DDFSTOP=$0.$FF, chipset=$0..$2, res=$0..$2.

Ultimately it's just for fun, but I'll let Ross decide the true boundary conditions. Personally I'm fine if somebody gets a good result by assuming DDFSTRT <= 0x10 <= DDFSTOP <= 0xf0 or whatever.

[a/b]

I'm using bad ddfstart/ddfstop upper byte (testing with start from $ff18 to $ffe0, and end from start to $ffe0, vs. original c code). Fmode I set to $fffc|<0-3>. The rest is expected to be valid (res/chipset 0-2).

[ross]

Originally Posted by paraj

so it just checks a few FMODE >3.

Wait, this is not the proper way.
FMODE lower two bits should always work, like the 'real' FMODE.

So a value = $8007 is valid and set 'fetch' or that which is =2.

[paraj]

Originally Posted by ross

Wait, this is not the proper way.
FMODE lower two bits should always work, like the 'real' FMODE.

So a value = $8007 is valid and set 'fetch' or that which is =2.

This should be checked properly by the test code (should check 0..5) if I didn't make another mistake.
Anyway, that's just how I tested my 58B version (testing now for $0.$FFFF FMODE since y'all be critical :P), not saying this is the proper way to judge it

[ross]

Originally Posted by paraj

This should be checked properly by the test code (should check 0..5)

Is it a typo? 0..3

[paraj]

Originally Posted by ross

Is it a typo? 0..3

What I meant: Test code checks FMODE=0, FMODE=1, ... FMODE=5. So it should check $8007 implicitly if code handles FMODE=2 the same as (FMODE&3). It wasn't meant to catch all cases and decide a winner, just do quick validation of my own code

[a/b]

Down to 56 bytes. This one I would classify as wicked.

[ross]

Originally Posted by a/b

Down to 56 bytes. This one I would classify as wicked.

I re-observed the 58 byte routine and there are so many similar values and bit positions that it makes you think that something can be 'recycled', but I can't think of anything obvious, so is it possible that you completely refactored the original routine?

Anyway I will soon give up and I will declare you the winner


Is the word 'wicked' that scares me, I dare not imagine what you have done