I'm sure this has been done before but it's a handy reminder that the carry is unaffected by logical instructions, increment, and decrement... this little routine takes an integer of any desired length (well, up to 256, but if you've got 2kbit integers, you're probably not using a 6502!) and negates it. X is a pointer to the integer on page zero; Y is the length of the integer in bytes.



A couple of signed arithmetic routines needed to change the sign of their operands before doing their sums; doing it this way saves a lot of space and doesn't take much longer than doing it inline.

Neil

I have visions of you doing maths on 256-byte-long integers.

I think you could use this instead of the EOR:

Doh, yes of course I can. Which saves an extra byte, too as well as being a few clocks faster... that's from too much thinking in hardware terms... first chip inverts it, second chip adds it, with the carry handily set

Neil

If you'd switch to big endian values it could be even smoother:



Actually I guess that for long values the big endian mode is a better choice

Ah, but the problem is that I'm too old and set in my ways to believe that big endian could ever be better, even if it clearly is

I think you can get little endian mode as good as big endian, if you use the rollover 'feature' of the zeropage addressing, if I'm not mistaken. Or even without that. Just use 'base-251' as address and count x from 251 up.

What do you think?

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Author of the GeckOS multitasking operating system, the usb65 stack, designer of the Micro-PET and many more 6502 content: http://6502.org/users/andre/

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x86?  We ain't got no x86.  We don't NEED no stinking x86!

Agreed on zp. I'm too deep into 6502 still

But if values are anywhere in memory you could just use

To invert a 32 bit / 4 byte value.

Right?

Edit: corrected off-by-one from 251 to 252

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Author of the GeckOS multitasking operating system, the usb65 stack, designer of the Micro-PET and many more 6502 content: http://6502.org/users/andre/

For a 32bit/4 bytes value the index you are looking for is 252 (4 iterations: 252...255).
Anyway you're falling almost certainly in a situation where you're crossing the page boundary, so 1 cycle will be added in two of those instructions; in the end you're just sparing the 2 bytes of the CPX #val.

It's a just matter of optimization for code size vs speed.

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Author of the GeckOS multitasking operating system, the usb65 stack, designer of the Micro-PET and many more 6502 content: http://6502.org/users/andre/

Thanks for the hint. I edited the post to the correct value.

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Author of the GeckOS multitasking operating system, the usb65 stack, designer of the Micro-PET and many more 6502 content: http://6502.org/users/andre/

Woz in 1976 would have done the same without the slightest hesitation.

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Got a kilobyte lying fallow in your 65xx's memory map? Sprinkle some VTL02C on it and see how it grows on you!

Mike B. (about me) (learning how to github)