How many key groups can you activate at one time? If I need to use keys from Groups 0,1,5,and 6, would you recommend that I use direct input/output, or just use getCSC?
I'm afraid I don't get your question. By groups, do you mean having something like: 01101110b as the group? I think that is allowed, but you will get weird results, and won't be able to know which key someone pressed. (ASMin28days has that example, I think). If you need to check all of those key groups, just resend the group to port 1. It will probably still be faster than GetCSC. All you would do is send the group, nop twice, and then check your keys. After you've checked them, send another group, and keep going.
kk. that answers. thanks.
Code:
this work??
Code:
KeyCheck:
ld a,%11111110
out ($01),a
nop
in a,($01)
bit 0,a
call z,DownPressed
bit 1,a
jr z,LeftPressed
bit 2,a
jr z,RightPressed
bit 3,a
jr z,UpPressed
ld a,%11111101
out ($01),a
nop
in a,($01)
bit 3,a
jr z,TimesPressed
bit 4,a
jr z,DivPressed
bit 5,a
jr z,ModPressed
bit 6,a
jr z,ClearPressed
ld a,%11011111
out ($01),a
nop
in a,($01)
bit 6,a
jr z,MathPressed
bit 7,a
jr z,AlphaPressed
ld a,%10111111
out ($01),a
nop
in a,($01)
bit 0,a
jr z,GraphPressed
bit 1,a
jr z,TracePressed
bit 2,a
jr z,ZoomPressed
bit 3,a
jr z,WindowPressed
bit 4,a
jr z,YEquPressed
bit 5,a
jr z,SecondPressed
bit 6,a
jr z,ModePressedthis work??
Assuming all of those numbers are correct, yes, that is the way you do it. Why do you calll z, the first time, and then jr z, the rest of the time? Also, you might want to do two nops, that is what Kerm told me to do a long time ago when I was trying to do direct input. A single nop might work for slower speeds, but if you bump it up to 15mhz, you will probably need two. Just my two cents.
I second everything that Tanner just said, namely:
>> 2 nops, not 1
>> Why the call z / jr z inconsistency?
>> Yes, that's the way to do it, and still much faster than GetCSC
You could, alternatively, come up with a table-based approach that would cut code size, but lose you a tiny bit of speed. Probably not worth it.
>> 2 nops, not 1
>> Why the call z / jr z inconsistency?
>> Yes, that's the way to do it, and still much faster than GetCSC
You could, alternatively, come up with a table-based approach that would cut code size, but lose you a tiny bit of speed. Probably not worth it.
ACagliano wrote:
The call is an error. Should be all jr z's
That's fair; I can only assume that that came from an incomplete conversion from all calls to all jumps? What about our other concerns?
As for the nop's, I put one in because of this line...
Should I have two?
As for doing a table, I'm interested in speed. Don't want to fall behind, since CALCnet will be active...
Z80 heaven wrote:
The keyboard isn't one of the slower drivers, so we won't need a delay. We'll just test to see if any keys were pressed.
Should I have two?
As for doing a table, I'm interested in speed. Don't want to fall behind, since CALCnet will be active...
souvik1997 wrote:
You _need_ two nops, as described by ASMin28days.
This. I've experimentally found that the two nops are indeed necessary over only having one. 
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