On Wed, 5 Jan 2000, Laurens Holst wrote:
> > > You can turn off the interrupts in VDP also, but this does not
> > >interrupt IRQs from another devices (assuming there is strange hardware
> > >connected in your machine).
> >
> > I think this is what I'm searching for... Could you provide me with more
> > information about that?
>
> I don't know exactly... take a look at the v9958 (or v9938) specs... in
> registers 0 and 1 there should be three bits called IE0, IE1 and IE2. Those
> can enable/disable the interrupts. One of them represents the line-interrupt
> (I think IE2), one represents the light pen interrupt (I think IE1), and one
> represents the 'normal' interrupt (I think IE0).
>
> Complement that one to disable the VDP interrupts.
The lightpen one is in register 0, bit 4. No matter what you set it to,
the lighpen will never give an interrupt on the MSX, since the lightpen is
not connected to the VDP. In register 1, bit 5 (IIRC), is the vertical
scanning interrupt flag. It controls if the VDP gives an interrupt every
time the screen has been built (when it starts with the lower border).
This is the usual interrupt which you might want to disable. There is one
last one, I thought it was in register 8 or 9, for the horizontal scanning
interrupt. If you set that right, the VDP gives interrupts when it starts
with the building of the line set in register &H18 (vdp(&H19) in basic).
Normally it is set to no interrupts, but it is very useful for
split-screen things.
Bye,
shevek
/***Use gcc to compile***Don't mind the warning/
int*a,k ,v[9];int*main (){int i,j,s=1, x,z ,c[
]={1,4,7,4,3,4,5,4 ,1,1,1 ,2,
3,3,3,4};for(i=0;( i++<9) &!k
;s=-s){k=0;scanf("%d",&z);v [--z ]=s;for(j=0;j
<8; j++){z=v[ c[j ]];k|=z==v [c[
j]- c[j+8]]&( z&&(v[ c[j]+c [j+
8]]==z));;;}}printf(" %d won\n",- s*k );}
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