[A83] Re: im 2

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[A83] Re: im 2


You have to use addresses like $9898 on the 83+/SE...The lower 8 bits for
the interrupt table you can't control.

Michael Vincent
Detached Solutions - www.detacheds.com
Radical Software - www.radicalsoft.org
----- Original Message -----
From: "Peter-Martijn Kuipers" <hyper@hysoft-automation.com>
To: <assembly-83@lists.ticalc.org>
Sent: Wednesday, September 05, 2001 11:07 AM
Subject: [A83] Re: im 2


>
> > Normally the least significant bit should indeed be 0, but on the 83-
this isn't the case. It is sometimes 1, so you can only use addresses like
$8787. So now I want to know if this is also the case on the 83+(SE).
>
> This I do not know, but tell me how you checked it out, and if you checked
> the address that was read from the pointer table, it might be that the
last bit
> is discarded.
>
>
>
>
> >
> >
> > > ----------------------------------------
> > > From: Peter-Martijn Kuipers <hyper@hysoft-automation.com>
> > >
> > > Fron your explanation, I understand this:
> > >
> > > mem_address_$HH00:
> > > $HH00: .dw pointer1
> > > $HH02: .dw pointer2
> > > $HH04: .dw pointer3
> > > $HH06: .dw pointer4
> > > $HHxx: .dw ...
> > >
> > > ;if you do:
> > > ld i,HH ; HH is highest bytes of mem table above.
> > > im 2
> > >
> > > ; then if an interrupt occurs, the interrupt device produces an 8 bit
number,
> > > ; e.g. 06h, whigh is used as the lower bytes. (the produced byte
itself,
> > > ; I'll refer to as $II.
> > > ; the address would then become $HH06. The word from that place is
read,
> > > ; and a jump takes place to the address the word at $HH06 references.
> > >
> > > ; Naturally, the lowest bit of $II would be 0 (producing an even
number),
> > > ; because a read from an uneven address would give an invalid pointer
address
> > > ; this could be interpreted as a 7-bit number.
> > >
> > > ; But when you say that an 8-bit number is produced, you mean that the
lowest
> > > ; bit (bit 0) of $II is sometimes 1 too, thus producing an odd
address???
> > > ; Or do you mean that it is an 8-bit byte with bit 0 on 0?
> > >
> > > --Peter Martijn
> > >
> > > >
> > > > According to the cpu manual, when you use interrupt mode 2, you
should create a pointer table somewhere in memory, then put in the I reg the
most significant byte. When an interrupt happens the Z80 takes the I
register and the interrupting device provides the lower 8 bits. Then it
jumps to the pointer stored at the address that is formed. The manual also
says that the lowest bit provided should always be 0 to create even numbers,
because pointers are 16 bit and therefore take 2 bytes. For now I've
discovered that this isn't the case on the 83-. The interrupting device
seems to generate an 8 bit numbers (not 7). But is this also on the 83+(SE)?
> > > >
> > > > --Tijl Coosemans
> > > > -----------------------------------------------------
> > > > Mail.be, Free WebMail and Virtual Office
> > > > http://www.mail.be
> > > >
> > > >
> > > >
> > >
> > >
> > >
> >
> > -----------------------------------------------------
> > Mail.be, Free WebMail and Virtual Office
> > http://www.mail.be
> >
> >
> >
>
>
>
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