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Z80 interrupt mode 2

by Jonathan Graham Harston

When an interupt occurs, the CPU jumps to the address stored in memory at I*256+databus. However, what happens to bit 0? If I=&FF and the data bus is &FF, is the vector fetched from &FFFE/&FFFF (ie, bit zero forced to 0) or from &FFFF/&0000 ? I have seen documentation, programs and (importantly!) emulators written declaring both to be true.

Zaks says bit 0 is forced to zero (p504). Leventhal says bit 0 is forced to zero (p12-5). Penfold says "bit 0 is always 0" without saying if it is forced to zero (p34), or the hardware should always *supply* a zero. Zilog is silent (p145).

With different emulators I have found some force bit 0 to zero, some do not. Because of the variation in different emulators' implementation of IM2, I have had to write odd interupt code to cope with them. This is what BBC BASIC for the ZX Spectrum uses:

.IRQFix                       ; I needs to be set to &FD
FDF7 E5         PUSH HL
FDF8 2A FE FF   LD HL,(&FFFE) ; Save HL and get IRQV
FDFB E3         EX (SP),HL    ; Restore HL and jump to IRQV
FDFC C9         RET
FDFD C3 F7 FD   JP &FDF7
FE00 FD         DEFB &FD
 
The consequence of this is that, with the databus at &FF:

* If bit zero is forced to zero, IRQs vector via &FDFE/&FDFF and thence to &FDF7.

* If bit zero is not forced to zero, IRQs vector via &FDFF/&FE00 and thence to &FDFD.

Consequently, setting &FDFD to &C9 would NOP out the interrupt code if bit zero is /not/ forced to zero. I tested this and got the following results:
Sinclair Spectrum 48K
 
Acorn BBC BASIC Version 2.20
(C) Copyright R.T.Russell 1983
>PRINT ~!&FDFD                       Confirm contents of memory
  FDFDF7C3
>10REPEAT PRINT TIME:VDU 11:UNTIL 0  Print TIME continously
>RUN                                 Initial test
    11666                            Continuously updated TIME displayed
 
Escape at line 20                    Press Escape to stop program
>?&FDFD=&C9:RUN                      Set RET and test
    11978                            Static display
                                     Also, no response to interrupt
                                     driven keyboard, so no response
                                     to Escape
Interrupt code is not being executed, which means the RET is being jumped to. The only way this can happen is if IRQs are being vectored to &FDFD, via &FDFE/&FE00, which can only be happening if bit zero is not being forced to zero.

Consequently, contrary to stated documentation, bit zero is /not/ forced to zero, an IM2 interupt jumps to I*256+databus, not to (I*256+databus) AND 0xFFFE. If you want to reply on an even address, you must use hardware to supply an even address.

--
J.G.Harston - mdfs.net/User/JGH


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