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 &FDThe consequence of this is that, with the databus at &FF:
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 EscapeInterrupt 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.