Technical 68K Question

[Shamus]

This may be a silly question, but then again, maybe not. I'm curious to know if any of you Jaguar developers out there know what the 68K in the Jaguar does when it hits an "official" illegal instruction (i.e., $4AFC). Does it fetch the next instruction or does it hang on it or does it halt the processor?

 

-- Shamus

[Stone]

I can't remember offhand, but I think it's quite a high-level interrupt (non-maskable?) which forces the 68k into its interrupt vector table to try and work out what do do next. If nobody's answered by tomorrow I'll look it up, we've got some of the original 68K books in the information centre at work

 

On a BJL system it crashes back to the homescreen (helpfully reporting Error 03) but I don't know what it would do on a production-level Jag...most likely either blackscreen, a reboot or the program freezing (leaving the display onscreen). My money would be one of the first two

 

Stone

[belboz]

It treats it like other exceptions. I.E. it goes to the vector table and pulls the address of the routine to handle the exception processing.

 

In the case of an illegal instruction it is vector number 4.

 

Which basically means the 68000 looks at address $c to pull the longword address of the routine to handle the exception.

 

Vector number 2 is a bus error, 3 is an address error.