dillon@CORY.BERKELEY.EDU (Matt Dillon) (03/14/88)
-Matt
dillon@CORY.BERKELEY.EDU (Matt Dillon) (03/14/88)
My first message seems to have died. My question is, essentially, If you are in the middle of software interrupt A, and a real interrupt comes along and Cause()'s another software interrupt (the SAME software interrupt A), when the interrupt ends, returning to the original software interrupt, and then the original software interrupt ends, will it be immediately re-invoked or will the Cause() call from the hard interrupt be ignored because that particular software interrupt was already in progress? -Matt
bryce@cbmvax.uucp (Bryce Nesbitt - SOFTWARE) (04/17/88)
In article <> dillon@CORY.BERKELEY.EDU (Matt Dillon) writes: | | If you are in the middle of software interrupt A, and a real interrupt | comes along and Cause()'s another software interrupt (the SAME | software interrupt A)... | ...will it be immediately re-invoked or will the Cause() | call from the hard interrupt be ignored because that particular | software interrupt was already in progress? A Cause() of an already active software interrupt is ignored. C'mon Matt, that one is easy to trace :-). |\_/| . ACK!, NAK!, EOT!, SOH! {o o} . Bryce Nesbitt (") BIX: mleeds (temporarily) U USENET: cbmvax!bryce@rutgers.EDU -or- rutgers!cbmvax!bryce Disclaimer: I'm not an official, and thus this is not an official opinion.