0 votes 0 votes A $CPU$ supports $4$ interrupts - $I1 , I2 , I3$ and $I4$. It supports priority of the interrupts. Nested interrupts are allowed if later interrupt has higher priority than the previous one. During a certain period of time, we observe the following sequence of entry and exit from the interrupt service routine. $I1$ start ……. $I2$ start ….. $I2$ end …. $I4$ start ….. $I3$ start ….. $I3$ end … $I4$ end .. $I1$ end What can we infer about the priority interrupt routines from the above sequence? $I3 > I4 > I2 ; I4 > I1$ $I2 > I1; I3 > I4 > I1$ $I1 > I2 > I3 > I4$ $I4 > I3 > I2 > I1$ GATE tbb-mockgate-3 co-and-architecture interrupts + – Bikram asked Feb 9, 2017 • retagged Sep 16, 2020 by ajaysoni1924 Bikram 403 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 1 votes 1 votes In the given sequence I1 is least priority interrupt as it ends at the last of Interrupt Service Routine. . I3 has higher priority than I4 as it ends before I4 . I2 has higher priority than I1 . Option B better describe the priority. Bikram answered Mar 14, 2017 Bikram comment Share Follow See all 0 reply Please log in or register to add a comment.