2 votes 2 votes In $8086$, the jump condition for the instruction $JNBE$ is? CF = 0 or ZF = 0 ZF = 0 and SF = 1 CF = 0 and ZF = 0 CF = 0 CO and Architecture isro2013 8086 non-gate + – makhdoom ghaya asked Apr 27, 2016 makhdoom ghaya 4.4k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 4 votes 4 votes OPTION C IS CORRECT http://marin.jb.free.fr/jumps/ asu answered Jun 14, 2016 selected Jun 27, 2016 by Arjun asu comment Share Follow See all 3 Comments See all 3 3 Comments reply shivanisrivarshini commented Jun 27, 2016 reply Follow Share option a ,c both are same na 0 votes 0 votes asu commented Jun 27, 2016 reply Follow Share a is or and c is AND 2 votes 2 votes shivanisrivarshini commented Jun 27, 2016 reply Follow Share yes sorry oversight 0 votes 0 votes Please log in or register to add a comment.
7 votes 7 votes Answer C is correct JNBE means Jump When condition is Not below AND Not Zero So, Zero flag should be False = ZF=0 Carry flag should be False = CF =0 naga praveen answered Jun 19, 2016 naga praveen comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes JNBE MEAN NEITHER BELOW NOR EQUAL EQUAL CAN BE CHECK IF RESULT IS NOT ZERO SO ZERO FLAG MUST BE 0 AND FOR BELOW CARRY SHOULD BE ZERO HENCE OPTION A Dexter answered May 5, 2016 Dexter comment Share Follow See all 0 reply Please log in or register to add a comment.