Show how the flag resister is affected by the addition of 38H and 2FH.

MOV BH, 38H ;BH = 3H

ADD BH, 2FH ;add 2F to BH, now BH = 67H

38		0011		1000
+ 2F		0010		1111
67		0110		0111

Flag register status

CF = 0 since there is no carry beyond d7

AF = 1 since there is a carry from d3 to d4

PF = 0 since there is an odd number of 1’s in the result

ZF = 0 since the result is not zero

SF = 0 since d7 of the result is zero.