We have Z80. This processor do following operation in U2:
200 - (-56). What will be values of following register flags:
S - the oldest bit of result
V - overflow
C - carry
Z - result is zero.
Now, we know that in U2 we have:
200-(-56) = 200+56 = 00000000(U2) Then, I know that:
S = 0
V = 1
Z = 1.
However, when it comes to
C I am not sure about it. Although, we see the carried bit, I think that in U2 we can ignore it, so
C = 0.
Am I right ?