Consider the following computation: 2^N (TWO TO THE POWER OF ‘N’, for Int. N>0), being executed on a processor with 32 bit internal, user and ALU registers. The registers rightmost bit is bit 0 and the leftmost bit(the sign) is bit 31. WHAT is the SMALLEST value of ‘N’ that will cause the Overflow Status Flag (V) to be set?
Seems like the calculation should be as simple as plugging values in for N but I don't quite understand if there are any other components to take into account, any help is appreciated.