The contents of a 8-bit register is initially 01110111. The register is shifted right six times with the serial input being 011001. The same register is then shifted left 4 times with the serial input being 0101. What is the final contents of the register?
I know what serial means or rather, I assumed that's what it meant. However the answer is dependant on if it's MSB or LSB first. There is also a third possibility - the bits are shifted the same way as the register. So, three possible answers due to an incomplete question.
Normally there is only 1 input, you can not shift with 2 inputs unless the register is designed for it.
If there is 1 input:
Shifting to the left clears the LSB bits being shifted and shifts the rest to the left.
If you want to shift with 2 inputs, that means that you have to supply the bits from the other end in which case the result is clear. However this requirement is not mentioned and ambiguous.
Ahah, we finally agree. It's an ambiguous question. As a computer programmer since the 70s I should be able to answer these type of questions but teachers keep asking incomplete questions.