A binary number is parallel-loaded into a shift register. The shift register is then commanded to shift right" for one clock pulse. How does the value of the shifted binary number compare to the number originally loaded in, assuming that the MSB is on the left flip-flop of the shift register?
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Just try it with a couple of numbers and you can find out by yourself:
Now try some more examples and you will see.
A logical shift right divides the original value by 2.
Correspondingly, a logical shift left multiplies by 2.