How this instrument works
Shifting a number's bits left by one place is the same as multiplying it by 2; shifting left by k places multiplies it by 2 raised to the k-th power. Shifting right works the opposite way, dividing by that same power of 2 and dropping any remainder — the bits that would fall off the right-hand edge are simply discarded rather than rounded.
This is one of the fastest operations a computer can perform, since a bit shift is a purely mechanical rearrangement rather than genuine multiplication or division circuitry — many programming languages expose it directly as its own operator for exactly that reason, even when what's ultimately wanted is multiplying or dividing by a power of two.
A shift amount of 0 leaves the number completely unchanged in both directions, since multiplying or dividing by 2⁰=1 does nothing at all — the smallest possible case this page can compute.
- Enter the starting number into the Number field.
- Enter how many bits to shift by into the Shift amount field.
- Read Left-shifted: the number multiplied by 2 raised to that shift amount.
- Read Right-shifted: the number divided by that same power of 2, with any remainder dropped.
Worked example — shifting 5 by 2 bits
Shifting 5 left by 2 bits multiplies it by 2²=4, giving 20 — each single leftward shift doubles the value, so two shifts quadruple it. Shifting 5 right by 2 bits instead divides it by 4 and drops the remainder, giving 1, since 5÷4=1.25 and only the whole part survives.
Shifting 8 left by 3 bits gives 8×2³=64; shifting 8 right by 3 bits gives ⌊8÷8⌋=1. A shift of 0 bits, applied to 100 in either direction, leaves it completely unchanged at 100, since 2⁰=1.
Questions
What does shifting a number left by one bit do?
It doubles the number — multiplying by 2¹=2. Shifting left by k bits multiplies by 2 raised to the k-th power, compounding that doubling k times over.
What does shifting a number right by one bit do?
It halves the number and discards any remainder — dividing by 2 and rounding down, rather than rounding to the nearest whole number.
Why is bit shifting considered fast for a computer?
Because it's a purely mechanical rearrangement of a number's own bits rather than genuine multiplication or division circuitry, making it one of the cheapest operations a processor can carry out.
Does information get lost when shifting right?
Yes — whatever bits would fall off the right-hand edge are simply discarded, which is exactly why a right shift always rounds DOWN (toward zero) rather than to the nearest whole value.
What happens with a shift amount of 0?
Nothing changes in either direction — multiplying or dividing by 2⁰=1 leaves the original number exactly as it was, the smallest possible case this calculator handles.