The barrel shifter

Q: What is the difference between a logical and an arithmetic shift?

They differ only on a right shift and only in what fills the vacated top bits. A logical right shift fills 0, which correctly divides an unsigned number. An arithmetic right shift copies the sign bit into the vacated bits, which divides a signed two's-complement number while keeping its sign. Using a logical shift on a negative value turns it large and positive.

Q: What is the difference between a shift register and a barrel shifter?

A shift register is sequential: it moves the stored bits one position per clock edge, so an N-position shift takes N clocks. A barrel shifter is combinational: it produces the whole shift by any amount in one pass with no clock, using a tree of muxes. The register streams data; the shifter computes a shift.

Q: Why is shifting the same as multiplying or dividing by a power of two?

Shifting a binary number left by k moves every bit into a higher place and brings in 0s, which multiplies it by 2^k, exactly as appending a 0 multiplies a decimal number by 10. Shifting right by k divides by 2^k. This needs no adder, only rewiring, so x * 8 becomes a left shift by 3.

Shifting a word in one combinational step

8 min read

A barrel shifter is a combinational circuit that shifts or rotates a word by any amount in a single step, built from a tree of multiplexer stages where stage k shifts by 0 or 2^k under one control bit; a logical shift fills the vacated bits with 0, an arithmetic right shift copies the sign bit, and a rotate wraps the bits around.

Builds onThe multiplexer Shift registers Two's complement

The shift register shifts one bit per clock: to move a byte four places you clock it four times. Often you need the shift now, in one pass of combinational logic, with the shift amount itself being data (as in x << y). That circuit is a shifter, and the general form built from multiplexers is a barrel shifter. It is a standard arithmetic building block, and a standard ALU operation, that our ALU does not yet have.

Shift by a constant is just rewiring

If the shift amount is fixed and known when you build the circuit, a shift costs no gates at all: it is pure wiring. To shift X = X3 X2 X1 X0 left by one, wire X0 to output bit 1, X1 to output bit 2, X2 to output bit 3, and feed a constant 0 into output bit 0. You have just relabelled which wire goes where. And shifting has a meaning in arithmetic: a **left shift by k multiplies by 2^k, and a right shift by k divides by 2^k**, exactly as appending a 0 in decimal multiplies by 10. That is exactly what powers shift-and-add multiplication, shown step by step in the interactive multiplication walkthrough.

0011 (3) shifted left by 2 = 1100 (12 = 3 × 4)

Three flavors: logical, arithmetic, rotate

Left shifts always bring in 0s at the bottom. Right shifts are where the flavors differ, and the difference is all about the vacated top bits:

Logical shift: fill the vacated bits with 0. Correct for unsigned numbers.
Arithmetic shift: on a right shift, copy the sign bit into the vacated top bits (sign-extension). Correct for signed two's-complement numbers. (Arithmetic left shift is the same as logical left: both fill 0 at the bottom.)
Rotate: no bits fall off; whatever leaves one end wraps around to the other. A rotate loses no information.

Here is why the arithmetic flavor exists. Take 1000, which is -8 in 4-bit two's complement (the top bit is the sign). Dividing by 2 should give -4. An arithmetic right shift by 1 copies the sign bit, giving 1100, which reads as -4: correct. A plain logical right shift would give 0100, which is +4: nonsense for a signed value. Copying the sign is exactly what keeps a negative number negative as you divide it.

value	operation	result	reads as
1000	arithmetic right 1	1100	-4
1000	logical right 1	0100	+4

1000 is -8 as a signed byte. An arithmetic right shift copies the sign bit and gives -4 (correct half); a logical right shift fills 0 and gives +4 (wrong for a signed value). Same input, same direction, different fill rule.

A shift is the cheapest arithmetic there is: multiplying or dividing by a power of two is just moving bits, no adder needed. That is why a compiler turns x * 8 into a left shift by 3, and an unsigned x / 2 into a right shift by 1. Signed division is close but not identical: an arithmetic right shift rounds toward negative infinity while C's / rounds toward zero, so a compiler adds a small correction for negative values. The flavor you pick still encodes whether the value is signed.

Variable shifts: the barrel shifter

When the shift amount is itself a signal (you do not know it until the circuit runs), constant rewiring is not enough: you need to choose, at run time, among all the possible shifts. The trick is to build the shift out of powers of two. Any shift from 0 to N-1 is a sum of 1, 2, 4, ... so lay the shifter out as a stack of stages, one per power of two:

Stage 0 shifts by 0 or 1, controlled by shift-amount bit sh0.
Stage 1 shifts by 0 or 2, controlled by bit sh1.
Stage 2 shifts by 0 or 4, controlled by bit sh2, and so on.
Each stage is a **row of 2:1 multiplexers**, one per bit: with its control bit 0 the bit passes straight through, with it 1 the bit takes the value from 2^k positions away.

So an N-bit barrel shifter is about log2(N) mux layers deep. To shift by 2, drive the amount sh1 sh0 = 10 (binary two): stage 0 passes through unchanged, stage 1 shifts by 2. Feed 0011 in and stage 1 produces 1100, the same 3 -> 12 as before, now selected by data rather than hard-wired. A 4-bit shifter needs just two stages to reach any shift from 0 to 3.

One stage of a barrel shifter: the shift-left-by-1 layer for a 4-bit value. Four 2:1 muxes share one control S. With S = 0 each Y equals its own X (no shift); with S = 1 each Y takes the bit one position lower (Y1 = X0, Y2 = X1, Y3 = X2) and Y0 is filled with 0. Stack a shift-by-2 stage on top and you can shift by any amount. Open it in the lab and toggle S to watch the whole word slide.

Contrast the two shift circuits you now know. A shift register is sequential: it moves one position per clock edge and stores the result. A barrel shifter is combinational: it delivers the entire shift in one pass, no clock, the output changing as soon as the inputs settle. The register is for streaming data along a wire; the barrel shifter is for computing x << y inside an ALU.

Common mistakes. Using a logical right shift on a signed value (a negative number turns large and positive); signed division must sign-extend with an arithmetic shift. Confusing rotate with shift (a rotate wraps and loses no bits; a shift drops bits off the end). Assuming a variable shifter must be clocked, when a barrel shifter is purely combinational and produces the shift in one pass. And forgetting that arithmetic left shift fills 0 just like a logical left shift (the sign trick is a right-shift thing), so a left shift can still overflow the sign bit.

Try it

Take the 4-bit value 0110 (6). Compute (a) a logical left shift by 1, (b) an arithmetic right shift by 2 treating it as a signed number, and (c) a rotate right by 1.

Shifting completes the arithmetic toolkit: with add, subtract, multiply, and shift you can build the multiply-and-divide-by-powers-of-two paths a real datapath leans on. A full ALU usually lists a shift among its selectable operations right next to add and AND; the barrel shifter is the block that supplies it. Next, division is the harder inverse, the same shift idea run as shift-and-subtract.

Frequently asked

What is a barrel shifter?

A barrel shifter is a combinational circuit that shifts or rotates a word by any amount in a single pass of logic, built from a tree of multiplexer stages where stage k shifts by 0 or 2^k under one bit of the shift amount. An N-bit barrel shifter is about log2(N) mux layers deep, and unlike a shift register it needs no clock.

What is the difference between a logical and an arithmetic shift?

They differ only on a right shift and only in what fills the vacated top bits. A logical right shift fills 0, which correctly divides an unsigned number. An arithmetic right shift copies the sign bit into the vacated bits, which divides a signed two's-complement number while keeping its sign. Using a logical shift on a negative value turns it large and positive.

What is the difference between a shift register and a barrel shifter?

A shift register is sequential: it moves the stored bits one position per clock edge, so an N-position shift takes N clocks. A barrel shifter is combinational: it produces the whole shift by any amount in one pass with no clock, using a tree of muxes. The register streams data; the shifter computes a shift.

Why is shifting the same as multiplying or dividing by a power of two?

Shifting a binary number left by k moves every bit into a higher place and brings in 0s, which multiplies it by 2^k, exactly as appending a 0 multiplies a decimal number by 10. Shifting right by k divides by 2^k. This needs no adder, only rewiring, so x * 8 becomes a left shift by 3.

Every lesson here builds toward one thing: a working CPU, from the transistor up.

Open the free lab →

Builds towardBinary division