The difference between logical shift right, arithmetic shift right, and rotate right

Question

I've been reading the classic Hacker's delight and I am having trouble understanding the difference between logical shift right,arithmetic shift right, and rotate right. Please excuse if the doubt seems too simple.

Solution 1

First remember that machine words are of fixed size. Say 4, and that your input is:

+---+---+---+---+
| a | b | c | d |
+---+---+---+---+

Then pushing everything one position to the left gives:

+---+---+---+---+
| b | c | d | X |
+---+---+---+---+

Question what to put as X?

1. with a shift put 0
2. with rotate put a

Now push everything one position to the right gives:

+---+---+---+---+
| X | a | b | c |
+---+---+---+---+

Question what to put as X?

1. with a logical shift put 0
2. with an arithmetic shift put a
3. with rotate put d

Roughly.

Logical shift correspond to (left-shift) multiplication by 2, (right-shift) integer division by 2.

Arithmetic shift is something related to 2's-complement representation of signed numbers. In this representation, the sign is the leftmost bit, then arithmetic shift preserves the sign (this is called sign extension).

Rotate has no ordinary mathematical meaning, and is almost an obsolete operation even in computers.

Solution 2

The difference is pretty much explained in the right-most column.

• Logical shift treats the number as a bunch of bits, and shifts in zeros. This is the >> operator in C.

• Arithmetic shift treats the number as a signed integer (in 2s complement), and "retains" the topmost bit, shifting in zeros if the topmost bit was 0, and ones if it was one. C's right-shift operator has implementation-defined behavior if the number being shifted is negative.
  
  For example, the binary number 11100101 (-27 in decimal, assuming 2s complement), when right-shifted 3 bits using logical shift, becomes 00011100 (decimal 28). This is clearly confusing. Using an arithmetic shift, the sign bit would be kept, and the result would become 11111100 (decimal -4, which is about right for -27 / 8).

• Rotation does neither, since topmost bits are replaced by lowermost bits. C does not have an operator to do rotation.

Solution 3

Logical right shift means shifting the bits to the right and MSB(most significant bit) becomes 0.

Example: Logical right shift of number 1 0 1 1 0 1 0 1 is 0 1 0 1 1 0 1 0.

Arithmetic right shift means shifting the bits to the right and MSB(most significant bit) is same as in the original number.

Example: Arithmetic right shift of number 1 0 1 1 0 1 0 1 is 1 1 0 1 1 0 1 0.

Comments

[Sourav Ghosh]

What this has to do with C?

[Lundin]

What's wrong with the wikipedia articles? What is it that you don't understand about them?

[Chandrahas Aroori]

I did not understand the difference between the arithmatic and logical shift.

[Chandrahas Aroori]

Can you explain arithmatic shift a little more clearly. And example please?

[phuclv]

@ChandrahasAroori there are tons of examples that you can find on google en.wikipedia.org/wiki/Bitwise_operation

[endolith]

"Shift" could be more specifically referred to as "logical shift", no?

[Jed]

Logical shift does not divide and multiply by 2. For example, -75 >>> 1 = 90. The arithmetic shift preserves the sign, thus multiplying and dividing by 2

[Jean-Baptiste Yunès]

I never said taht logical shift are for 2's complement numbers, only arithmetic shift.

[Peter Cordes]

Useful to note that arithmetic right shift rounds towards -Infinity while ordinary signed division (in C) truncated towards 0. So compiling x/2 for signed x requires some sign-bit trickery on top of an SAR instruction.

[KMC]

"11111100 (decimal ~4)" .. isn't that negative of 00000011 or -3 instead ?

[Bbake Waikhom]

The definition of Logical shift and Arithmetic shift at the end is incorrect. The arithmetic shift is the one that's used to divide or multiply by 2, not the Logical shift. Please correct.