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

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.

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.