Question

Can you please explain the below lines, with some good examples.

A left arithmetic shift by n is
equivalent to multiplying by 2ⁿ
(provided the value does not
overflow).

And:

A right arithmetic
shift by n of a two’s complement value
is equivalent to dividing by 2ⁿ and
rounding toward negative infinity. If
the binary number is treated as ones’
complement, then the same right-shift
operation results in division by 2ⁿ
and rounding toward zero.

Answer 1

I will explain what happens in a base that we’re more familiar with: 10.

In base 10, let’s say you have a number N=123. Now, you “shift” this number to the left k=3 positions, filling the emptied digits with 0. So you get X=123000.

Note that X = N * 10^k.

The case with base 2 is analogous.

 Example 1 (base 10)   |  Example 2 (base 2)
                       |
 N        =    123     |  N       =   110101 (53 in base 10)
 k        =      3     |  k       =        2 (in base 10)
 N << k   = 123000     |  N << k  = 11010100 (212 in base 10)
                       |
 10^k     =   1000     |  2^k     =      100 (in base 2; 4 in base 10)
 N * 10^k = 123000     |  N * 2^k = 11010100 (53 * 4 = 212 in base 10)
                       |

The case with right shift is simply a mirror of the process, and is also analogous in base 10. For example, if I have 123456 in base 10, and I “shift” right 3 positions, I get 123. This is 123456 / 1000 (integer division), where 1000 = 10³.