Question

I am working through a problem which I was able to solve, all but for the last piece—I am not sure how one can do multiplication using bitwise operators:

0*8 = 0

1*8 = 8

2*8 = 16

3*8 = 24

4*8 = 32

Is there an approach to solve this?

Answer 1

To multiply by any value of 2 to the power of N (i.e., 2^N), shift the bits N times to the left.

0000 0001 = 1

times 4 = (2^2 => N = 2) = 2 bit shift : 0000 0100 = 4

times 8 = (2^3 -> N = 3) = 3 bit shift : 0010 0000 = 32

etc..

To divide, shift the bits to the right.

The bits are whole 1 or 0 – you can’t shift by a part of a bit, thus if the number you’re multiplying by is does not factor a whole value of N.
I.e.,

since: 17 = 16  + 1
thus:  17 = 2^4 + 1

therefore: x * 17 = (x * 16) + x in other words 17 x's

Thus to multiply by 17, you have to do a 4 bit shift to the left, and then add the original number again:

==> x * 17 = (x * 16) + x
==> x * 17 = (x * 2^4) + x
==> x * 17 = (x shifted to left by 4 bits) + x

so let x = 3 = 0000 0011

times 16 = (2^4 => N = 4) = 4 bit shift : 0011 0000 = 48

plus the x (0000 0011)

I.e.,

    0011 0000  (48)
+   0000 0011   (3)
=============
    0011 0011  (51)

Charles Petzold has written a fantastic book ‘Code’ that will explain all of this and more in the easiest of ways. I thoroughly recommend this.