How to perform multiplication, using bitwise operators?

Question

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

0*8 = 0  1*8 = 8  2*8 = 16   3*8 = 24   4*8 = 32 

Can you please recommend 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 ie.

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) 

ie.

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

Edit: Update to the original answer. 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.