CODEWITHSUNDEEP
c++bit-manipulation
Jack Smith
Jack Smith

Reputation: 763

Bitshifting to multiply an integer by 10

Simple question, but I can't quite seem to figure it out:

If i have an integer, say 12, and I perform the following bit-manipulation on it:

int i = 12;  
i = (i << 3) + (i << 1);

I end up with 120 (12*10). This is the case with any number.

Can someone explain to me, succinctly, why it is that this works? (I'm obviously missing something quite rudimentary when it comes to bitshifting).

Upvotes: 21

Views: 15031

Answers (6)

NominSim
NominSim

Reputation: 8511

Left bitshift is the same (usually) as multiplying by power's of two. i.e. << 1 is equivalent to *(2^1) , << 2 is equivalent to *(2^2) and so on...

If you substitute that into your example you can see why your result is multiplied by 10:

int i = 12;
i = (i * 2^3) + (i * 2^1);
= { i = (i * 8) + (i * 2);}
= { i = 8i + 2i; }
= { i = 10i; }

Upvotes: 1

SirPentor
SirPentor

Reputation: 2044

i << 3 is equivalent to i * 8. i << 1 is equivalent to i * 2.

The distribute property tells us that:

x = i * 10
x = i * (8 + 2)
x = 8i + 2i

Upvotes: 5

TheJuice
TheJuice

Reputation: 4484

Shifting left by 3 places is equal to multiplying by 8, shifting by 1 places is equal to multiplying by 2 so you are doing

i = i * 8 + i * 2

Upvotes: 1

Oliver Charlesworth
Oliver Charlesworth

Reputation: 272447

Rewrite the bitshifts as multiplications by powers-of-2, and all should become clear.

Upvotes: -1

Puppy
Puppy

Reputation: 146900

Express as multiplication.

i = (i << 3) + (i << 1);
i = (i * 8) + (i * 2);
i = 8i + 2i
i = 10i

Upvotes: 33

lezebulon
lezebulon

Reputation: 7994

You are basically doing:

i = i*2^3 + i*2

Upvotes: 6

Related Questions