How to convert last 3 digits of number into 0

Question

How to convert last 3 digits of number into 0

example 3444678 to 3444000

I can do like

(int)(3444678/1000) * 1000= 3444000

But division and multiplication could be costly...

Any other solution????

Answer 1

You could try

n - (n % 1000)

but the modulus operator might be as costly as a division. In any case, this sounds an awful lot like a micro-optimization. Is this really your bottleneck?

Answer 2

A shift trick, then:

n >>= 3;
n -= n % (5 * 5 * 5);
n <<= 3;

Is it faster? Doubtful.

But here's a fun fact: gcc doesn't use division/modulus for this:

n -= n % 1000;

It multiplies by some crazy number (274877907) and does some other stuff which is presumably faster.

The moral of this story: the more obvious the purpose of your code is to the compiler, the more likely it is that the compiler will optimise it in a way you'd never think of. If the code is easier for humans to understand, that's another bonus.