Thread: Finding next largest integer multiple

Hi all. I recently stumbled across the following code:

Code:

do 
{
     some_operation();
} while (++val & 7);

What this (supposedly??) does is continuously call some_operation and increment val until val is an integer multiple of 8. I understand binary and use of the bitwise-operator, I just can't convince myself that this is 'true' for and works properly for all integer m such where (m*8) < val < (m+1)*8 . Is this some kind of trick only possible because 8 is a power of 2? The formula clearly does not work if we do (++val & 4) until val is a multiple of 5, as this test will fail for decimal values 1, 2, and 3..

Moreover, I need to expand this functionality for any integer n > 1, not just n = 8. Obviously the trivial solution is

Code:

do
{
    some_operation();
} while( (++val % m) != 0 );

But I have had it burned into my head that the modulo operator is expensive. Is there a better, more efficient solution for this problem?

Originally Posted by Is there a better, more efficient solution for this problem?

Is there a better, more efficient solution for this problem?

Maybe it is sufficient to compute ((val / n) + 1) * n, assuming non-negative integer division, and that you really do want the next largest multiple of n even if val is perfectly divisible by n.