Rounding up to the nearest multiple of a number

Solution 1:

This works for positive numbers, not sure about negative. It only uses integer math.

int roundUp(int numToRound, int multiple)
{
    if (multiple == 0)
        return numToRound;

    int remainder = numToRound % multiple;
    if (remainder == 0)
        return numToRound;

    return numToRound + multiple - remainder;
}

Edit: Here's a version that works with negative numbers, if by "up" you mean a result that's always >= the input.

int roundUp(int numToRound, int multiple)
{
    if (multiple == 0)
        return numToRound;

    int remainder = abs(numToRound) % multiple;
    if (remainder == 0)
        return numToRound;

    if (numToRound < 0)
        return -(abs(numToRound) - remainder);
    else
        return numToRound + multiple - remainder;
}

Solution 2:

Without conditions:

int roundUp(int numToRound, int multiple) 
{
    assert(multiple);
    return ((numToRound + multiple - 1) / multiple) * multiple;
}

This works like rounding away from zero for negative numbers

EDIT: Version that works also for negative numbers

int roundUp(int numToRound, int multiple) 
{
    assert(multiple);
    int isPositive = (int)(numToRound >= 0);
    return ((numToRound + isPositive * (multiple - 1)) / multiple) * multiple;
}

Tests

If multiple is a power of 2 (faster in ~3.7 times http://quick-bench.com/sgPEZV9AUDqtx2uujRSa3-eTE80)

int roundUp(int numToRound, int multiple) 
{
    assert(multiple && ((multiple & (multiple - 1)) == 0));
    return (numToRound + multiple - 1) & -multiple;
}

Tests

Solution 3:

This works when factor will always be positive:

int round_up(int num, int factor)
{
    return num + factor - 1 - (num + factor - 1) % factor;
}

Edit: This returns round_up(0,100)=100. Please see Paul's comment below for a solution that returns round_up(0,100)=0.

Solution 4:

This is a generalization of the problem of "how do I find out how many bytes n bits will take? (A: (n bits + 7) / 8).

int RoundUp(int n, int roundTo)
{
    // fails on negative?  What does that mean?
    if (roundTo == 0) return 0;
    return ((n + roundTo - 1) / roundTo) * roundTo; // edit - fixed error
}

Solution 5:

int roundUp(int numToRound, int multiple)
{
 if(multiple == 0)
 {
  return 0;
 }
 return ((numToRound - 1) / multiple + 1) * multiple;  
}

And no need to mess around with conditions