C++: Rounding up to the nearest multiple of a number

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;
}

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

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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

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.