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

尽管:

对负数不成立 不会工作，如果numRound +多个溢出

建议使用无符号整数，这已经定义了溢出行为。

您将得到一个异常是multiple == 0，但在这种情况下，这不是一个定义良好的问题。

没有条件:

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

这就像对负数进行舍入一样

同样适用于负数的版本:

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

测试

如果倍数是2的幂(快3.7倍)

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

测试

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

没有必要在条件上添乱

这里有一个超级简单的解决方案来展示优雅的概念。它主要用于网格快照。

(伪代码)

nearestPos = Math.Ceil( numberToRound / multiple ) * multiple;

这就是我要做的:

#include <cmath>

int roundUp(int numToRound, int multiple)
{
    // if our number is zero, return immediately
   if (numToRound == 0)
        return multiple;

    // if multiplier is zero, return immediately
    if (multiple == 0)
        return numToRound;

    // how many times are number greater than multiple
    float rounds = static_cast<float>(numToRound) / static_cast<float>(multiple);

    // determine, whether if number is multiplier of multiple
    int floorRounds = static_cast<int>(floor(rounds));

    if (rounds - floorRounds > 0)
        // multiple is not multiplier of number -> advance to the next multiplier
        return (floorRounds+1) * multiple;
    else
        // multiple is multiplier of number -> return actual multiplier
        return (floorRounds) * multiple;
}

代码可能不是最优的，但比起枯燥的性能，我更喜欢干净的代码。

这是对“如何求出n位需要多少字节?”(A:(n位+ 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
}