Rounding up to the nearest multiple of a number

Round to nearest multiple that happens to be a power of 2

unsigned int round(unsigned int value, unsigned int multiple){
    return ((value-1u) & ~(multiple-1u)) + multiple;
}

This can be useful for when allocating along cachelines, where the rounding increment you want is a power of two, but the resulting value only needs to be a multiple of it. On gcc the body of this function generates 8 assembly instructions with no division or branches.

round(  0,  16) ->   0
round(  1,  16) ->  16
round( 16,  16) ->  16
round(257, 128) -> 384 (128 * 3)
round(333,   2) -> 334

I found an algorithm which is somewhat similar to one posted above:

int[(|x|+n-1)/n]*[(nx)/|x|], where x is a user-input value and n is the multiple being used.

It works for all values x, where x is an integer (positive or negative, including zero). I wrote it specifically for a C++ program, but this can basically be implemented in any language.

Here's my solution based on the OP's suggestion, and the examples given by everyone else. Since most everyone was looking for it to handle negative numbers, this solution does just that, without the use of any special functions, i.e. abs, and the like.

By avoiding the modulus and using division instead, the negative number is a natural result, although it's rounded down. After the rounded down version is calculated, then it does the required math to round up, either in the negative or positive direction.

Also note that no special functions are used to calculate anything, so there is a small speed boost there.

int RoundUp(int n, int multiple)
{
    // prevent divide by 0 by returning n
    if (multiple == 0) return n;

    // calculate the rounded down version
    int roundedDown = n / multiple * multiple;

    // if the rounded version and original are the same, then return the original
    if (roundedDown == n) return n;

    // handle negative number and round up according to the sign
    // NOTE: if n is < 0 then subtract the multiple, otherwise add it
    return (n < 0) ? roundedDown - multiple : roundedDown + multiple;
}

I think this should help you. I have written the below program in C.

# include <stdio.h>
int main()
{
  int i, j;
  printf("\nEnter Two Integers i and j...");
  scanf("%d %d", &i, &j);
  int Round_Off=i+j-i%j;
  printf("The Rounded Off Integer Is...%d\n", Round_Off);
  return 0;
}

Endless possibilities, for signed integers only:

n + ((r - n) % r)

I think this works:

int roundUp(int numToRound, int multiple) {
    return multiple? !(numToRound%multiple)? numToRound : ((numToRound/multiple)+1)*multiple: numToRound;
}

Here is a super simple solution to show the concept of elegance. It's basically for grid snaps.

(pseudo code)

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