How do I find the next multiple of 10 of any integer?

Question

Dynamic integer will be any number from 0 to 150.

i.e. - number returns 41, need to return 50. If number is 10 need to return 10. Number is 1 need to return 10.

Answer 1

n + (((9 - (n % 10)) + 1) % 10)

Answer 2

in C, one-liner:

int inline roundup10(int n) {
  return ((n - 1) / 10 + 1) * 10;
}

Answer 3

You can do this by performing integer division by 10 rounding up, and then multiplying the result by 10.

To divide A by B rounding up, add B - 1 to A and then divide it by B using "ordinary" integer division

Q = (A + B - 1) / B 

So, for your specific problem the while thing together will look as follows

A = (A + 9) / 10 * 10

This will "snap" A to the next greater multiple of 10.

The need for the division and for the alignment comes up so often that normally in my programs I'd have macros for dividing [unsigned] integers with rounding up

#define UDIV_UP(a, b) (((a) + (b) - 1) / (b))

and for aligning an integer to the next boundary

#define ALIGN_UP(a, b) (UDIV_UP(a, b) * (b))

which would make the above look as

A = ALIGN_UP(A, 10);

P.S. I don't know whether you need this extended to negative numbers. If you do, care should be taken to do it properly, depending on what you need as the result.

Answer 4

What about ((n + 9) / 10) * 10 ?

Yields 0 => 0, 1 => 10, 8 => 10, 29 => 30, 30 => 30, 31 => 40

Answer 5

How about using integer math:

N=41
N+=9   // Add 9 first to ensure rounding.
N/=10  // Drops the ones place
N*=10  // Puts the ones place back with a zero

Answer 6

Be aware that answers based on the div and mod operators ("/" and "%") will not work for negative numbers without an if-test, because C and C++ implement those operators incorrectly for negative numbers. (-3 mod 5) is 2, but C and C++ calculate (-3 % 5) as -3.

You can define your own div and mod functions. For example,

int mod(int x, int y) {
  // Assert y > 0
  int ret = x % y;
  if(ret < 0) {
    ret += y;
  }
  return ret;
}