Round Integral numbers in C#

Question

I've been searching about this question for a few hours but I didn't find my answer so I ask it:

I'm looking for a method or something to round 25,599,999 to 25,000,000 or 25,599,999 to 30,000,000:

 int num1 = 25599999;
 Console.WriteLine(RoundDown(num1, 6 /* This is the places (numbers from end that must be 0) */)) // Should return 25,000,000;

or Round Up:

 int num1 = 25599999;
 Console.WriteLine(RoundUp(num1, 6 /* This is the places (numbers from end that must be 0) */)) // Should return 30,000,000;

Note: I'm not looking for a way to round decimal numbers.

Search for "round to nearest multiple" there's plenty of answers — pinkfloydx33, Sep 24 '18 at 08:46
What did you try? Show your code. I don't think you can find this kind of customization directly. — Gaurang Dave, Sep 24 '18 at 08:46
I'm not looking for rounding decimal numbers. @Dmitry Bychenko — Arad, Sep 24 '18 at 08:47
Don't vote down, read the question and keep in mind that I'm not stupid to ask questions that has been answered many times. — Arad, Sep 24 '18 at 08:48
Just to be clear, if you had `20,000,001` and you wanted to round UP using `8` significant digits, you'd want the result to be `30,000,000`? — Matthew Watson, Sep 24 '18 at 08:51
Possible duplicate of [Rounding integers to nearest multiple of 10](https://stackoverflow.com/questions/15154457/rounding-integers-to-nearest-multiple-of-10) — aloisdg, Sep 24 '18 at 08:51

score 8 · Accepted Answer · edited Sep 24 '18 at 08:57

8

int RoundDown(int num, int digitsToRound)
{
    double tmp = num / Math.Pow(10, digitsToRound);

    return (int)(Math.Floor(tmp) * Math.Pow(10, digitsToRound));
}

int RoundUp(int num, int digitsToRound)
{
    double tmp = num / Math.Pow(10, (digitsToRound + 1)) + 1;

    return (int)(Math.Floor(tmp) * Math.Pow(10, (digitsToRound + 1)));
}

edited Sep 24 '18 at 08:57

Gaurang Dave

3,752
2
13
31

answered Sep 24 '18 at 08:52

Goran Ćojanović

136
4

Thank you so much! This is exactly what I was looking for. – Arad Sep 24 '18 at 08:56

score 1 · Answer 2 · answered Sep 24 '18 at 09:28

Methods for Rounding up and down for expected trailing zero count:

    public static int RoundUp(int number, int trailingZeroes)
    {
        var divider = (int)Math.Pow(10, trailingZeroes);
        return (number / divider + (number % divider > 0 ? 1 : 0)) * divider;
    }

    public static int RoundDown(int number, int trailingZeroes)
    {
        var divider = (int)Math.Pow(10, trailingZeroes);
        return number / divider * divider;
    }

This would however return for:

Console.WriteLine(RoundUp(25599999, 6));

Result will be 26 000 000. If you want it to be rounded to 30 000 000 then you need to call it with 7 trailing zeroes since in your example you are rounding to 7 trailing zeroes:

Console.WriteLine(RoundUp(25599999, 7));

Gereon99 · Answer 3 · 2018-09-24T09:28:37.507

0

Take a look at:

Math.Round

Example usage: Rounding a Floating-Point Value

Might be transferable to non-floating point values. Maybe that helps you?

edited Sep 24 '18 at 09:28

answered Sep 24 '18 at 08:58

Gereon99

592
6
13

Adrien G. · Answer 4 · 2018-09-24T08:51:54.943

-1

You can try these functions:

public void RoundUp(int n){
    int up = ((n / 1000000) + 1) * 1000000;
    return up;
}

public void RoundDown(int n){
    int down = (n / 1000000) * 1000000;
    return down;
}

This only works with millions number (like 2 000 000, 25 252 325, ...).

Hope it helps!

edited Sep 24 '18 at 08:51

answered Sep 24 '18 at 08:49

Adrien G.

371
1
14

Yep sorry, i'm used to another language. I'm changing that – Adrien G. Sep 24 '18 at 08:51

Round Integral numbers in C#

4 Answers4