Math Round to always upper integer

Question

I need to find a division of two integers and round it to next upper integer

e.g x=7/y=5 = 2; here x and y always greater than 0

This is my current code

 int roundValue = x % y > 0? x / y + 1: x / y;

Is there any better way to do this?

Duplicate? http://stackoverflow.com/questions/921180/how-can-i-ensure-that-a-division-of-integers-is-always-rounded-up/926806 — chrisaut, Aug 01 '11 at 11:12

score 51 · Accepted Answer · edited Aug 29 '18 at 13:36

51

You could use Math.Ceiling... but that will require converting to/from double values.

Another alternative is to use Math.DivRem to do both parts at the same time.

public static int DivideRoundingUp(int x, int y)
{
    // TODO: Define behaviour for negative numbers
    int remainder;
    int quotient = Math.DivRem(x, y, out remainder);
    return remainder == 0 ? quotient : quotient + 1;
}

edited Aug 29 '18 at 13:36

Kols

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answered Aug 01 '11 at 10:35

Jon Skeet

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Thanks @Jone please add _out remainder_ as parameter to complete your answer – Damith Aug 01 '11 at 10:43

score 34 · Answer 2 · answered Aug 01 '11 at 10:49

34

Try (int)Math.Ceiling(((double)x) / y)

answered Aug 01 '11 at 10:49

Evren Kuzucuoglu

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51

score 12 · Answer 3 · answered Mar 18 '14 at 13:06

12

All solutions looks too hard. For upper value of x/y, use this one

( x + y - 1 ) / y

answered Mar 18 '14 at 13:06

Milan Matějka

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'add another group, but subtract one in case you already had a full group' - great idea. seems there's no problem so long as ints are positive. thanks for sharing! – jacoblambert Jun 03 '16 at 12:24

score 4 · Answer 4 · answered Aug 01 '11 at 10:54

dunno what's better way or how to define a better way (if in terms of performance you have to run tests to see which will be faster), but here's my solution:

int roundValue = x / y + Convert.ToInt32(x%y>0);

p.s.

still have to deal somehow with neg. numbers... IMO this is the simplest.

score 3 · Answer 5 · answered Apr 18 '18 at 09:07

3

+0.5 will aways round to the higher.

answered Apr 18 '18 at 09:07

Atanas Atanasov

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score 1 · Answer 6 · edited Aug 01 '11 at 11:15

1

Use ceil() function. It gives the upper value.

edited Aug 01 '11 at 11:15

Soner Gönül

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answered Aug 01 '11 at 10:35

Shubham agarwal

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[ceil()](https://learn.microsoft.com/en-us/cpp/c-runtime-library/reference/ceil-ceilf-ceill?view=vs-2019) is part of C++. Equivalent for C# is [Math.Ceiling()](https://learn.microsoft.com/en-us/dotnet/api/system.math.ceiling?view=netframework-4.8) – mins Aug 24 '19 at 10:56

score 0 · Answer 7 · edited Sep 30 '22 at 08:43

0

It's better to use MidpointRounding from Math.Round

In your case:

Math.Round(value, MidpointRounding.AwayFromZero);

see more: https://learn.microsoft.com/ru-ru/dotnet/api/system.math.round?view=net-6.0#system-math-round(system-double-system-midpointrounding)

edited Sep 30 '22 at 08:43

Suraj Rao

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answered Sep 30 '22 at 08:35

Федор Лицкевич

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Math Round to always upper integer

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