Round a decimal to the nearest quarter in C#

Question

Is there a simple way in c# to round a decimal to the nearest quarter i.e. x.0, x.25, x.50 x.75 for example 0.21 would round to 0.25, 5.03 would round to 5.0

Thanks

Answer 1

Alternatively, you can use UltimateRoundingFunction given in this blog: http://rajputyh.blogspot.in/2014/09/the-ultimate-rounding-function.html

//amountToRound => input amount
//nearestOf => .25 if round to quater, 0.01 for rounding to 1 cent, 1 for rounding to $1
//fairness => btween 0 to 0.9999999___.
//            0 means floor and 0.99999... means ceiling. But for ceiling, I would recommend, Math.Ceiling
//            0.5 = Standard Rounding function. It will round up the border case. i.e. 1.5 to 2 and not 1.
//            0.4999999... non-standard rounding function. Where border case is rounded down. i.e. 1.5 to 1 and not 2.
//            0.75 means first 75% values will be rounded down, rest 25% value will be rounded up.
decimal UltimateRoundingFunction(decimal amountToRound, decimal nearstOf, decimal fairness)
{
    return Math.Floor(amountToRound / nearstOf + fairness) * nearstOf;
}

Call below for standard rounding. i.e. 1.125 will be rounded to 1.25

UltimateRoundingFunction(amountToRound, 0.25m, 0.5m);

Call below for rounding down border values. i.e. 1.125 will be rounded to 1.00

UltimateRoundingFunction(amountToRound, 0.25m, 0.4999999999999999m);

So called "Banker's Rounding" is not possible with UltimateRoundingFunction, you have to go with paxdiablo's answer for that support :)

Answer 2

Multiply it by four, round it as you need to an integer, then divide it by four again:

x = Math.Round (x * 4, MidpointRounding.ToEven) / 4;

The various options for rounding, and their explanations, can be found in this excellent answer here :-)