Equivalent of Math.Min & Math.Max for Dates?

Question

What\'s the quickest and easiest way to get the Min (or Max) value between two dates? Is there an equivalent to Math.Min (& Math.Max) for dates?

I want to do som

Answer 1

How about:

public static T Min<T>(params T[] values)
{
    if (values == null) throw new ArgumentNullException("values");
    var comparer = Comparer<T>.Default;
    switch(values.Length) {
        case 0: throw new ArgumentException();
        case 1: return values[0];
        case 2: return comparer.Compare(values[0],values[1]) < 0
               ? values[0] : values[1];
        default:
            T best = values[0];
            for (int i = 1; i < values.Length; i++)
            {
                if (comparer.Compare(values[i], best) < 0)
                {
                    best = values[i];
                }
            }
            return best;
    }        
}
// overload for the common "2" case...
public static T Min<T>(T x, T y)
{
    return Comparer<T>.Default.Compare(x, y) < 0 ? x : y;
}

Works with any type that supports IComparable<T> or IComparable.

Actually, with LINQ, another alternative is:

var min = new[] {x,y,z}.Min();

Answer 2

Now that we have LINQ, you can create an array with your two values (DateTimes, TimeSpans, whatever) and then use the .Max() extension method.

var values = new[] { Date1, Date2 }; 
var max = values.Max(); 

It reads nice, it's as efficient as Max can be, and it's reusable for more than 2 values of comparison.

The whole problem below worrying about .Kind is a big deal... but I avoid that by never working in local times, ever. If I have something important regarding times, I always work in UTC, even if it means more work to get there.

Answer 3

public static class DateTool
{
    public static DateTime Min(DateTime x, DateTime y)
    {
        return (x.ToUniversalTime() < y.ToUniversalTime()) ? x : y;
    }
    public static DateTime Max(DateTime x, DateTime y)
    {
        return (x.ToUniversalTime() > y.ToUniversalTime()) ? x : y;
    }
}

This allows the dates to have different 'kinds' and returns the instance that was passed in (not returning a new DateTime constructed from Ticks or Milliseconds).

[TestMethod()]
    public void MinTest2()
    {
        DateTime x = new DateTime(2001, 1, 1, 1, 1, 2, DateTimeKind.Utc);
        DateTime y = new DateTime(2001, 1, 1, 1, 1, 1, DateTimeKind.Local);

        //Presumes Local TimeZone adjustment to UTC > 0
        DateTime actual = DateTool.Min(x, y);
        Assert.AreEqual(x, actual);
    }

Note that this test would fail East of Greenwich...

Answer 4

There is no overload for DateTime values, but you can get the long value Ticks that is what the values contain, compare them and then create a new DateTime value from the result:

new DateTime(Math.Min(Date1.Ticks, Date2.Ticks))

(Note that the DateTime structure also contains a Kind property, that is not retained in the new value. This is normally not a problem; if you compare DateTime values of different kinds the comparison doesn't make sense anyway.)

Answer 5

// Two different dates
var date1 = new Date(2013, 05, 13); 
var date2 = new Date(2013, 04, 10) ;
// convert both dates in milliseconds and use Math.min function
var minDate = Math.min(date1.valueOf(), date2.valueOf());
// convert minDate to Date
var date = new Date(minDate); 

http://jsfiddle.net/5CR37/

Answer 6

How about a DateTime extension method?

public static DateTime MaxOf(this DateTime instance, DateTime dateTime)
{
    return instance > dateTime ? instance : dateTime;
}

Usage:

var maxDate = date1.MaxOf(date2);