Random date in C#

Question

I'm looking for some succinct C# 3 code to generate a random date between Jan 1 1995 and the current date.

I'm thinking some solution that utilizes Enumerable.Range somehow may make this more succinct.

Answer 1

Start with a fixed date object (Jan 1, 1995), and add a random number of days with AddDays (obviusly, pay attention not surpassing the current date).

Answer 2

    DateTime RandomDay()
    {
        DateTime start = new DateTime(1995, 1, 1);
        Random gen = new Random();

        int range = ((TimeSpan)(DateTime.Today - start)).Days;           
        return start.AddDays(gen.Next(range));
    }

For better performance if this will be called repeatedly, create the start and gen (and maybe even range) variables outside of the function.

Answer 3

This is in slight response to Joel's comment about making a slighly more optimized version. Instead of returning a random date directly, why not return a generator function which can be called repeatedly to create a random date.

        Func<DateTime> RandomDayFunc()
        {
            DateTime start = new DateTime(1995, 1, 1); 
            Random gen = new Random(); 
            int range = ((TimeSpan)(DateTime.Today - start)).Days; 
            return () => start.AddDays(gen.Next(range));
        }

Answer 4

Well, if you gonna present alternate optimization, we can also go for an iterator:

 static IEnumerable<DateTime> RandomDay()
 {
    DateTime start = new DateTime(1995, 1, 1);
    Random gen = new Random();
    int range = ((TimeSpan)(DateTime.Today - start)).Days;
    while (true)
     yield return  start.AddDays(gen.Next(range));        
}

you could use it like this:

int i=0;
foreach(DateTime dt in RandomDay())
{
    Console.WriteLine(dt);
    if (++i == 10)
     break;
}

Answer 5

// Chosen by a random dice roll. Guaranteed to be random.
return DateTime.Parse("10/07/2001");

That's the problem with random numbers, you can never be sure.