Why doesn't the Random.Next method in .NET support floats/double?

Question

So maybe I want to have values between -1.0 and 1.0 as floats. It's clumsy having to write this and bolt it on top using extension methods IMO.

Answer 1

There's a .NextDouble() method as well that does almost exactly what you want- just 0 to 1.0 instead of -1.0 to 1.0.

Answer 2

The vast majority of random number usage I've seen (in fact all I've ever seen, although apparently in 3 decades of programming I haven't seen much) is generating random integers in a specified range.

Hence Random.Next seems very convenient to me.

Answer 3

Because most of the time, we want an int within a range, so we were provided with methods to do that. Many languages only support a random method that returns a double between 0.0 and 1.0. C# provides this functionality in the .NextDouble() method.

This is decent enough design, as it makes the common easy (Rolling a die - myRandom.Next(1,7);) and everthing else possible. - myRandom.NextDouble() * 2.0 - 1.0;

Answer 4

It's clumsy having to write this and bolt it on top using extension methods

Not really - extension methods are succinct and I'm glad the feature is there for things like this.

Answer 5

You can use the Random.NextDouble() method which will produces a random value between 0.0 and 1.0 as @Joel_Coehorn's suggested

I just want to add that, extending the range to -1.0 and 1.0 is a simple calculation

var r = new Random();

var randomValue = r.NextDouble() * 2.0 - 1.0

Or to generalize it, to extend NextDouble() result to any range (a, b), you can do this:

var randomValue = r.NextDouble() * (b - a) - a

Answer 6

On a side note, I'd have to point out that, at last, the random-number generation algorithm in a framework is not something clumsy (like cstdlib's rand()) but actually a good implementation (Park&Miller, IIRC).

Answer 7

This method isn't that hard to add, seems like the perfect usage of an extension method.

public double NextSignedDouble(this Random r)
{
    return (r.Next(0, 2) == 0) ? r.NextDouble() : (r.NextDouble() * -1);
}

Answer 8

public static double NextDouble(this Random rnd, double min, double max){
    return rnd.NextDouble() * (max - min) + min;
}

Call with:

double x = rnd.NextDouble(-1, 1);

to get a value in the range -1 <= x < 1