Question

If I wanted to generate a random number for all possible numbers an Int32 could contain would the following code be a reasonable way of doing so? Is there any reason why it may not be a good idea? (ie. a uniform distribution at least as good as Random.Next() itself anyway)

    public static int NextInt(Random Rnd) //-2,147,483,648 to 2,147,483,647
    {
         int AnInt;
         AnInt = Rnd.Next(System.Int32.MinValue, System.Int32.MaxValue);
         AnInt += Rnd.Next(2);
         return AnInt;
    }

Answer 1

Your proposed solution will slightly skew the distribution. The minValue and maxValue will occur less frequently than the interior values. As an example, assume that int has a MinValue of -2 and a MaxValue of 1. Here are the possible initial values, with each followed by the resulting values after the Random(2):

-2: -2 -1
-1: -1  0
 0:  0  1

half of the negative -2 values will get modified up to -1, and only half of 0 will get modified up to 1. So the values -2 and 1 will occur less frequently than -1 and 0.

Damien’s solution is good. Another choice would be:

if (Random(2) == 0) {
    return Random(int.MinValue, 0);
} else {
    return 1 + Random(-1, int.MaxValue);
}

another solution, similar to Damiens approach, and faster than the previous one would be

 int i = r.Next(ushort.MinValue, ushort.MaxValue + 1) << 16;
 i |= r.Next(ushort.MinValue, ushort.MaxValue + 1);