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Random number in long range, is this the way?

Asked 2011-07-11T14:21:47.807
71

Can somebody verify this method. I need a long type number inside a range of two longs. I use the .NET Random.Next(min, max) function which return int's. Is my reasoning correct if I simply divide the long by 2, generate the random number and finally multiply it by 2 again? Or am I too enthusiastic... I understand that my random resolution will decrease but are there any other mistakes which will lead to no such a random number.

long min = st.MinimumTime.Ticks;    //long is Signed 64-bit integer
long max = st.MaximumTime.Ticks;
int minInt = (int) (min / 2);      //int is Signed 64-bit integer
int maxInt = (int) (max / 2);      //int is Signed 64-bit integer

Random random = new Random();
int randomInt = random.Next(minInt, maxInt);
long randomLong = (randomInt * 2);
c#-4.0 random
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2 Answers

112

Why don't you just generate two random Int32 values and make one Int64 out of them?

long LongRandom(long min, long max, Random rand) {
    long result = rand.Next((Int32)(min >> 32), (Int32)(max >> 32));
    result = (result << 32);
    result = result | (long)rand.Next((Int32)min, (Int32)max);
    return result;
}

Sorry, I forgot to add boundaries the first time. Added min and max params. You can test it like that:

long r = LongRandom(100000000000000000, 100000000000000050, new Random());

Values of r will lie in the desired range.

EDIT: the implementation above is flawed. It's probably worth it to generate 4 16-bit integers rather than 2 32-bit ones to avoid signed-unsigned problems. But at this point the solution loses its elegancy, so I think it's best to stick with Random.NextBytes version:

long LongRandom(long min, long max, Random rand) {
    byte[] buf = new byte[8];
    rand.NextBytes(buf);
    long longRand = BitConverter.ToInt64(buf, 0);

    return (Math.Abs(longRand % (max - min)) + min);
}

It looks pretty well in terms of value distribution (judging by very simple tests I ran).

answered 2011-07-11T14:28:35.497
-1

Is there anything wrong with using this simple approach?

        long min = 10000000000001;
        long max = 99999999999999;
        Random random = new Random();
        long randomNumber = min + random.Next() % (max - min);

d

answered 2013-06-29T08:15:48.160

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