Random number generator only generating one random

问题:

I have the following function:

//Function to get random number
public static int RandomNumber(int min, int max)
{
    Random random = new Random();
    return random.Next(min, max);
}

How I call it:

byte[] mac = new byte[6];
for (int x = 0; x < 6; ++x)
    mac[x] = (byte)(Misc.RandomNumber((int)0xFFFF, (int)0xFFFFFF) % 256);

If I step that loop with the debugger during runtime I get different values (which is what I want). However, if I put a breakpoint two lines below that code, all members of the "mac" array have equal value.

Why does that happen?

回答1:

Every time you do new Random() it is initialized using the clock. This means that in a tight loop you get the same value lots of times. You should keep a single Random instance and keep using Next on the same instance.

//Function to get a random number 
private static readonly Random random = new Random(); 
private static readonly object syncLock = new object(); 
public static int RandomNumber(int min, int max)
{
    lock(syncLock) { // synchronize
        return random.Next(min, max);
    }
}

---

Edit (see comments): why do we need a lock here?

Basically, Next is going to change the internal state of the Random instance. If we do that at the same time from multiple threads, you could argue "we've just made the outcome even more random", but what we are actually doing is potentially breaking the internal implementation, and we could also start getting the same numbers from different threads, which might be a problem - and might not. The guarantee of what happens internally is the bigger issue, though; since Random does not make any guarantees of thread-safety. Thus there are two valid approaches:

• synchronize so that we don't access it at the same time from different threads
• use different Random instances per thread

Either can be fine; but mutexing a single instance from multiple callers at the same time is just asking for trouble.

The lock achieves the first (and simpler) of these approaches; however, another approach might be:

private static readonly ThreadLocal<Random> appRandom
     = new ThreadLocal<Random>(() => new Random());

this is then per-thread, so you don't need to synchronize.

回答2:

For ease of re-use throughout your application a static class may help.

public static class StaticRandom
{
    private static int seed;

    private static ThreadLocal<Random> threadLocal = new ThreadLocal<Random>
        (() => new Random(Interlocked.Increment(ref seed)));

    static StaticRandom()
    {
        seed = Environment.TickCount;
    }

    public static Random Instance { get { return threadLocal.Value; } }
}

You can use then use static random instance with code such as

StaticRandom.Instance.Next(1, 100);

回答3:

Mark's solution can be quite expensive since it needs to synchronize everytime.

We can get around the need for synchronization by using the thread-specific storage pattern:

public class RandomNumber : IRandomNumber
{
    private static readonly Random Global = new Random();
    [ThreadStatic] private static Random _local;

    public int Next(int max)
    {
        var localBuffer = _local;
        if (localBuffer == null) 
        {
            int seed;
            lock(Global) seed = Global.Next();
            localBuffer = new Random(seed);
            _local = localBuffer;
        }
        return localBuffer.Next(max);
    }
}

Measure the two implementations and you should see a significant difference.

回答4:

My answer from here:

Just reiterating the right solution:

namespace mySpace
{
    public static class Util
    {
        private static rnd = new Random();
        public static int GetRandom()
        {
            return rnd.Next();
        }
    }
}

So you can call:

var i = Util.GetRandom();

all throughout.

If you strictly need a true stateless static method to generate random numbers, you can rely on a Guid.

public static class Util
{
    public static int GetRandom()
    {
        return Guid.NewGuid().GetHashCode();
    }
}

It's going to be a wee bit slower, but can be much more random than Random.Next, at least from my experience.

But not:

new Random(Guid.NewGuid().GetHashCode()).Next();

The unnecessary object creation is going to make it slower especially under a loop.

And never:

new Random().Next();

Not only it's slower (inside a loop), its randomness is... well not really good according to me..

回答5:

I would rather use the following class to generate random numbers:

byte[] random;
System.Security.Cryptography.RNGCryptoServiceProvider prov = new System.Security.Cryptography.RNGCryptoServiceProvider();
prov.GetBytes(random);

回答6:

1) As Marc Gravell said, try to use ONE random-generator. It's always cool to add this to the constructor: System.Environment.TickCount.

2) One tip. Let's say you want to create 100 objects and suppose each of them should have its-own random-generator (handy if you calculate LOADS of random numbers in a very short period of time). If you would do this in a loop (generation of 100 objects), you could do this like that (to assure fully-randomness):

int inMyRandSeed;

for(int i=0;i<100;i++)
{
   inMyRandSeed = System.Environment.TickCount + i;
   .
   .
   .
   myNewObject = new MyNewObject(inMyRandSeed);  
   .
   .
   .
}

// Usage: Random m_rndGen = new Random(inMyRandSeed);

Cheers.

回答7:

Every time you execute

Random random = new Random (15);

It does not matter if you execute it millions of times, you will always use the same seed.

If you use

Random random = new Random ();

You get different random number sequence, if a hacker guesses the seed and your algorithm is related to the security of your system - your algorithm is broken. I you execute mult. In this constructor the seed is specified by the system clock and if several instances are created in a very short period of time (milliseconds) it is possible that they may have the same seed.

If you need safe random numbers you must use the class

System.Security.Cryptography.RNGCryptoServiceProvider

public static int Next(int min, int max)
{
    if(min >= max)
    {
        throw new ArgumentException("Min value is greater or equals than Max value.");
    }
    byte[] intBytes = new byte[4];
    using(RNGCryptoServiceProvider rng = new RNGCryptoServiceProvider())
    {
        rng.GetNonZeroBytes(intBytes);
    }
    return  min +  Math.Abs(BitConverter.ToInt32(intBytes, 0)) % (max - min + 1);
}

Usage:

int randomNumber = Next(1,100);

回答8:

just declare the Random class variable like this:

    Random r = new Random();
    // ... Get three random numbers.
    //     Here you'll get numbers from 5 to 9
    Console.WriteLine(r.Next(5, 10));

if you want to get different random number each time from your list then use

r.Next(StartPoint,EndPoint) //Here end point will not be included

Each time by declaring Random r = new Random() once.

回答9:

There are a lot of solutions, here one: if you want only number erase the letters and the method receives a random and the result length.

public String GenerateRandom(Random oRandom, int iLongitudPin)
{
    String sCharacters = "123456789ABCDEFGHIJKLMNPQRSTUVWXYZ123456789";
    int iLength = sCharacters.Length;
    char cCharacter;
    int iLongitudNuevaCadena = iLongitudPin; 
    String sRandomResult = "";
    for (int i = 0; i < iLongitudNuevaCadena; i++)
    {
        cCharacter = sCharacters[oRandom.Next(iLength)];
        sRandomResult += cCharacter.ToString();
    }
    return (sRandomResult);
}