如何在C＃中生成0到1之间的随机数？

Question

I want to get the random number between 1 and 0. However, I'm getting 0 every single time. Can someone explain me the reason why I and getting 0 all the time? This is the code I have tried.

Random random = new Random();
int test = random.Next(0, 1);
Console.WriteLine(test);
Console.ReadKey();

Answer 1

According to the documentation , Next returns an integer random number between the (inclusive) minimum and the (exclusive) maximum:

Return Value

A 32-bit signed integer greater than or equal to minValue and less than maxValue; that is, the range of return values includes minValue but not maxValue. If minValue equals maxValue, minValue is returned.

The only integer number which fulfills

0 <= x < 1

is 0 , hence you always get the value 0 . In other words, 0 is the only integer that is within the half-closed interval [0, 1) .

So, if you are actually interested in the integer values 0 or 1 , then use 2 as upper bound:

var n = random.Next(0, 2);

If instead you want to get a decimal between 0 and 1, try:

var n = random.NextDouble();

Hope this helps :-)

Answer 2

You could, but you should do it this way:

double test = random.NextDouble();

If you wanted to get random integer ( 0 or 1), you should set upper bound to 2, because it is exclusive

int test = random.Next(0, 2);

Answer 3

Every single answer on this page regarding doubles is wrong, which is sort of hilarious because everyone is quoting the documentation. If you generate a double using NextDouble(), you will not get a number between 0 and 1 inclusive of 1, you will get a number from 0 to 1 exclusive of 1.

To get a double, you would have to do some trickery like this:

public double NextRandomRange(double minimum, double maximum)
{
     Random rand = new Random();
     return rand.NextDouble() * (maximum - minimum) + minimum;
}

and then call

NextRandomRange(0,1 + Double.Epsilon);

Seems like that would work, doesn't it? 1 + Double.Epsilon should be the next biggest number after 1 when working with doubles, right? This is how you would solve the problem with ints.

Wellllllllllllllll.........

I suspect that this will not work correctly, since the underlying code will be generating a few bytes of randomness, and then doing some math tricks to fit it in the expected range. The short answer is that Logic that applies to ints doesn't quite work the same when working with floats.

Lets look, shall we? ( https://referencesource.microsoft.com/#mscorlib/system/random.cs,e137873446fcef75 )

  /*=====================================Next=====================================
  **Returns: A double [0..1)
  **Arguments: None
  **Exceptions: None
  ==============================================================================*/
  public virtual double NextDouble() {
    return Sample();
  }

What the hell is Sample()?

  /*====================================Sample====================================
  **Action: Return a new random number [0..1) and reSeed the Seed array.
  **Returns: A double [0..1)
  **Arguments: None
  **Exceptions: None
  ==============================================================================*/
  protected virtual double Sample() {
      //Including this division at the end gives us significantly improved
      //random number distribution.
      return (InternalSample()*(1.0/MBIG));
  }

Ok, starting to get somewhere. MBIG btw, is Int32.MaxValue(2147483647 or 2^31-1), making the division work out to:

InternalSample()*0.0000000004656612873077392578125;

Ok, what the hell is InternalSample()?

  private int InternalSample() {
      int retVal;
      int locINext = inext;
      int locINextp = inextp;

      if (++locINext >=56) locINext=1;
      if (++locINextp>= 56) locINextp = 1;

      retVal = SeedArray[locINext]-SeedArray[locINextp];

      if (retVal == MBIG) retVal--;          
      if (retVal<0) retVal+=MBIG;

      SeedArray[locINext]=retVal;

      inext = locINext;
      inextp = locINextp;

      return retVal;
  }

Well...that is something. But what is this SeedArray and inext crap all about?

  private int inext;
  private int inextp;
  private int[] SeedArray = new int[56];

So things start to fall together. Seed array is an array of ints that is used for generating values from. If you look at the init function def, you see that there is a whole lot of bit addition and trickery being done to randomize an array of 55 values with initial quasi-random values.

  public Random(int Seed) {
    int ii;
    int mj, mk;

    //Initialize our Seed array.
    //This algorithm comes from Numerical Recipes in C (2nd Ed.)
    int subtraction = (Seed == Int32.MinValue) ? Int32.MaxValue : Math.Abs(Seed);
    mj = MSEED - subtraction;
    SeedArray[55]=mj;
    mk=1;
    for (int i=1; i<55; i++) {  //Apparently the range [1..55] is special (All hail Knuth!) and so we're skipping over the 0th position.
      ii = (21*i)%55;
      SeedArray[ii]=mk;
      mk = mj - mk;
      if (mk<0) mk+=MBIG;
      mj=SeedArray[ii];
    }
    for (int k=1; k<5; k++) {
      for (int i=1; i<56; i++) {
    SeedArray[i] -= SeedArray[1+(i+30)%55];
    if (SeedArray[i]<0) SeedArray[i]+=MBIG;
      }
    }
    inext=0;
    inextp = 21;
    Seed = 1;
  }

Ok, going back to InternalSample(), we can now see that random doubles are generated by taking the difference of two scrambled up 32 bit ints, clamping the result into the range of 0 to 2147483647 - 1 and then multiplying the result by 1/2147483647. More trickery is done to scramble up the list of seed values as it uses values, but that is essentially it.

(It is interesting to note that the chance of getting any number in the range is roughly 1/r EXCEPT for 2^31-2, which is 2 * (1/r)! So if you think some dumbass coder is using RandNext() to generate numbers on a video poker machine, you should always bet on 2^32-2! This is one reason why we don't use Random for anything important...)

so, if the output of InternalSample() is 0 we multiply that by 0.0000000004656612873077392578125 and get 0, the bottom end of our range. if we get 2147483646, we end up with 0.9999999995343387126922607421875, so the claim that NextDouble produces a result of [0,1) is...sort of right? It would be more accurate to say it is int he range of [0,0.9999999995343387126922607421875].

My suggested above solution would fall on its face, since double.Epsilon = 4.94065645841247E-324, which is WAY smaller than 0.0000000004656612873077392578125 (the amount you would add to our above result to get 1).

Ironically, if it were not for the subtraction of one in the InternalSample() method:

if (retVal == MBIG) retVal--;

we could get to 1 in the return values that come back. So either you copy all the code in the Random class and omit the retVal-- line, or multiply the NextDouble() output by something like 1.0000000004656612875245796924106 to slightly stretch the output to include 1 in the range. Actually testing that value gets us really close, but I don't know if the few hundred million tests I ran just didn't produce 2147483646 (quite likely) or there is a floating point error creeping into the equation. I suspect the former. Millions of test are unlikely to yield a result that has 1 in 2 billion odds.

NextRandomRange(0,1.0000000004656612875245796924106); // try to explain where you got that number during the code review...

TLDR? Inclusive ranges with random doubles is tricky...

Answer 4

You are getting zero because Random.Next(a,b) returns number in range [a, b), which is greater than or equal to a, and less than b.

If you want to get one of the {0, 1}, you should use:

var random = new Random();
var test = random.Next(0, 2);

Answer 5

Because you asked for a number less than 1 .

The documentation says:

Return Value
A 32-bit signed integer greater than or equal to minValue and less than maxValue; that is, the range of return values includes minValue but not maxValue. If minValue equals maxValue, minValue is returned.

Answer 6

Rewrite the code like this if you are targeting 0.0 to 1.0

Random random = new Random();

double test = random.NextDouble();

Console.WriteLine(test);

Console.ReadKey();