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The problem is that you are creating instances of the Random class too close in time.
When you create a Random object, it's seeded with a value from the system clock. If you create Random instances too close in time, they will all be seeded with the same random sequence.
Create a single Random object and pass its reference to the constructor when you create instances of the "a" class, instead of creating one Random object for each "a" instance.
Random.Next returns always the same values, Returns a random integer. The following example makes repeated calls to the Next method to generate a specific number of random numbers requested by� Random.Next returns always the same values [duplicate] Ask Question Asked 9 years, 2 months ago. Active 9 years, 2 months ago. Viewed 6k times 3. This question
Use a single, static Random number generator for all instances of the class.
class a
{
private static Random rnd;
static a() {
rnd = new Random();
}
private void Count()
{
int r1 = rnd.Next(-1, 2);
int r2 = rnd.Next(-1, 2);
}
}
Note the change to give you numbers in the range -1,1 rather than -1,0
Random.Next Method (System), If I put in a break point and then step through the code it will return different numbers. I'm assuming the random function is seeded based on time� The problem is that Random.Next returns the same "random" numbers for each instance. When the results for the first instance are 0 and -1, the same ones will be returned from following instances. When the results for the first instance are 0 and -1, the same ones will be returned from following instances.
You're creating a new instance of Random very close together (your loop is very tight) so each instance is effectively using the same seed value.
A better approach would be to create one instance and pass that to your Count method.
You problably know this next bit, but I'll include it here for completeness:
The MSDN has the details on this, but basically your problem is the Random.Next method you're using generates:
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.
because of this your calls will return -1 or 0.
Random.Next returning same number, public virtual int Next ();. Return Value: This method returns the 32-bit signed integer which is greater than or equal to 0 and less than MaxValue. Example:. Recommend：c# - Random.Next() returns always the same value. cate] 4 answers I want to get a random instance of an ArrayList, but my code always returns the same value. public static void randomEvent(){ Random rnd = new Random(); int which = rnd.Next(1,events.Count);
You include a random instance for each A instance. It sounds like they're all getting the same default seed value. You probably want to make a static random for all A instances and use it repeatedly, or alternatively provide a seed value to the Random() instance in the A constructor.
C#, As mentioned, this will almost always return false, and will thus have to scan the Next(i + 1); int value = result[k]; result[k] = result[i]; result[i] = value; } return result; top < Int32.MaxValue; top++) { // May strike a duplicate. int value = random. Suppose you have to extract 5 numbers in the 1..100 range. The odds of getting one or more duplicates are: p=(100^5 - 100*99*98*97*96)/100^5 ≅ 9.7 % So, in about ten percent of the cases you are going to get dulpicates (if you observe a greater frequency then either there is a bug in your code or (less probably) the random generator is incorrect.
Generate random numbers without repetitions, Suppose you have to extract 5 numbers in the 1..100 range. The odds of getting one or more duplicates are: Hide Copy Code. p=(100^5� Recommend：c# - Random.Next() returns always the same value. cate] 4 answers I want to get a random instance of an ArrayList, but my code always returns the same value. public static void randomEvent(){ Random rnd = new Random(); int which = rnd.Next(1,events.Count);
Why does C# random return same value?, Next() method returns a random number in C#. C# Random class provides functionality to generate random numbers in C#. and keep using Next on the same instance. private readonly Random _random = new Random� Random.Next generates a random number whose value ranges from 0 to less than Int32.MaxValue. To generate a random number whose value ranges from 0 to some other positive number, use the Random.Next(Int32) method overload. To generate a random number within a different range, use the Random.Next(Int32, Int32) method overload. Notes to Inheritors
Generate Random Number And Random String In C#, Random has a Next method and can be used as a field. C# program that causes repeated random numbers class Program { static void Main() { Test(); Test (); Test(); } static void Test() So the first argument is always lower than the second. A typical "random" number generator cannot return a truly random number. Random.Next returns always the same values. 0. LINQ group by and select random element from each group. Related. 2496. What are the correct version numbers for C#? 755.
Comments I forgot as well, I used to have the same issue back in the day making Bingo Cartons for a club, and back then, I used the worst trick ever known to man: Pausing the thread for 2 Ms. Inexperienced and crazy... Crazy enough, I have a class that creates random names with a static Random declaration on top of everything. i think with this range it can also return 1 and they only want 0 and -1? @Svante edited the question to make the interval open to match the code sample. The original question specified a closed interval, though not using precise language. I think the code was incorrect and am going to restore the question to specify a closed interval. @tvanfosson Can you tell me why this works? I don't understand how making this statics gives you randomness. I know it works, but not why. They are still very close in time. Thanks. @johnny when it's created as an instance variable, the constructor for Random is executed twice in quick succession. Since it uses the current time as the seed, this can result in both random number generators using the same sequence of pseudorandom numbers and thus they both return the same values in lock step. Using a single random number generator, initialized once in the static constructor for the class, returns different numbers from the single pseudorandom sequence giving the behavior that you would expect from a random number generator.
