Answers for "how to get random in c#"

C#
37

c# random number

Random rnd = new Random();
int number  = rnd.Next(1, 10);
Posted by: Guest on September-16-2020
1

c# random

Random rnd = new Random();
int lowerBound = 10;
int upperBound = 11;
int[] range = new int[10];
for (int ctr = 1; ctr <= 1000000; ctr++) {
   Double value = rnd.NextDouble() * (upperBound - lowerBound) + lowerBound;
   range[(int) Math.Truncate((value - lowerBound) * 10)]++;
}

for (int ctr = 0; ctr <= 9; ctr++) {
   Double lowerRange = 10 + ctr * .1;
   Console.WriteLine("{0:N1} to {1:N1}: {2,8:N0}  ({3,7:P2})",
                     lowerRange, lowerRange + .1, range[ctr],
                     range[ctr] / 1000000.0);
}

// The example displays output like the following:
//       10.0 to 10.1:   99,929  ( 9.99 %)
//       10.1 to 10.2:  100,189  (10.02 %)
//       10.2 to 10.3:   99,384  ( 9.94 %)
//       10.3 to 10.4:  100,240  (10.02 %)
//       10.4 to 10.5:   99,397  ( 9.94 %)
//       10.5 to 10.6:  100,580  (10.06 %)
//       10.6 to 10.7:  100,293  (10.03 %)
//       10.7 to 10.8:  100,135  (10.01 %)
//       10.8 to 10.9:   99,905  ( 9.99 %)
//       10.9 to 11.0:   99,948  ( 9.99 %)
Posted by: Guest on December-28-2020

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