Is there any builtin random function ???

Start with Random variablename = new Random(); Generate the random numbers with the Next() method. The first argument is the starting number. The second argument is the ending number + 1. newnum = randomvariable.Next(start, finish + 1) Here is some sample code to study: int x, first, last; string entry; Random randnum = new Random(); for (x = 1; x <= 20; x++) Console.Write("{0} ", randnum.Next(1, 11)); Console.WriteLine("\n"); Console.Write("Enter starting number: "); entry = Console.ReadLine(); first = Convert.ToInt32(entry); Console.Write("Enter ending number: "); entry = Console.ReadLine(); last = Convert.ToInt32(entry) + 1; Console.WriteLine(); for (x = 1; x <= 20; x++) Console.Write("{0} ", randnum.Next(first, last)); Console.WriteLine("\n");

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joy.simpson21y ago
Start with Random variablename = new Random(); Generate the random numbers with the Next() method. The first argument is the starting number. The second argument is the ending number + 1. newnum = randomvariable.Next(start, finish + 1) Here is some sample code to study: int x, first, last; string entry; Random randnum = new Random(); for (x = 1; x <= 20; x++) Console.Write("{0} ", randnum.Next(1, 11)); Console.WriteLine("\n"); Console.Write("Enter starting number: "); entry = Console.ReadLine(); first = Convert.ToInt32(entry); Console.Write("Enter ending number: "); entry = Console.ReadLine(); last = Convert.ToInt32(entry) + 1; Console.WriteLine(); for (x = 1; x <= 20; x++) Console.Write("{0} ", randnum.Next(first, last)); Console.WriteLine("\n");