how to construct a class? in c# console application

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1.00/5 (2 votes)

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i confuse about class.. how to construct this code to class? in C#console appli

C#

static void Main(string[] args)
{

Random generator = new Random();
int Random = generator.Next(1032, 2353);

Console.WriteLine(Random);
}
}

---------------------------------------------

C#

static void Main(string[] args)
{
          Console.WriteLine("RandomCode:");
          string code = string.Intern (Console.ReadLine());

          Random generator = new Random();
          generator.Code = code;
}


//my class

class Random
{
    public Random() { }


    private string code;
    public string Code
    {
        get { return code; }
        set { code = value; }
    }

}

I dont know how to put this to a class.

C#

Random generator = new Random();
int Random = generator.Next(1032, 2353);

Console.WriteLine(Random);

-------

Posted 3-Mar-14 23:24pm

jone,jake

Updated 4-Mar-14 0:34am

phil.o

v5

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Comments

Maciej Los 4-Mar-14 5:26am

What?

OriginalGriff 4-Mar-14 5:34am

DON'T SHOUT. Using all capitals is considered shouting on the internet, and rude (using all lower case is considered childish). Use proper capitalisation if you want to be taken seriously.

jone,jake 4-Mar-14 5:43am

KEYBOARD PROBLEM SIR...SORRY.

OriginalGriff 4-Mar-14 5:55am

AND AGAIN?
Yeah, I believe you...

jone,jake 4-Mar-14 6:05am

help me with my code sir,..:)
im just 3yr. high school..

OriginalGriff 4-Mar-14 6:13am

Doesn't give you the right to be rude, I'm afraid.
When you start off by shouting at people, you do not encourage them to help you...

jone,jake 4-Mar-14 6:17am

Sorry everyone....

1 solution

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Thomas Nielsen - getCore · Answer 1 · 2014-03-04T00:17:00

Solution 1

C#

using System;

public static class Randomizer
   {
       private static Random RND
       {
           get { return _rnd ?? (_rnd = new Random()); }
       }
       private static Random _rnd ;

       public static int GetNext()
       {
           return RND.Next();
       }

       public static int GetNext(int max)
       {
           return RND.Next(max);
       }

       public static int GetNext(int min, int max)
       {
           return RND.Next(min, max);
       }
   }

   class Program
   {
       static void Main(string[] args)
       {
           Console.WriteLine("Next random is: {0}", Randomizer.GetNext(1032, 2353));
           Console.WriteLine("Next random is: {0}", Randomizer.GetNext(1032, 2353));
           Console.WriteLine("Next random is: {0}", Randomizer.GetNext(1032, 2353));
           Console.ReadKey();
           return;

Posted 4-Mar-14 0:17am

Thomas Nielsen - getCore

Updated 4-Mar-14 2:41am

v2

Comments

jone,jake 4-Mar-14 6:32am

the type or namespace 'Random' could not be found(are you missing a using directive or an assembly....

what is missing???

Thomas Nielsen - getCore 4-Mar-14 8:40am

using System;

i've added my usings to an updated solution

jone,jake 4-Mar-14 9:31am

i get it..tnx:)

jone,jake 4-Mar-14 9:42am

how about using loop sir?
random number using for loop?

Thomas Nielsen - getCore 5-Mar-14 5:31am

i am unsure what you want to loop, forever? a random number of times?

jone,jake 7-Mar-14 1:29am

sample of a pincode number sir.

Thomas Nielsen - getCore 7-Mar-14 3:07am

simple old fashioned for loop should do it, say you wanted hundred for(idx = 0;idx < 100;idx++){for(ix = 0;ix < 4;ix++){ //Do your get next random numer construct} //do what you want with individual sequence}
Just remember that you mustn't new the Random object up every time or you'll get the same sequence..