CODEWITHSUNDEEP
c#
Michael M
Michael M

Reputation: 141

C# Error- Small console app

I'm still trying to get the hang of this. The second part of my Main method will not execute. I believe I've called it correctly. But, obviously I didn't. A little help would be greatly appreciated!

using System;
using static System.Console;
using System.Threading;

namespace mellon_Assignment2
{
    class Getting2KnowUapp
    {
        static void Main()
        {
            WriteLine("The current time is: " + DateTime.Now);
            Thread.Sleep(2000);

            AboutMe Me = new AboutMe();

        }
    }
}



using System;
using static System.Console;
using System.Threading;

namespace mellon_Assignment2
{
    class AboutMe
    {
        public void DisplayInfo()
        {
            WriteLine("My Name\tAssignment 2");
            Thread.Sleep(1500);
            WriteLine("ITDEV110\tIntro to Object-oriented Programming");
            Thread.Sleep(1500);
            WriteLine("Professor\tSeptember 18th");
            Thread.Sleep(1500);    
        }    
    }
}

Upvotes: 1

Views: 58

Answers (3)

user8578091
user8578091

Reputation:

public void DisplayInfo() is it's own method and has to be called directly after initialization of the class AboutMe.

If you want the DisplayInfo() method to fire immediately upon initialization of AboutMe then simply add a constructor for AboutMe like so. 

class AboutMe {
    public AboutMe() {
      DisplayInfo();
    }    

     public void DisplayInfo() {
     ...
     }
}

Then you can call: AboutMe myvariable = new AboutMe();

Upvotes: 0

Miek
Miek

Reputation: 1228

Echoing the other replies, you aren't invoking the method on your class.

If you want it to occur when you create a new instance, you could move it into the constructor.

To do that, change: 

public void DisplayInfo()

to

 public AboutMe()

Upvotes: 0

NicoRiff
NicoRiff

Reputation: 4883

You need to call DisplaInfo method. You are only creating the object and doing nothing with it:

AboutMe Me = new AboutMe();
Me.DisplayInfo();

Upvotes: 1

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