Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

C# C# Basics (Retired) Perfect Variable Scope

I do not understand "The name `output' does not exist in the current context" in the editor ,

What it is I am not doing correct here?

Program.cs
using System;

namespace Treehouse.CodeChallenges
{
    class Program
    {
        static void Main()
        {            
           string input = Console.ReadLine();

            if (input == "quit")
            {
                string output = "Goodbye.";
            }
            else
            {
                string output = "You entered " + input + "." ;
            }

            Console.WriteLine(output);
        }
    }
}

1 Answer

Rune Andreas Nielsen
Rune Andreas Nielsen
5,354 Points

Hi Yoshoda.

You need to make a variable named "output" outside of the scope.

The problem is that you define the "output" variable inside the "if"/"else" scopes, and the variable will be "destroyed" when the scope/if statement is finished, this means you won't be able to access it in the Console.WriteLine(output); On the last line.

using System;

namespace Treehouse.CodeChallenges
{
    class Program
    {
        static void Main()
        {            
           string input = Console.ReadLine();
           string output = "";


            if (input == "quit")
            {
                output = "Goodbye.";
            }
            else
            {
                output = "You entered " + input + "." ;
            }

            Console.WriteLine(output);
        }
    }
}