Inconsistent accessibility: property type

Question

What is wrong with this code ?

public ScreenManager _ScreenManager
        {
            get { return screenManager; }
            internal set { screenManager = value; }
        }
        ScreenManager screenManager;

and I get this error:

Inconsistent accessibility: field type 'ScreenSystem.ScreenManager' is less accessible than field 'ScreenSystem.Screen.ScreenManager'

Answer 1

I'm going to assume the type ScreenManager is internal while the class containing your property is public.

The compiler is saying that a publicly-accessible field (ScreenSystem.Screen.ScreenManager) is of a type (ScreenSystem.ScreenManager) that isn't publicly-accessible.

Your field should generally be private anyway. And you might be missing a public in front of your class ScreenSystem.ScreenManager (it defaults to internal IIRC).

Answer 2

It means that you have ScreenManager set up as private or protected yet you are are trying to make a property public that relies on ScreenManager. You can't do that.

Make ScreenManager public or make _ScreenManager the same accessibility as ScreenManager.

Answer 3

Here is your scenario:

namespace ScreenSystem
{
    internal class ScreenManager
    {
        public string Test { get; set; }
    }
}

namespace ScreenSystem
{
    public class Screen
    {
        public ScreenManager Manager
        {
            get; internal set;
        }
    }
}

Compiler Output
Inconsistent accessibility: property type 'ScreenSystem.ScreenManager' is less accessible than property 'ScreenSystem.Screen.Manager'

Here are the solutions, depending on what you're trying to do:

• Make the Screen class internal
• Make the ScreenManager class public
• Make the ScreenManager class public and the Screen class internal
• Make the Screen.Manager property internal (and remove the internal set accessor)

Either one of the above will compile without errors. It really just depends on what you're trying to achieve.