可访问性不一致：属性类型 [英] Inconsistent accessibility: property type

问题描述

什么是错的代码？

 公开的ScreenManager _ScreenManager 
 {
 {返回的ScreenManager ; } 
内部集合{=的ScreenManager价值; } 
} 
的ScreenManager屏幕管理; 
  

和我得到这个错误：

 可访问性不一致：字段类型'ScreenSystem.ScreenManager'比少外地访问ScreenSystem.Screen.ScreenManager'
  

解决方案

我要承担键入屏幕显示器是内部而类包含你的财产是公共。

编译器说，可公开访问的字段（ ScreenSystem。 Screen.ScreenManager ）是一类的（ ScreenSystem.ScreenManager ）不是公开访问。

您现场一般应私人反正。而且你可能忽略了公共在全班同学面前 ScreenSystem.ScreenManager （默认为内部 IIRC）。

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'

解决方案

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).

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