Why does Java specify that the access specifier for an overriding method can allow more, but not less, access than the overridden method?开发者_Go百科 For example, a protected instance method in the s
I am reading some Java text and the text says that we can only apply public or default access modifier for class and interface. Therefore, it is a compiling error i开发者_运维知识库f we declare:
I have a hierarchy of classes: class Page { } class ChildPage1 extends Page { private static final String URL;
Eclipse is giving me an error when defining a top level class with the package modifier, the error is :
OK, well, I am a beginner, so.. yeah, this may be a very stupid question. I read that if I declare variable or object, without mentioniong the access-modifier (public, private, etc.) than it\'s automa
consider the following: class X { public: X(int i) { cout << \"X(int i)\" << endl; } X(const X& x) { cout << \"X(const X& x)\" << endl; }
In C#, we can not change access modifier while overriding a method from base class. e.g. Class Base { **protected** string foo()
In C# you can mark a class as internal so that it is开发者_开发技巧 only accessible from within the same package. Is there anything similar in Java?You can create package-private classes by omitting t
Can we have different access modifier for get and set in a property开发者_运维技巧?Yes, you can, however it is subject to the rule that your getter/setter cannot have a less restricted access modifier
I know you can do this public String<T> Get { get; private set; } The problem is I want the set method to开发者_C百科 be accessible from a different class within the same DLL (ie: internal)
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