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Bookmark or Share- programming languages - What is early and late binding? - Software ...In this example,a->f() will actually call void A::f(), because it's early (or statically) bound, and so the program at runtime thinks it's just a pointer to an A type variable, whereas a->g() will actually call void B::g(), because the compiler, seeing g() is virtual, injects code to look up the address of the correct function to call at runtime.
- Why does Java compiler decide whether you can call a method based on ...Your example is in some way a special case. In a non trivial program you can typically not determine the class of the object a certain reference points to better than "it is of the type the reference was declared for or a subtype of it".
- keywords - What are some practical uses of the "new" modifier in C# ...A co-worker and I were looking at the behavior of the new keyword in C# as it applies to the concept of hiding. From the documentation: Use the new modifier to explicitly hide a member inherited from a base class. To hide an inherited member, declare it in the derived class using the same name, and modify it with the new modifier.
- Should curly braces appear on their own line? [closed]However, I want else to start on a new line because if and else belong together, visually. If there's a bracket in front of the else, it's much more difficult to spot what belongs to what. disqualifies itself. We all know what bad things can happen if you leave out the brackets and forget about it.
- How can I explain the A a = new B ();? - Software Engineering Stack ...//in a main IA ia = new B(); ia.doA(); Am I correct in stating that the the runtime object of ia is a B class and it has inherited all the super classes methods and attributes? The printout from within the imaginary main is B class doA as the method doA in B will hide the same methods in its super classes?
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