What is deadly diamond problem in Java?

Deadly diamond of death is a problem which occurs with the inheritance of classes. … The ambiguity is, classes B and C inherits from class A, and class D inherits from the classes B and C. If there is a method in class A, both the B and C classes and overriding the method and D doesn’t override it.

What is deadly diamond of death problem in Java?

The “diamond problem” (sometimes referred to as the “Deadly Diamond of Death”) is an ambiguity that arises when two classes B and C inherit from A, and class D inherits from both B and C.

What is Java diamond problem?

The diamond problem is a common problem in Java when it comes to inheritance. … Multi-level inheritance allows a child class to inherit properties from a class that can inherit properties from some other classes. For example, class C can inherit its property from B class which itself inherits from A class.

What is the diamond problem in Java how can it be removed?

The diamond problem can be resolve manually. In this case, resolve the conflict manually by using the super keyword within the StudentRecord class to explicitly mention which method the definition wants to use. The super keyword uses to call the method of parent class/interface.

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What is the diamond problem and how is it solved?

Virtual inheritance solves the classic “Diamond Problem”. It ensures that the child class gets only a single instance of the common base class. In other words, the Snake class will have only one instance of the LivingThing class. The Animal and Reptile classes share this instance.

What is Java encapsulation?

Encapsulation in Java is a mechanism of wrapping the data (variables) and code acting on the data (methods) together as a single unit. … Therefore, it is also known as data hiding. To achieve encapsulation in Java − Declare the variables of a class as private.

Why do we use super in Java?

The super keyword in java is a reference variable that is used to refer parent class objects. … Basically this form of super is used to initialize superclass variables when there is no constructor present in superclass.

Can we achieve abstraction without encapsulation?

The object is the abstract form of the real-world and its details are hidden using encapsulation. Thus encapsulation is required for abstraction.

Can we extend 2 classes in Java?

Two classes are not allowed, but a class can extend two interfaces in Java. This language allows extending two or more interfaces in a class. … So, if you want to extend multiple inheritances, it would be better to use the interface. See the example below.

What is Diamond Problem in Java 8?

Java 8 brought a major change where interfaces can provide default implementation for its methods. Java designers kept in mind the diamond problem of inheritance while making this big change. … Any method inherited from a class or a superclass is given higher priority over any default method inherited from an interface.

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How will you achieve encapsulation?

Encapsulation in Java can be achieved by: Declaring the variables of a class as private. Providing public setter and getter methods to modify and view the variables values.

What is diamond operator in Java?

Diamond Operator: Diamond operator was introduced in Java 7 as a new feature. The main purpose of the diamond operator is to simplify the use of generics when creating an object. It avoids unchecked warnings in a program and makes the program more readable.

Is overriding possible in Java?

Instance methods can be overridden only if they are inherited by the subclass. A method declared final cannot be overridden. A method declared static cannot be overridden but can be re-declared. If a method cannot be inherited, then it cannot be overridden.

How do you avoid the diamond problem?

The solution to the diamond problem is to use the virtual keyword. We make the two parent classes (who inherit from the same grandparent class) into virtual classes in order to avoid two copies of the grandparent class in the child class.

Why Java doesn’t support multiple inheritance but C++ does?

Since interface in java can only declare the signature of methods without implementing them, the problem does not exists if multiple interface are derived. In conclusion, in order to avoid the problem Java forbids directly multiple inheritance, and allows only multiple implementation of interface.