Understand variable scopes, use local variable type inference, apply encapsulation, and make objects immutable.
Implement inheritance, including abstract and sealed classes. Override methods, including that of Object class. Implement polymorphism and differentiate object type versus reference type. Perform type casting, identify object types using instanceof operator and pattern matching.
Create and use interfaces, identify functional interfaces, and utilize private, static, and default interface methods.
1. The correct answer is C.
Explanation:
3 followed by 1.
3 followed by 1 due to x being in scope, attempting to access y outside of its declaring block (the if block) will cause a compile-time error.3 followed by 1 and an undefined value for y.
y in this case) outside of their scope. The notion of an “undefined value for y” is not applicable here; the compiler will simply not compile the code.y is accessed outside of its scope.
y is declared inside the if block, and thus, it is only accessible within that block. Trying to access it outside of its scope, as done in the last System.out.println(y);, causes a compile-time error, specifically saying that y cannot be found.y.
y, hence a runtime exception regarding y is out of the question.2. The correct answers are C and D.
Explanation:
double x, double y;
double x, y;.int i = 0, String s = "hello";
int and String in this case) in a single statement.float f1 = 3.14, f2 = 6.28f;
float in this case) in a single statement, and it’s also fine to initialize them with values in the same statement.char a = 'A', b, c = 'C';
char in this case), and initialize some, all, or none of them in the same statement.3. The correct answers are B and D.
Explanation:
var can be used to declare both local variables within methods and instance variables within classes.
var cannot be used to declare instance variables. It is specifically restricted to local variables within methods, constructors, or initializer blocks, as using var for fields would reduce the clarity of a class’s public API.var is restricted to local variables within methods, constructors, or initializer blocks.
var is intended for local variable type inference, significantly reducing the verbosity of Java code in scenarios where the compiler can easily determine the type of the local variable from its initializer. Its use is restricted to ensure clarity and prevent ambiguity in more complex constructs like class fields or method parameters.var can be used to declare method parameters.
var cannot be used to declare method parameters. This limitation ensures that method signatures remain explicit in their type requirements, a critical aspect of a class’s contract with its callers.var enhances readability by inferring types where it’s clear from the context, but it’s not allowed in method signatures to maintain clarity.
var is primarily used to improve code readability by reducing the need for explicit type declarations where the type can be inferred from the context, it is not allowed in method signatures. This restriction ensures that the types of parameters in methods are always explicitly defined, aiding in the readability and maintainability of public APIs.var can be used to declare class (static) variables.
var is not permissible for declaring class (static) variables. The rationale behind this restriction aligns with the goal of maintaining explicit type declarations in the class’s structure, ensuring the class’s design remains clear and unambiguous to both the compiler and developers.4. The correct answers are A and C.
Explanation:
extends keyword is used in Java to create a subclass that inherits from a superclass.
protected and public members of their superclass. This accessibility allows subclasses to leverage and extend the functionality provided by the superclass while maintaining encapsulation of private members.protected and public members of their superclass directly.
extends keyword is indeed used to define a subclass that inherits properties and behaviors from a single superclass, establishing an is-a relationship between the subclass and the superclass. This is a fundamental concept in Java’s implementation of inheritance.private members of its superclass.
private members of its superclass. Instead, it can access them through public or protected accessors provided by the superclass. This encapsulation principle ensures a controlled interaction with the superclass’s state.5. The correct answers are A and C.
Explanation:
"Dog eats" when executed.
Dog class has provided an implementation for the eat method, which is abstract in the superclass Animal. Since myAnimal is of type Animal but instantiated as a Dog, it will call the overridden eat method in the Dog class, printing "Dog eats".Animal class can be instantiated.
Animal class is abstract and cannot be instantiated. Attempting to create an instance of Animal directly (new Animal()) would result in a compilation error.eat method from the Dog class will cause a compilation error.
Dog extends the abstract class Animal and Animal has an abstract eat method, Dog must provide an implementation for eat. Failing to do so will prevent the code from compiling because Dog would also be considered abstract.Cat class is necessary for the code to compile and run.
Cat class is not referenced in the main method or anywhere else in the provided code snippet. Thus, it is unnecessary for the compilation and execution of the given code segment.6. The correct answers are B and C.
Explanation:
Person class must override the getSpeed method.
Person class is not required to override the getSpeed method because it is a default method in the Runnable interface. Default methods provide an implementation that can be used or overridden by implementing classes, but overriding is not mandatory.distance variable in the Walkable interface is implicitly public, static, and final.
public, static, and final. This means the distance variable in the Walkable interface is a constant and must be initialized at the point of declaration. It is accessible with the interface name, like Walkable.distance.Person object can call the getSpeed method without any implementation in the Person class.
Runnable interface provides a default implementation for the getSpeed method, a Person object can call the getSpeed method without any additional implementation in the Person class itself. The default implementation from the interface will be used.Runnable interface causes a compilation error due to a naming conflict with java.lang.Runnable.
Runnable interface declared in the code snippet and java.lang.Runnable exist in different packages. There is no compilation error unless there’s an attempt to import both in the same file without using a fully qualified name. Plus, this situation does not directly relate to the functionality or declaration of interfaces per the exam’s focus.7. The correct answer is A.
Explanation:
Shape class is correctly defined as a sealed class, allowing only specified classes to extend it.
Shape class is declared as a sealed class, which means it can be extended only by the classes it explicitly permits through the permits clause. In this case, Shape permits Circle and Square to extend it, and both classes are correctly defined as permitted subclasses.Square class does not correctly extend the Shape class because it is not marked as final.
final if they are non-sealed. The keyword non-sealed explicitly allows the Square class to extend the sealed Shape class without being final, indicating it can be further extended.Circle class can be further extended by other classes.
Circle class is declared as final, which means it cannot be extended further and aligning with the constraints of extending a sealed class where the permitted subclass can be final, sealed, or non-sealed.area method in the Shape class must provide a default implementation.
Shape are not required to provide implementations for their abstract methods. The purpose of an abstract class is to define a template that its subclasses will follow, which includes implementing any abstract methods declared in the abstract class.8. The correct answer is D.
Explanation:
static context of the class, allowing access to static methods and fields.
this keyword does not refer to the static context of the class. It specifically refers to the current instance of the class. Static methods and fields belong to the class itself and are not part of any instance, so they cannot be accessed through this.this keyword does not store the return value of a method. It is used within an instance method or a constructor to refer to the current object the method or constructor is being invoked upon.this can be omitted (for example, when accessing instance fields or methods without any naming conflict), its use is necessary for situations like constructor chaining (this() call) or when the method parameter names shadow the instance field names. In such scenarios, this clarifies to which variable the code is referring.this keyword in Java is used to refer to the current object—the object whose instance variable, method, or constructor is being called. You can see its usage in line 5 to call another constructor within the same class, in line 14 to differentiate between the method parameter size and the instance variable size, and in the updateWidget method to access the instance variable size. This usage demonstrates this as a way to refer explicitly to properties or methods of the current object.9. The correct answers are A and B.
Explanation:
super keyword is used in the Dog constructor to call the superclass constructor.
Dog constructor, super(name); is used to call the superclass (Animal) constructor with the name parameter. This is necessary to initialize the name field inherited from the Animal class in the Dog instance.eat method in the Dog class uses super to invoke the superclass’s eat method.
eat method in the Dog class calls super.eat(); to invoke the eat method defined in the superclass (Animal). This allows the Dog class to extend the functionality of the eat method beyond what is defined in the superclass, demonstrating method overriding and use of super to access the overridden method.super.eat(); call in the Dog class’s eat method will prevent the Dog class from compiling.
super.eat(); call from the Dog class’s eat method would not prevent the class from compiling. It would simply mean that the Dog class’s eat method no longer calls the superclass’s eat method, altering the program’s behavior but not its compilability.super keyword can be used to access static methods from the superclass.
super can indeed be used to access superclass methods, it’s not specifically used or necessary for accessing static methods. Static methods belong to the class, not to instances, and should be invoked using the class name. super is used primarily for instance methods and constructors.10. The correct answer is D.
Explanation:
"Car driving at speed: 60".
drive method in the Car class has a different parameter type (long) than the method in the Vehicle class (int). Due to the difference in parameter types, the Car class’s drive method does not override but rather overloads the Vehicle class’s drive method. Since the method is called on a Vehicle reference, the Vehicle class’s drive method is invoked.drive method cannot be called using a Vehicle reference.
Vehicle defines the drive method correctly.drive method in the Car class does not properly override the drive method in the Vehicle class.
@Override annotation does not cause a compile-time error here because it is not strictly enforced in terms of method overloading (changing the parameter type creates a new method signature, making this a valid overload)."Vehicle driving at speed: 60" because the drive method in the Car class is an overload, not an override.
drive method in the Car class has a different signature from the drive method in the Vehicle class due to the parameter type (int vs. long). Therefore, the drive method in the Car class overloads the superclass method rather than overriding it. When a Vehicle reference calls the drive method with an int argument, it invokes the Vehicle class’s drive method, not the Car class’s method.11. The correct answer is E.
Explanation:
"Apple flavor" followed by "Red".
flavor method will indeed print "Apple flavor" due to polymorphism (the Apple class overrides the flavor method of Fruit), the code will not compile if the color() method is called on a Fruit reference. This is because the color method is not part of the Fruit class’s interface."Fruit flavor".
flavor method would print "Apple flavor" because of the overridden method in the Apple class, not "Fruit flavor". However, the presence of the color() method call would still prevent compilation.color().
Apple is not a valid type of Fruit.
Apple is a valid type of Fruit due to inheritance (Apple extends Fruit). This relationship allows an Apple object to be referenced by a Fruit variable.color method is not defined in the Fruit class.
color method is only defined in the Apple class and not in the Fruit class. Since the reference type of myFruit is Fruit, which does not have a color method, attempting to call myFruit.color() will result in a compilation error. This illustrates a key principle of polymorphism: the type of the reference (not the object) determines what methods can be called.12. The correct answers are B and D.
Explanation:
((Dog)anotherAnimal).bark();
anotherAnimal to Dog without checking its actual type first. Since anotherAnimal is an instance of Animal (not Dog), attempting this cast will compile, but it will cause a ClassCastException at runtime.if (anotherAnimal instanceof Dog) ((Dog)anotherAnimal).bark();
instanceof to check whether anotherAnimal is an instance of Dog before attempting the cast and calling bark(). In this case, since anotherAnimal is not an instance of Dog, the check prevents the cast and method call, avoiding a ClassCastException.((Cat)animal).meow();
animal to Cat and attempts to call meow(). Since animal is actually an instance of Dog, this cast will compile but will result in a ClassCastException at runtime.if (anotherAnimal instanceof Cat) ((Cat)anotherAnimal).meow();
anotherAnimal is an instance of Cat before casting it to Cat and calling meow().13. The correct answer is A.
Explanation:
"String with Java: Hello Java!" followed by "Integer greater than 10: 15".
instanceof operator for both String and Integer types. The pattern matching feature checks if input is an instance of String or Integer and binds it to a variable (s for String and i for Integer) within the scope of the if and else if blocks. The logical operator && is correctly used to further conditionally check properties of the variables (s.contains("Java") and i > 10). Thus, the method process prints output for inputs that are a String containing "Java" and an Integer greater than 10, respectively."String with Java: Hello Java!" because integers are not supported with pattern matching.
Integer. The code does support integers and performs additional checks using pattern matching correctly.instanceof cannot be combined with logical operators like &&.
instanceof can indeed be combined with logical operators like && for additional checks in the same conditional statement, as demonstrated in the code snippet.true.
true. The code correctly prints specific messages only for the inputs that match the given conditions.14. The correct answer is E.
Explanation:
setName, setPrice, and setStock methods public would enhance the class’s encapsulation.
Product class’s fields are private.
private fields is a fundamental aspect of encapsulation. It prevents external classes from directly accessing and modifying the object’s state, thus enforcing encapsulation.private setters that contain validation logic. This ensures that the object’s state is correctly managed and validated upon creation.Product class should have package-private getters to improve encapsulation.
Product class demonstrates proper encapsulation practices by making its fields private and controlling access to them through public getters and private setters. The setters include validation logic, ensuring that only valid states are assigned to the fields. This design pattern ensures that the internal state of Product instances is both protected and correctly managed.15. The correct answers are A and E.
Explanation:
SavingsAccount class cannot access the balance field directly due to its private access modifier in the Account class.
balance field to maintain encapsulation.getBalance method should be public to allow SavingsAccount to access the account balance.
getBalance public would increase its visibility unnecessarily. protected is sufficient for subclass access, and this change is not required for SavingsAccount to function correctly, making this statement incorrect.deposit method in the Account class should be marked as final to prevent overriding.
deposit as final would prevent it from being overridden in subclasses, which is not a requirement or suggestion indicated by the given code. The decision to make a method final should be based on the specific design needs rather than a general principle of encapsulation.interestRate field in the SavingsAccount class violates encapsulation principles by being private.
private access modifier for interestRate in SavingsAccount is an example of proper encapsulation. It restricts access to the field from outside the class, which is aligned with encapsulation principles, making this option incorrect.Account class correctly encapsulates the balance field, and SavingsAccount adheres to encapsulation by accessing balance through getBalance and deposit.
Account class uses private access for the balance field to encapsulate its state, providing protected and package-private methods (getBalance and deposit) for controlled access and modification. SavingsAccount respects this encapsulation by using these methods to interact with the balance field, demonstrating a proper understanding and application of encapsulation principles. This design allows SavingsAccount to leverage functionality provided by Account without breaking encapsulation, which is a key objective in object-oriented design.16. The correct answer is C.
Explanation:
Contact object is mutable because the Address class is not final.
Address class does not directly impact the immutability of the Contact object. The Contact class ensures its immutability by not providing setters and by making deep copies of mutable objects, such as Address, both in the constructor and the getter.Contact object is immutable, but only because it does not provide setters.
Contact are final and the defensive copying strategy.Contact object is immutable, and it properly prevents leakage of mutable internal state through defensive copying.
Contact class is immutable because it meets all criteria for immutability: the class is declared as final (preventing subclassing), all its fields are private and final, and it does not provide any setters. Also, it implements defensive copying for the mutable Address field to ensure that the internal state cannot be altered by external changes to Address objects passed in or returned. This prevents the leakage of its mutable internal state.Contact object is mutable because the Address object can be changed via the getAddress method.
Contact object’s immutability is maintained through defensive copying. The getAddress method returns a new Address instance each time it is called, ensuring that the original Address object’s state cannot be altered from outside the Contact object.Contact object is immutable but fails to prevent access to its mutable internal state.
Contact object does implement a strategy to prevent access to its mutable internal state: it uses defensive copying for the Address object in both the constructor and the getter method, which ensures that the internal state remains unchanged from outside modifications.Do you like what you read? Would you consider?
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