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| Section | Objectives |
|---|---|
| Java Concurrency | - Use virtual threads - Create and manage threads - Work with concurrent collections and executors |
| Annotations and JDBC | - Use built-in and custom annotations - Execute SQL operations and process results - Connect to databases using JDBC |
| Exception Handling | - Handle checked and unchecked exceptions - Create custom exceptions - Use try-with-resources |
| Object-Oriented Programming | - Implement encapsulation and abstraction - Create and use classes, interfaces, enums, and records - Apply inheritance and polymorphism |
| Modules and Packaging | - Package and deploy applications - Create and use Java modules - Manage dependencies and exports |
| Handling Date, Time, Text, Numeric and Boolean Values | - Use String, StringBuilder, and related APIs - Work with primitive wrappers and number formatting - Use date, time, duration, period, and localization APIs |
| Java I/O and NIO | - Process file system resources - Use Path, Files, and related APIs - Read and write files |
| Controlling Program Flow | - Work with records and sealed classes - Apply decision statements and loops - Use switch expressions and pattern matching |
| Functional Programming | - Process data using Stream API - Use lambda expressions and method references - Work with functional interfaces |
| Collections and Generics | - Apply generics and bounded types - Use List, Set, Map, Queue, and Deque implementations - Use sequenced collections and maps |
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NEW QUESTION # 42
Given:
java
var _ = 3;
var $ = 7;
System.out.println(_ + $);
What is printed?
Answer: B
Explanation:
* The var keyword and identifier rules:
* The var keyword is used for local variable type inference introduced inJava 10.
* However,Java does not allow _ (underscore) as an identifiersinceJava 9.
* If we try to use _ as a variable name, the compiler will throw an error:
pgsql
error: as of release 9, '_' is a keyword, and may not be used as an identifier
* The $ symbol as an identifier:
* The $ characteris a valid identifierin Java.
* However, since _ is not allowed, the codefails to compile before even reaching $.
Thus,the correct answer is "Compilation fails."
References:
* Java SE 21 - var Local Variable Type Inference
* Java SE 9 - Restrictions on _ Identifier
NEW QUESTION # 43
Given:
java
public class ExceptionPropagation {
public static void main(String[] args) {
try {
thrower();
System.out.print("Dom Perignon, ");
} catch (Exception e) {
System.out.print("Chablis, ");
} finally {
System.out.print("Saint-Emilion");
}
}
static int thrower() {
try {
int i = 0;
return i / i;
} catch (NumberFormatException e) {
System.out.print("Rose");
return -1;
} finally {
System.out.print("Beaujolais Nouveau, ");
}
}
}
What is printed?
Answer: D
Explanation:
* Analyzing the thrower() Method Execution
java
int i = 0;
return i / i;
* i / i evaluates to 0 / 0, whichthrows ArithmeticException (/ by zero).
* Since catch (NumberFormatException e) doesnot matchArithmeticException, it is skipped.
* The finally block always executes, printing:
nginx
Beaujolais Nouveau,
* The exceptionpropagates backto main().
* Handling the Exception in main()
java
try {
thrower();
System.out.print("Dom Perignon, ");
} catch (Exception e) {
System.out.print("Chablis, ");
} finally {
System.out.print("Saint-Emilion");
}
* Since thrower() throws ArithmeticException, it is caught by catch (Exception e).
* "Chablis, "is printed.
* Thefinally block always executes, printing "Saint-Emilion".
* Final Output
nginx
Beaujolais Nouveau, Chablis, Saint-Emilion
Thus, the correct answer is:Beaujolais Nouveau, Chablis, Saint-Emilion
References:
* Java SE 21 - Exception Handling
* Java SE 21 - finally Block Execution
NEW QUESTION # 44
Given:
java
public class SpecialAddition extends Addition implements Special {
public static void main(String[] args) {
System.out.println(new SpecialAddition().add());
}
int add() {
return --foo + bar--;
}
}
class Addition {
int foo = 1;
}
interface Special {
int bar = 1;
}
What is printed?
Answer: B
Explanation:
1. Why does the compilation fail?
* The interface Special contains bar as int bar = 1;.
* In Java, all interface fields are implicitly public, static, and final.
* This means that bar is a constant (final variable).
* The method add() contains bar--, which attempts to modify bar.
* Since bar is final, it cannot be modified, causing acompilation error.
2. Correcting the Code
To make the code compile, bar must not be final. One way to fix this:
java
class SpecialImpl implements Special {
int bar = 1;
}
Or modify the add() method:
java
int add() {
return --foo + bar; // No modification of bar
}
Thus, the correct answer is:Compilation fails.
References:
* Java SE 21 - Interfaces
* Java SE 21 - Final Variables
NEW QUESTION # 45
Given:
java
package vehicule.parent;
public class Car {
protected String brand = "Peugeot";
}
and
java
package vehicule.child;
import vehicule.parent.Car;
public class MiniVan extends Car {
public static void main(String[] args) {
Car car = new Car();
car.brand = "Peugeot 807";
System.out.println(car.brand);
}
}
What is printed?
Answer: A
Explanation:
In Java,protected memberscan only be accessedwithin the same packageor bysubclasses, but there is a key restriction:
* A protected member of a superclass is only accessible through inheritance in a subclass but not through an instance of the superclass that is declared outside the package.
Why does compilation fail?
In the MiniVan class, the following line causes acompilation error:
java
Car car = new Car();
car.brand = "Peugeot 807";
* The brand field isprotectedin Car, which means it isnot accessible via an instance of Car outside the vehicule.parent package.
* Even though MiniVan extends Car, itcannotaccess brand using a Car instance (car.brand) because car is declared as an instance of Car, not MiniVan.
* The correct way to access brand inside MiniVan is through inheritance (this.brand or super.brand).
Corrected Code
If we change the MiniVan class like this, it will compile and run successfully:
java
package vehicule.child;
import vehicule.parent.Car;
public class MiniVan extends Car {
public static void main(String[] args) {
MiniVan minivan = new MiniVan(); // Access via inheritance
minivan.brand = "Peugeot 807";
System.out.println(minivan.brand);
}
}
This would output:
nginx
Peugeot 807
Key Rule from Oracle Java Documentation
* Protected membersof a class are accessible withinthe same packageand tosubclasses, butonly through inheritance, not through a superclass instance declared outside the package.
References:
* Java SE 21 & JDK 21 - Controlling Access to Members of a Class
* Java SE 21 & JDK 21 - Inheritance Rules
NEW QUESTION # 46
Given:
java
interface SmartPhone {
boolean ring();
}
class Iphone15 implements SmartPhone {
boolean isRinging;
boolean ring() {
isRinging = !isRinging;
return isRinging;
}
}
Choose the right statement.
Answer: B
Explanation:
In this code, the SmartPhone interface declares a method ring() with a boolean return type. The Iphone15 class implements the SmartPhone interface and provides an implementation for the ring() method.
However, in the Iphone15 class, the ring() method is declared without the public access modifier. In Java, when a class implements an interface, it must provide implementations for all the interface's methods with the same or a more accessible access level. Since interface methods are implicitly public, the implementing methods in the class must also be public. Failing to do so results in a compilation error.
Therefore, the Iphone15 class does not compile because the ring() method is not declared as public.
NEW QUESTION # 47
......
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