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Oracle 1z0-830 Exam Syllabus Topics:

SectionWeightObjectives
Topic 1: Working with Arrays and Collections12%- Declare, instantiate, initialize, use arrays and multidimensional arrays
- Collections Framework: List, Set, Map, Deque, Queue, sorting, searching
Topic 2: Functional Programming and Streams15%- Stream API: create, intermediate/terminal operations, parallel streams, grouping, partitioning
- Lambda expressions, functional interfaces, method references
- Optional class, primitive streams
Topic 3: Advanced Features and Annotations3%- Generics, type parameters, wildcards, type erasure
- Annotations, built-in annotations, custom annotations
Topic 4: Handling Exceptions8%- Exception hierarchy, try-catch-finally, multi-catch, try-with-resources
- Create and use custom exceptions, throw, throws
Topic 5: Modules and Packaging5%- Create and use JAR files, modular and non-modular builds
- Module system: module-info.java, exports, requires, provides, uses
Topic 6: Java I/O and Localization5%- Resource bundles, locale, formatting messages, numbers, dates
- File I/O, NIO.2, streams, readers/writers, serialization
Topic 7: Handling Date, Time, Text, Numeric and Boolean Values12%- Use Date-Time API: LocalDate, LocalTime, LocalDateTime, Period, Duration, Instant, ZonedDateTime
- Manipulate text, text blocks, String, StringBuilder and StringBuffer
- Use primitives and wrapper classes, evaluate expressions and apply type conversions
Topic 8: Controlling Program Flow10%- Decision constructs: if-else, switch expressions and statements, pattern matching
- Loops: for, enhanced for, while, do-while, break, continue, return
Topic 9: Concurrency and Multithreading10%- Synchronization, locks, concurrent collections, thread safety
- Thread lifecycle, Runnable, Callable, ExecutorService, virtual threads
Topic 10: Using Object-Oriented Concepts20%- Enums, nested classes, local variable type inference
- Overloading, overriding, Object class methods, immutable objects
- Classes, records, objects, constructors, initializers, methods, fields, encapsulation
- Inheritance, abstract classes, sealed classes, interfaces, polymorphism

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Oracle Java SE 21 Developer Professional Sample Questions (Q50-Q55):

NEW QUESTION # 50
Which of the following statements are correct?

Answer: A

Explanation:
1. private Access Modifier
* The private access modifiercan only be used for inner classes(nested classes).
* Top-level classes cannot be private.
* Example ofinvaliduse:
java
private class MyClass {} // Compilation error
* Example ofvaliduse (for inner class):
java
class Outer {
private class Inner {}
}
2. protected Access Modifier
* Top-level classes cannot be protected.
* protectedonly applies to members (fields, methods, and constructors).
* Example ofinvaliduse:
java
protected class MyClass {} // Compilation error
* Example ofvaliduse (for methods/fields):
java
class Parent {
protected void display() {}
}
3. public Access Modifier
* Atop-level class can be public, butonly one public class per file is allowed.
* Example ofvaliduse:
java
public class MyClass {}
* Example ofinvaliduse:
java
public class A {}
public class B {} // Compilation error: Only one public class per file
4. final Modifier
* finalcan be used with classes, but not all kinds of classes.
* Interfaces cannot be final, because they are meant to be implemented.
* Example ofinvaliduse:
java
final interface MyInterface {} // Compilation error
Thus,none of the statements are fully correct, making the correct answer:None References:
* Java SE 21 - Access Modifiers
* Java SE 21 - Class Modifiers


NEW QUESTION # 51
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: C

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 # 52
Given:
java
String textBlock = """
j \
a \t
v \s
a \
""";
System.out.println(textBlock.length());
What is the output?

Answer: C

Explanation:
In this code, a text block is defined using the """ syntax introduced in Java 13. Text blocks allow for multiline string literals, preserving the format as written in the code.
Text Block Analysis:
The text block is defined as:
java
String textBlock = """
j \
a \t
contentReference[oaicite:0]{index=0}


NEW QUESTION # 53
How would you create a ConcurrentHashMap configured to allow a maximum of 10 concurrent writer threads and an initial capacity of 42?
Which of the following options meets this requirement?

Answer: E

Explanation:
In Java, the ConcurrentHashMap class provides several constructors that allow for the customization of its initial capacity, load factor, and concurrency level. To configure a ConcurrentHashMap with an initial capacity of 42 and a concurrency level of 10, you can use the following constructor:
java
public ConcurrentHashMap(int initialCapacity, float loadFactor, int concurrencyLevel) Parameters:
* initialCapacity: The initial capacity of the hash table. This is the number of buckets that the hash table will have when it is created. In this case, it is set to 42.
* loadFactor: A measure of how full the hash table is allowed to get before it is resized. The default value is 0.75, but in this case, it is set to 0.88.
* concurrencyLevel: The estimated number of concurrently updating threads. This is used as a hint for internal sizing. In this case, it is set to 10.
Therefore, to create a ConcurrentHashMap with an initial capacity of 42, a load factor of 0.88, and a concurrency level of 10, you can use the following code:
java
var concurrentHashMap = new ConcurrentHashMap<>(42, 0.88f, 10);
Option Evaluations:
* A. var concurrentHashMap = new ConcurrentHashMap(42);: This constructor sets the initial capacity to 42 but uses the default load factor (0.75) and concurrency level (16). It does not meet the requirement of setting the concurrency level to 10.
* B. None of the suggestions.: This is incorrect because option E provides the correct configuration.
* C. var concurrentHashMap = new ConcurrentHashMap();: This uses the default constructor, which sets the initial capacity to 16, the load factor to 0.75, and the concurrency level to 16. It does not meet the specified requirements.
* D. var concurrentHashMap = new ConcurrentHashMap(42, 10);: This constructor sets the initial capacity to 42 and the load factor to 10, which is incorrect because the load factor should be a float value between 0 and 1.
* E. var concurrentHashMap = new ConcurrentHashMap(42, 0.88f, 10);: This correctly sets the initial capacity to 42, the load factor to 0.88, and the concurrency level to 10, meeting all the specified requirements.
Therefore, the correct answer is option E.


NEW QUESTION # 54
Given:
java
String colors = "red\n" +
"green\n" +
"blue\n";
Which text block can replace the above code?

Answer: A

Explanation:
* Understanding Multi-line Strings in Java (""" Text Blocks)
* Java 13 introducedtext blocks ("""), allowing multi-line stringswithout needing explicit \n for new lines.
* In a text block,each line is preserved as it appears in the source code.
* Analyzing the Options
* Option A: \ (Backslash Continuation)
* The backslash (\) at the end of a lineprevents a new line from being added, meaning:
nginx
red green blue
* Incorrect.
* Option B: \s (Whitespace Escape)
* \s represents asingle space,not a new line.
* The output would be:
nginx
red green blue
* Incorrect.
* Option C: \t (Tab Escape)
* \t inserts atab, not a new line.
* The output would be:
nginx
red green blue
* Incorrect.
* Option D: Correct Text Block
java
String colors = """
red
green
blue
""";
* Thispreserves the new lines, producing:
nginx
red
green
blue
* Correct.
Thus, the correct answer is:"String colors = """ red green blue """."
References:
* Java SE 21 - Text Blocks
* Java SE 21 - String Formatting


NEW QUESTION # 55
......

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