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1z1-830 Dumps PDF 2026 Program Your Preparation EXAM SUCCESS [Q44-Q63]

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1z1-830 Dumps PDF 2026 Program Your Preparation EXAM SUCCESS

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NEW QUESTION # 44
Which methods compile?

  • A. ```java public List<? extends IOException> getListExtends() { return new ArrayList<Exception>(); } csharp
  • B. ```java
    public List<? super IOException> getListSuper() {
    return new ArrayList<FileNotFoundException>();
    }
  • C. ```java public List<? super IOException> getListSuper() { return new ArrayList<Exception>(); } csharp
  • D. ```java
    public List<? extends IOException> getListExtends() {
    return new ArrayList<FileNotFoundException>();
    }

Answer: C,D

Explanation:
In Java generics, wildcards are used to relax the type constraints of generic types. The extends wildcard (<?
extends Type>) denotes an upper bounded wildcard, allowing any type that is a subclass of Type. Conversely, the super wildcard (<? super Type>) denotes a lower bounded wildcard, allowing any type that is a superclass of Type.
Option A:
java
public List<? super IOException> getListSuper() {
return new ArrayList<Exception>();
}
Here, List<? super IOException> represents a list that can hold IOException objects and objects of its supertypes. Since Exception is a superclass of IOException, ArrayList<Exception> is compatible with List<?
super IOException>. Therefore, this method compiles successfully.
Option B:
java
public List<? extends IOException> getListExtends() {
return new ArrayList<FileNotFoundException>();
}
In this case, List<? extends IOException> represents a list that can hold objects of IOException and its subclasses. Since FileNotFoundException is a subclass of IOException, ArrayList<FileNotFoundException> is compatible with List<? extends IOException>. Thus, this method compiles successfully.
Option C:
java
public List<? extends IOException> getListExtends() {
return new ArrayList<Exception>();
}
Here, List<? extends IOException> expects a list of IOException or its subclasses. However, Exception is a superclass of IOException, not a subclass. Therefore, ArrayList<Exception> is not compatible with List<?
extends IOException>, and this method will not compile.
Option D:
java
public List<? super IOException> getListSuper() {
return new ArrayList<FileNotFoundException>();
}
In this scenario, List<? super IOException> expects a list that can hold IOException objects and objects of its supertypes. Since FileNotFoundException is a subclass of IOException, ArrayList<FileNotFoundException> is not compatible with List<? super IOException>, and this method will not compile.
Therefore, the methods in options A and B compile successfully, while those in options C and D do not.


NEW QUESTION # 45
Given:
java
Stream<String> strings = Stream.of("United", "States");
BinaryOperator<String> operator = (s1, s2) -> s1.concat(s2.toUpperCase()); String result = strings.reduce("-", operator); System.out.println(result); What is the output of this code fragment?

  • A. -UnitedStates
  • B. -UnitedSTATES
  • C. -UNITEDSTATES
  • D. United-STATES
  • E. United-States
  • F. UnitedStates
  • G. UNITED-STATES

Answer: B

Explanation:
In this code, a Stream of String elements is created containing "United" and "States". A BinaryOperator<String> named operator is defined to concatenate the first string (s1) with the uppercase version of the second string (s2). The reduce method is then used with "-" as the identity value and operator as the accumulator.
The reduce method processes the elements of the stream as follows:
* Initial Identity Value: "-"
* First Iteration:
* Accumulator Operation: "-".concat("United".toUpperCase())
* Result: "-UNITED"
* Second Iteration:
* Accumulator Operation: "-UNITED".concat("States".toUpperCase())
* Result: "-UNITEDSTATES"
Therefore, the final result stored in result is "-UNITEDSTATES", and the output of theSystem.out.println (result); statement is -UNITEDSTATES.


NEW QUESTION # 46
Which of the following statements are correct?

  • A. You can use 'public' access modifier with all kinds of classes
  • B. None
  • C. You can use 'private' access modifier with all kinds of classes
  • D. You can use 'protected' access modifier with all kinds of classes
  • E. You can use 'final' modifier with all kinds of classes

Answer: B

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 # 47
Given:
java
Map<String, Integer> map = Map.of("b", 1, "a", 3, "c", 2);
TreeMap<String, Integer> treeMap = new TreeMap<>(map);
System.out.println(treeMap);
What is the output of the given code fragment?

  • A. {b=1, c=2, a=3}
  • B. {a=3, b=1, c=2}
  • C. {a=1, b=2, c=3}
  • D. {c=1, b=2, a=3}
  • E. {b=1, a=3, c=2}
  • F. {c=2, a=3, b=1}
  • G. Compilation fails

Answer: B

Explanation:
In this code, a Map named map is created using Map.of with the following key-value pairs:
* "b": 1
* "a": 3
* "c": 2
The Map.of method returns an immutable map containing these mappings.
Next, a TreeMap named treeMap is instantiated by passing the map to its constructor:
java
TreeMap<String, Integer> treeMap = new TreeMap<>(map);
The TreeMap constructor with a Map parameter creates a new tree map containing the same mappings as the given map, ordered according to the natural ordering of its keys. In Java, the natural ordering for String keys is lexicographical order.
Therefore, the TreeMap will store the entries in the following order:
* "a": 3
* "b": 1
* "c": 2
When System.out.println(treeMap); is executed, it outputs the TreeMap in its natural order, resulting in:
r
{a=3, b=1, c=2}
Thus, the correct answer is option F: {a=3, b=1, c=2}.


NEW QUESTION # 48
Which StringBuilder variable fails to compile?
java
public class StringBuilderInstantiations {
public static void main(String[] args) {
var stringBuilder1 = new StringBuilder();
var stringBuilder2 = new StringBuilder(10);
var stringBuilder3 = new StringBuilder("Java");
var stringBuilder4 = new StringBuilder(new char[]{'J', 'a', 'v', 'a'});
}
}

  • A. stringBuilder4
  • B. stringBuilder3
  • C. stringBuilder1
  • D. stringBuilder2
  • E. None of them

Answer: A

Explanation:
In the provided code, four StringBuilder instances are being created using different constructors:
* stringBuilder1: new StringBuilder()
* This constructor creates an empty StringBuilder with an initial capacity of 16 characters.
* stringBuilder2: new StringBuilder(10)
* This constructor creates an empty StringBuilder with a specified initial capacity of 10 characters.
* stringBuilder3: new StringBuilder("Java")
* This constructor creates a StringBuilder initialized to the contents of the specified string "Java".
* stringBuilder4: new StringBuilder(new char[]{'J', 'a', 'v', 'a'})
* This line attempts to create a StringBuilder using a char array. However, the StringBuilder class does not have a constructor that accepts a char array directly. The available constructors are:
* StringBuilder()
* StringBuilder(int capacity)
* StringBuilder(String str)
* StringBuilder(CharSequence seq)
Since a char array does not implement the CharSequence interface, and there is no constructor that directly accepts a char array, this line will cause a compilation error.
To initialize a StringBuilder with a char array, you can convert the char array to a String first:
java
var stringBuilder4 = new StringBuilder(new String(new char[]{'J', 'a', 'v', 'a'})); This approach utilizes the String constructor that accepts a char array, and then passes the resulting String to the StringBuilder constructor.


NEW QUESTION # 49
Given:
java
public class Test {
static int count;
synchronized Test() {
count++;
}
public static void main(String[] args) throws InterruptedException {
Runnable task = Test::new;
Thread t1 = new Thread(task);
Thread t2 = new Thread(task);
t1.start();
t2.start();
t1.join();
t2.join();
System.out.println(count);
}
}
What is the given program's output?

  • A. It's always 1
  • B. It's either 1 or 2
  • C. It's always 2
  • D. Compilation fails
  • E. It's either 0 or 1

Answer: D

Explanation:
In this code, the Test class has a static integer field count and a constructor that is declared with the synchronized modifier. In Java, the synchronized modifier can be applied to methods to control access to critical sections, but it cannot be applied directly to constructors. Attempting to declare a constructor as synchronized will result in a compilation error.
Compilation Error Details:
The Java Language Specification does not permit the use of the synchronized modifier on constructors.
Therefore, the compiler will produce an error indicating that the synchronized modifier is not allowed in this context.
Correct Usage:
If you need to synchronize the initialization of instances, you can use a synchronized block within the constructor:
java
public class Test {
static int count;
Test() {
synchronized (Test.class) {
count++;
}
}
public static void main(String[] args) throws InterruptedException {
Runnable task = Test::new;
Thread t1 = new Thread(task);
Thread t2 = new Thread(task);
t1.start();
t2.start();
t1.join();
t2.join();
System.out.println(count);
}
}
In this corrected version, the synchronized block within the constructor ensures that the increment operation on count is thread-safe.
Conclusion:
The original program will fail to compile due to the illegal use of the synchronized modifier on the constructor. Therefore, the correct answer is E: Compilation fails.


NEW QUESTION # 50
Which two of the following aren't the correct ways to create a Stream?

  • A. Stream stream = Stream.ofNullable("a");
  • B. Stream stream = Stream.empty();
  • C. Stream stream = new Stream();
  • D. Stream stream = Stream.of();
  • E. Stream stream = Stream.of("a");
  • F. Stream<String> stream = Stream.builder().add("a").build();
  • G. Stream stream = Stream.generate(() -> "a");

Answer: C,F

Explanation:
In Java, the Stream API provides several methods to create streams. However, not all approaches are valid.


NEW QUESTION # 51
Given:
java
var deque = new ArrayDeque<>();
deque.add(1);
deque.add(2);
deque.add(3);
deque.add(4);
deque.add(5);
System.out.print(deque.peek() + " ");
System.out.print(deque.poll() + " ");
System.out.print(deque.pop() + " ");
System.out.print(deque.element() + " ");
What is printed?

  • A. 1 1 2 3
  • B. 1 5 5 1
  • C. 1 1 1 1
  • D. 1 1 2 2
  • E. 5 5 2 3

Answer: A

Explanation:
* Understanding ArrayDeque Behavior
* ArrayDeque<E>is a double-ended queue (deque), working as aFIFO (queue) and LIFO (stack).
* Thedefault behaviorisqueue-like (FIFO)unless explicitly used as a stack.
* Step-by-Step Execution
java
var deque = new ArrayDeque<>();
deque.add(1);
deque.add(2);
deque.add(3);
deque.add(4);
deque.add(5);
* Deque after additions# [1, 2, 3, 4, 5]
* Operations Breakdown
* deque.peek()# Returns thehead(first element)without removal.
makefile
Output: 1
* deque.poll()# Removes and returns thehead.
go
Output: 1, Deque after poll # `[2, 3, 4, 5]`
* deque.pop()#Same as removeFirst(); removes and returns thehead.
perl
Output: 2, Deque after pop # `[3, 4, 5]`
* deque.element()# Returns thehead(same as peek(), but throws an exception if empty).
makefile
Output: 3
* Final Output
1 1 2 3
Thus, the correct answer is:1 1 2 3
References:
* Java SE 21 - ArrayDeque
* Java SE 21 - Queue Operations


NEW QUESTION # 52
Given:
java
System.out.print(Boolean.logicalAnd(1 == 1, 2 < 1));
System.out.print(Boolean.logicalOr(1 == 1, 2 < 1));
System.out.print(Boolean.logicalXor(1 == 1, 2 < 1));
What is printed?

  • A. truetruefalse
  • B. falsetruetrue
  • C. Compilation fails
  • D. truetruetrue
  • E. truefalsetrue

Answer: E

Explanation:
In this code, three static methods from the Boolean class are used: logicalAnd, logicalOr, and logicalXor.
Each method takes two boolean arguments and returns a boolean result based on the respective logical operation.
Evaluation of Each Statement:
* Boolean.logicalAnd(1 == 1, 2 < 1)
* Operands:
* 1 == 1 evaluates to true.
* 2 < 1 evaluates to false.
* Operation:
* Boolean.logicalAnd(true, false) performs a logical AND operation.
* The result is false because both operands must be true for the AND operation to return true.
* Output:
* System.out.print(false); prints false.
* Boolean.logicalOr(1 == 1, 2 < 1)
* Operands:
* 1 == 1 evaluates to true.
* 2 < 1 evaluates to false.
* Operation:
* Boolean.logicalOr(true, false) performs a logical OR operation.
* The result is true because at least one operand is true.
* Output:
* System.out.print(true); prints true.
* Boolean.logicalXor(1 == 1, 2 < 1)
* Operands:
* 1 == 1 evaluates to true.
* 2 < 1 evaluates to false.
* Operation:
* Boolean.logicalXor(true, false) performs a logical XOR (exclusive OR) operation.
* The result is true because exactly one operand is true.
* Output:
* System.out.print(true); prints true.
Combined Output:
Combining the outputs from each statement, the final printed result is:
nginx
falsetruetrue


NEW QUESTION # 53
A module com.eiffeltower.shop with the related sources in the src directory.
That module requires com.eiffeltower.membership, available in a JAR located in the lib directory.
What is the command to compile the module com.eiffeltower.shop?

  • A. bash
    CopyEdit
    javac -source src -p lib/com.eiffel.membership.jar -d out -m com.eiffeltower.shop
  • B. css
    CopyEdit
    javac -path src -p lib/com.eiffel.membership.jar -d out -m com.eiffeltower.shop
  • C. css
    CopyEdit
    javac --module-source-path src -p lib/com.eiffel.membership.jar -s out -m com.eiffeltower.shop
  • D. css
    CopyEdit
    javac --module-source-path src -p lib/com.eiffel.membership.jar -d out -m com.eiffeltower.shop

Answer: D

Explanation:
Comprehensive and Detailed In-Depth Explanation:
Understanding Java Module Compilation (javac)
Java modules are compiled using the javac command with specific options to specify:
* Where the source files are located (--module-source-path)
* Where required dependencies (external modules) are located (-p / --module-path)
* Where the compiled output should be placed (-d)
Breaking Down the Correct Compilation Command
css
CopyEdit
javac --module-source-path src -p lib/com.eiffel.membership.jar -d out -m com.eiffeltower.shop
* --module-source-path src # Specifies the directory where module sources are located.
* -p lib/com.eiffel.membership.jar # Specifies the module path (JAR dependency in lib).
* -d out # Specifies the output directory for compiled .class files.
* -m com.eiffeltower.shop # Specifies the module to compile (com.eiffeltower.shop).


NEW QUESTION # 54
Given:
java
public static void main(String[] args) {
try {
throw new IOException();
} catch (IOException e) {
throw new RuntimeException();
} finally {
throw new ArithmeticException();
}
}
What is the output?

  • A. ArithmeticException
  • B. IOException
  • C. RuntimeException
  • D. Compilation fails

Answer: A

Explanation:
In this code, the try block throws an IOException. The catch block catches this exception and throws a new RuntimeException. Regardless of exceptions thrown in the try or catch blocks, the finally block is always executed. In this case, the finally block throws an ArithmeticException.
When an exception is thrown in a finally block, it overrides any previous exceptions that were thrown in the try or catch blocks. Therefore, the ArithmeticException thrown in the finally block is the exception that propagates out of the method. As a result, the program terminates with an ArithmeticException.


NEW QUESTION # 55
Given:
java
DoubleStream doubleStream = DoubleStream.of(3.3, 4, 5.25, 6.66);
Predicate<Double> doublePredicate = d -> d < 5;
System.out.println(doubleStream.anyMatch(doublePredicate));
What is printed?

  • A. false
  • B. true
  • C. Compilation fails
  • D. 3.3
  • E. An exception is thrown at runtime

Answer: C

Explanation:
In this code, there is a type mismatch between the DoubleStream and the Predicate<Double>.
* DoubleStream: A sequence of primitive double values.
* Predicate<Double>: A functional interface that operates on objects of type Double (the wrapper class), not on primitive double values.
The DoubleStream class provides a method anyMatch(DoublePredicate predicate), where DoublePredicate is a functional interface that operates on primitive double values. However, in the code, a Predicate<Double> is used instead of a DoublePredicate. This mismatch leads to a compilation error because anyMatch cannot accept a Predicate<Double> when working with a DoubleStream.
To correct this, the predicate should be defined as a DoublePredicate to match the primitive double type:
java
DoubleStream doubleStream = DoubleStream.of(3.3, 4, 5.25, 6.66);
DoublePredicate doublePredicate = d -> d < 5;
System.out.println(doubleStream.anyMatch(doublePredicate));
With this correction, the code will compile and print true because there are elements in the stream (e.g., 3.3 and 4.0) that are less than 5.


NEW QUESTION # 56
Given:
java
Deque<Integer> deque = new ArrayDeque<>();
deque.offer(1);
deque.offer(2);
var i1 = deque.peek();
var i2 = deque.poll();
var i3 = deque.peek();
System.out.println(i1 + " " + i2 + " " + i3);
What is the output of the given code fragment?

  • A. 1 2 2
  • B. 2 2 2
  • C. 1 1 2
  • D. 2 1 1
  • E. 1 1 1
  • F. An exception is thrown.
  • G. 2 2 1
  • H. 2 1 2
  • I. 1 2 1

Answer: A

Explanation:
In this code, an ArrayDeque named deque is created, and the integers 1 and 2 are added to it using the offer method. The offer method inserts the specified element at the end of the deque.
* State of deque after offers:[1, 2]
The peek method retrieves, but does not remove, the head of the deque, returning 1. Therefore, i1 is assigned the value 1.
* State of deque after peek:[1, 2]
* Value of i1:1
The poll method retrieves and removes the head of the deque, returning 1. Therefore, i2 is assigned the value
1.
* State of deque after poll:[2]
* Value of i2:1
Another peek operation retrieves the current head of the deque, which is now 2, without removing it.
Therefore, i3 is assigned the value 2.
* State of deque after second peek:[2]
* Value of i3:2
The System.out.println statement then outputs the values of i1, i2, and i3, resulting in 1 1 2.


NEW QUESTION # 57
What do the following print?
java
import java.time.Duration;
public class DividedDuration {
public static void main(String[] args) {
var day = Duration.ofDays(2);
System.out.print(day.dividedBy(8));
}
}

  • A. PT6H
  • B. PT0D
  • C. PT0H
  • D. It throws an exception
  • E. Compilation fails

Answer: A

Explanation:
In this code, a Duration object day is created representing a duration of 2 days using the Duration.ofDays(2) method. The dividedBy(long divisor) method is then called on this Duration object with the argument 8.
The dividedBy(long divisor) method returns a copy of the original Duration divided by the specified value. In this case, dividing 2 days by 8 results in a duration of 0.25 days. In the ISO-8601 duration format used by Java's Duration class, this is represented as PT6H, which stands for a period of 6 hours.
Therefore, the output of the System.out.print statement is PT6H.


NEW QUESTION # 58
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?

  • A. Peugeot 807
  • B. Compilation fails.
  • C. An exception is thrown at runtime.
  • D. Peugeot

Answer: B

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 # 59
Given:
java
record WithInstanceField(String foo, int bar) {
double fuz;
}
record WithStaticField(String foo, int bar) {
static double wiz;
}
record ExtendingClass(String foo) extends Exception {}
record ImplementingInterface(String foo) implements Cloneable {}
Which records compile? (Select 2)

  • A. WithStaticField
  • B. ExtendingClass
  • C. ImplementingInterface
  • D. WithInstanceField

Answer: A,C

Explanation:
In Java, records are a special kind of class designed to act as transparent carriers for immutabledata. They automatically provide implementations for equals(), hashCode(), and toString(), and their fields are final and private by default.
* Option A: ExtendingClass
* Analysis: Records in Java implicitly extend java.lang.Record and cannot extend any other class because Java does not support multiple inheritance. Attempting to extend another class, such as Exception, will result in a compilation error.
* Conclusion: Does not compile.
* Option B: WithInstanceField
* Analysis: Records do not allow the declaration of instance fields outside of their components.
The declaration of double fuz; is not permitted and will cause a compilation error.
* Conclusion: Does not compile.
* Option C: ImplementingInterface
* Analysis: Records can implement interfaces. In this case, ImplementingInterface implements Cloneable, which is valid.
* Conclusion: Compiles successfully.


NEW QUESTION # 60
Given:
java
CopyOnWriteArrayList<String> list = new CopyOnWriteArrayList<>();
list.add("A");
list.add("B");
list.add("C");
// Writing in one thread
new Thread(() -> {
list.add("D");
System.out.println("Element added: D");
}).start();
// Reading in another thread
new Thread(() -> {
for (String element : list) {
System.out.println("Read element: " + element);
}
}).start();
What is printed?

  • A. It throws an exception.
  • B. It prints all elements, including changes made during iteration.
  • C. It prints all elements, but changes made during iteration may not be visible.
  • D. Compilation fails.

Answer: C

Explanation:
* Understanding CopyOnWriteArrayList
* CopyOnWriteArrayList is a thread-safe variant of ArrayList whereall mutative operations (add, set, remove, etc.) create a new copy of the underlying array.
* This meansiterations will not reflect modifications made after the iterator was created.
* Instead of modifying the existing array, a new copy is created for modifications, ensuring that readers always see a consistent snapshot.
* Thread Execution Behavior
* Thread 1 (Writer Thread)adds "D" to the list.
* Thread 2 (Reader Thread)iterates over the list.
* The reader thread gets a snapshot of the listbefore"D" is added.
* The output may look like:
mathematica
Read element: A
Read element: B
Read element: C
Element added: D
* "D" may not appear in the output of the reader threadbecause the iteration occurs on a snapshot before the modification.
* Why doesn't it print all elements including changes?
* Since CopyOnWriteArrayList doesnot allow changes to be visible during iteration, the reader threadwill not see "D"if it started iterating before "D" was added.
Thus, the correct answer is:"It prints all elements, but changes made during iteration may not be visible." References:
* Java SE 21 - CopyOnWriteArrayList


NEW QUESTION # 61
Given:
java
var counter = 0;
do {
System.out.print(counter + " ");
} while (++counter < 3);
What is printed?

  • A. 1 2 3
  • B. 1 2 3 4
  • C. 0 1 2 3
  • D. 0 1 2
  • E. Compilation fails.
  • F. An exception is thrown.

Answer: D

Explanation:
* Understanding do-while Execution
* A do-while loopexecutes at least oncebefore checking the condition.
* ++counter < 3 increments counterbeforeevaluating the condition.
* Step-by-Step Execution
* Iteration 1:counter = 0, print "0", then ++counter becomes 1, condition 1 < 3 istrue.
* Iteration 2:counter = 1, print "1", then ++counter becomes 2, condition 2 < 3 istrue.
* Iteration 3:counter = 2, print "2", then ++counter becomes 3, condition 3 < 3 isfalse, so loop exits.
* Final Output
0 1 2
Thus, the correct answer is:0 1 2
References:
* Java SE 21 - Control Flow Statements
* Java SE 21 - do-while Loop


NEW QUESTION # 62
What do the following print?
java
public class DefaultAndStaticMethods {
public static void main(String[] args) {
WithStaticMethod.print();
}
}
interface WithDefaultMethod {
default void print() {
System.out.print("default");
}
}
interface WithStaticMethod extends WithDefaultMethod {
static void print() {
System.out.print("static");
}
}

  • A. default
  • B. static
  • C. nothing
  • D. Compilation fails

Answer: B

Explanation:
In this code, we have two interfaces and a class with a main method:
* WithDefaultMethod Interface:
* Declares a default method print() that outputs "default".
* WithStaticMethod Interface:
* Extends WithDefaultMethod.
* Declares a static method print() that outputs "static".
* DefaultAndStaticMethods Class:
* Contains the main method, which calls WithStaticMethod.print().
Key Points:
* Static Methods in Interfaces:
* Static methods in interfaces are not inherited by implementing or extending classes or interfaces.
They belong solely to the interface in which they are declared.
* Default Methods in Interfaces:
* Default methods can be inherited by implementing classes, but they cannot be overridden by static methods in subinterfaces.
Execution Flow:
* The main method calls WithStaticMethod.print().
* This invokes the static method print() defined in the WithStaticMethod interface, which outputs "static".
Therefore, the program compiles successfully and prints static.


NEW QUESTION # 63
......

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