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Oracle Java SE 21 Developer Professional Sample Questions (Q65-Q70):

NEW QUESTION # 65
Given:
java
Integer frenchRevolution = 1789;
Object o1 = new String("1789");
Object o2 = frenchRevolution;
frenchRevolution = null;
Object o3 = o2.toString();
System.out.println(o1.equals(o3));
What is printed?

A. Compilation fails.
B. A NullPointerException is thrown.
C. A ClassCastException is thrown.
D. true
E. false

Answer: D

Explanation:
* Understanding Variable Assignments
java
Integer frenchRevolution = 1789;
Object o1 = new String("1789");
Object o2 = frenchRevolution;
frenchRevolution = null;
* frenchRevolution is an Integer with value1789.
* o1 is aString with value "1789".
* o2 storesa reference to frenchRevolution, which is an Integer (1789).
* frenchRevolution = null;only nullifies the reference, but o2 still holds the Integer 1789.
* Calling toString() on o2
java
Object o3 = o2.toString();
* o2 refers to an Integer (1789).
* Integer.toString() returns theString representation "1789".
* o3 is assigned "1789" (String).
* Evaluating o1.equals(o3)
java
System.out.println(o1.equals(o3));
* o1.equals(o3) isequivalent to:
java
"1789".equals("1789")
* Since both areequal strings, the output is:
arduino
true
Thus, the correct answer is:true
References:
* Java SE 21 - Integer.toString()
* Java SE 21 - String.equals()

NEW QUESTION # 66
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. {c=2, a=3, b=1}
B. Compilation fails
C. {b=1, a=3, c=2}
D. {a=3, b=1, c=2}
E. {b=1, c=2, a=3}
F. {c=1, b=2, a=3}
G. {a=1, b=2, c=3}

Answer: D

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 # 67
Given:
java
ExecutorService service = Executors.newFixedThreadPool(2);
Runnable task = () -> System.out.println("Task is complete");
service.submit(task);
service.shutdown();
service.submit(task);
What happens when executing the given code fragment?

A. It prints "Task is complete" twice and throws an exception.
B. It exits normally without printing anything to the console.
C. It prints "Task is complete" twice, then exits normally.
D. It prints "Task is complete" once, then exits normally.
E. It prints "Task is complete" once and throws an exception.

Answer: E

Explanation:
In this code, an ExecutorService is created with a fixed thread pool of size 2 using Executors.
newFixedThreadPool(2). A Runnable task is defined to print "Task is complete" to the console.
The sequence of operations is as follows:
* service.submit(task);
This submits the task to the executor service for execution. Since the thread pool has a size of 2 and no other tasks are running, this task will be executed promptly, printing "Task is complete" to the console.
* service.shutdown();
This initiates an orderly shutdown of the executor service. In this state, the service stops accepting new tasks

NEW QUESTION # 68
Given:
java
DoubleSummaryStatistics stats1 = new DoubleSummaryStatistics();
stats1.accept(4.5);
stats1.accept(6.0);
DoubleSummaryStatistics stats2 = new DoubleSummaryStatistics();
stats2.accept(3.0);
stats2.accept(8.5);
stats1.combine(stats2);
System.out.println("Sum: " + stats1.getSum() + ", Max: " + stats1.getMax() + ", Avg: " + stats1.getAverage()); What is printed?

A. Compilation fails.
B. Sum: 22.0, Max: 8.5, Avg: 5.0
C. An exception is thrown at runtime.
D. Sum: 22.0, Max: 8.5, Avg: 5.5

Answer: D

Explanation:
The DoubleSummaryStatistics class in Java is part of the java.util package and is used to collect and summarize statistics for a stream of double values. Let's analyze how the methods work:
* Initialization and Data Insertion
* stats1.accept(4.5); # Adds 4.5 to stats1.
* stats1.accept(6.0); # Adds 6.0 to stats1.
* stats2.accept(3.0); # Adds 3.0 to stats2.
* stats2.accept(8.5); # Adds 8.5 to stats2.
* Combining stats1 and stats2
* stats1.combine(stats2); merges stats2 into stats1, resulting in one statistics summary containing all values {4.5, 6.0, 3.0, 8.5}.
* Calculating Output Values
* Sum= 4.5 + 6.0 + 3.0 + 8.5 = 22.0
* Max= 8.5
* Average= (22.0) / 4 = 5.5
Thus, the output is:
yaml
Sum: 22.0, Max: 8.5, Avg: 5.5
References:
* Java SE 21 & JDK 21 - DoubleSummaryStatistics
* Java SE 21 - Streams and Statistical Operations

NEW QUESTION # 69
Given:
java
var array1 = new String[]{ "foo", "bar", "buz" };
var array2[] = { "foo", "bar", "buz" };
var array3 = new String[3] { "foo", "bar", "buz" };
var array4 = { "foo", "bar", "buz" };
String array5[] = new String[]{ "foo", "bar", "buz" };
Which arrays compile? (Select 2)

A. array5
B. array4
C. array3
D. array2
E. array1

Answer: A,E

Explanation:
In Java, array initialization can be performed in several ways, but certain syntaxes are invalid and will cause compilation errors. Let's analyze each declaration:
* var array1 = new String[]{ "foo", "bar", "buz" };
This is a valid declaration. The var keyword allows the compiler to infer the type from the initializer. Here, new String[]{ "foo", "bar", "buz" } creates an anonymous array of String with three elements. The compiler infers array1 as String[]. This syntax is correct and compiles successfully.
* var array2[] = { "foo", "bar", "buz" };
This declaration is invalid. While var can be used for type inference, appending [] after var is not allowed.
The correct syntax would be either String[] array2 = { "foo", "bar", "buz" }; or var array2 = new String[]{
"foo", "bar", "buz" };. Therefore, this line will cause a compilation error.
* var array3 = new String[3] { "foo", "bar", "buz" };
This declaration is invalid. In Java, when specifying the size of the array (new String[3]), you cannot simultaneously provide an initializer. The correct approach is either to provide the size without an initializer (new String[3]) or to provide the initializer without specifying the size (new String[]{ "foo", "bar", "buz" }).
Therefore, this line will cause a compilation error.
* var array4 = { "foo", "bar", "buz" };
This declaration is invalid. The array initializer { "foo", "bar", "buz" } can only be used in an array declaration when the type is explicitly provided. Since var relies on type inference and there's no explicit type provided here, this will cause a compilation error. The correct syntax would be String[] array4 = { "foo",
"bar", "buz" };.
* String array5[] = new String[]{ "foo", "bar", "buz" };
This is a valid declaration. Here, String array5[] declares array5 as an array of String. The initializer new String[]{ "foo", "bar", "buz" } creates an array with three elements. This syntax is correct and compiles successfully.
Therefore, the declarations that compile successfully are array1 and array5.
References:
* Java SE 21 & JDK 21 - Local Variable Type Inference
* Java SE 21 & JDK 21 - Arrays

NEW QUESTION # 70
......

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