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Oracle Java SE 21 Developer Professional Sample Questions (Q59-Q64):
NEW QUESTION # 59
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?
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 # 60
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: E
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 # 61
Given:
java
public class BoomBoom implements AutoCloseable {
public static void main(String[] args) {
try (BoomBoom boomBoom = new BoomBoom()) {
System.out.print("bim ");
throw new Exception();
} catch (Exception e) {
System.out.print("boom ");
}
}
@Override
public void close() throws Exception {
System.out.print("bam ");
throw new RuntimeException();
}
}
What is printed?
Answer: D
Explanation:
* Understanding Try-With-Resources (AutoCloseable)
* BoomBoom implements AutoCloseable, meaning its close() method isautomatically calledat the end of the try block.
* Step-by-Step Execution
* Step 1: Enter Try Block
java
try (BoomBoom boomBoom = new BoomBoom()) {
System.out.print("bim ");
throw new Exception();
}
* "bim " is printed.
* Anexception (Exception) is thrown, butbefore it is handled, the close() method is executed.
* Step 2: close() is Called
java
@Override
public void close() throws Exception {
System.out.print("bam ");
throw new RuntimeException();
}
* "bam " is printed.
* A new RuntimeException is thrown, but it doesnot override the existing Exception yet.
* Step 3: Exception Handling
java
} catch (Exception e) {
System.out.print("boom ");
}
* The catch (Exception e)catches the original Exception from the try block.
* "boom " is printed.
* Final Output
nginx
bim bam boom
* Theoriginal Exception is caught, not the RuntimeException from close().
* TheRuntimeException from close() is ignoredbecause thecatch block is already handling Exception.
Thus, the correct answer is:bim bam boom
References:
* Java SE 21 - Try-With-Resources
* Java SE 21 - AutoCloseable Interface
NEW QUESTION # 62
Given:
java
Optional<String> optionalName = Optional.ofNullable(null);
String bread = optionalName.orElse("Baguette");
System.out.print("bread:" + bread);
String dish = optionalName.orElseGet(() -> "Frog legs");
System.out.print(", dish:" + dish);
try {
String cheese = optionalName.orElseThrow(() -> new Exception());
System.out.println(", cheese:" + cheese);
} catch (Exception exc) {
System.out.println(", no cheese.");
}
What is printed?
Answer: A
Explanation:
Understanding Optional.ofNullable(null)
* Optional.ofNullable(null); creates an empty Optional (i.e., it contains no value).
* Optional.of(null); would throw a NullPointerException, but ofNullable(null); safely creates an empty Optional.
Execution of orElse, orElseGet, and orElseThrow
* orElse("Baguette")
* Since optionalName is empty, "Baguette" is returned.
* bread = "Baguette"
* Output:"bread:Baguette"
* orElseGet(() -> "Frog legs")
* Since optionalName is empty, "Frog legs" is returned from the lambda expression.
* dish = "Frog legs"
* Output:", dish:Frog legs"
* orElseThrow(() -> new Exception())
* Since optionalName is empty, an exception is thrown.
* The catch block catches this exception and prints ", no cheese.".
Thus, the final output is:
makefile
bread:Baguette, dish:Frog legs, no cheese.
References:
* Java SE 21 & JDK 21 - Optional
* Java SE 21 - Functional Interfaces
NEW QUESTION # 63
Which two of the following aren't the correct ways to create a Stream?
Answer: C,D
Explanation:
In Java, the Stream API provides several methods to create streams. However, not all approaches are valid.
NEW QUESTION # 64
......
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