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2025 Professional 1z1-830: Java SE 21 Developer Professional Test Assessment
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Oracle Java SE 21 Developer Professional Sample Questions (Q31-Q36):
NEW QUESTION # 31
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. stringBuilder3
- B. stringBuilder4
- C. None of them
- D. stringBuilder2
- E. stringBuilder1
Answer: B
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 # 32
Given:
java
public class OuterClass {
String outerField = "Outer field";
class InnerClass {
void accessMembers() {
System.out.println(outerField);
}
}
public static void main(String[] args) {
System.out.println("Inner class:");
System.out.println("------------");
OuterClass outerObject = new OuterClass();
InnerClass innerObject = new InnerClass(); // n1
innerObject.accessMembers(); // n2
}
}
What is printed?
- A. Compilation fails at line n2.
- B. Compilation fails at line n1.
- C. Nothing
- D. markdown
Inner class:
------------
Outer field - E. An exception is thrown at runtime.
Answer: B
Explanation:
* Understanding Inner Classes in Java
* Aninner class (non-static nested class)requires an instance of the outer classbefore it can be instantiated.
* Incorrect instantiationof the inner class at n1:
java
InnerClass innerObject = new InnerClass(); // Compilation error
* Since InnerClass is anon-staticinner class, itmust be created from an instance of OuterClass.
* Correct Way to Instantiate the Inner Class
java
OuterClass outerObject = new OuterClass();
OuterClass.InnerClass innerObject = outerObject.new InnerClass(); // Correct
* Thiscorrectly associatesthe inner class with an instance of OuterClass.
* Why Does Compilation Fail?
* The error occurs atline n1because InnerClass is beinginstantiated incorrectly.
Thus, the correct answer is:Compilation fails at line n1.
References:
* Java SE 21 - Nested and Inner Classes
* Java SE 21 - Accessing Outer Class Members
NEW QUESTION # 33
Given:
java
List<Long> cannesFestivalfeatureFilms = LongStream.range(1, 1945)
.boxed()
.toList();
try (var executor = Executors.newVirtualThreadPerTaskExecutor()) {
cannesFestivalfeatureFilms.stream()
.limit(25)
.forEach(film -> executor.submit(() -> {
System.out.println(film);
}));
}
What is printed?
- A. Numbers from 1 to 25 randomly
- B. Numbers from 1 to 25 sequentially
- C. Numbers from 1 to 1945 randomly
- D. An exception is thrown at runtime
- E. Compilation fails
Answer: A
Explanation:
* Understanding LongStream.range(1, 1945).boxed().toList();
* LongStream.range(1, 1945) generates a stream of numbersfrom 1 to 1944.
* .boxed() converts the primitive long values to Long objects.
* .toList() (introduced in Java 16)creates an immutable list.
* Understanding Executors.newVirtualThreadPerTaskExecutor()
* Java 21 introducedvirtual threadsto improve concurrency.
* Executors.newVirtualThreadPerTaskExecutor()creates a new virtual thread per submitted task
, allowing highly concurrent execution.
* Execution Behavior
* cannesFestivalfeatureFilms.stream().limit(25) # Limits the stream to thefirst 25 numbers(1 to
25).
* .forEach(film -> executor.submit(() -> System.out.println(film)))
* Each film is printed inside a virtual thread.
* Virtual threads execute asynchronously, meaning numbers arenot guaranteed to print sequentially.
* Output will contain numbers from 1 to 25, but their order is random due to concurrent execution.
* Possible Output (Random Order)
python-repl
3
1
5
2
4
7
25
* The ordermay differ in each rundue to concurrent execution.
Thus, the correct answer is:"Numbers from 1 to 25 randomly."
References:
* Java SE 21 - Virtual Threads
* Java SE 21 - Executors.newVirtualThreadPerTaskExecutor()
NEW QUESTION # 34
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. IOException
- B. RuntimeException
- C. Compilation fails
- D. ArithmeticException
Answer: D
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 # 35
Given:
java
List<String> frenchAuthors = new ArrayList<>();
frenchAuthors.add("Victor Hugo");
frenchAuthors.add("Gustave Flaubert");
Which compiles?
- A. Map<String, List<String>> authorsMap4 = new HashMap<String, ArrayList<String>>(); java authorsMap4.put("FR", frenchAuthors);
- B. var authorsMap3 = new HashMap<>();
java
authorsMap3.put("FR", frenchAuthors); - C. Map<String, List<String>> authorsMap5 = new HashMap<String, List<String>>(); java authorsMap5.put("FR", frenchAuthors);
- D. Map<String, ArrayList<String>> authorsMap1 = new HashMap<>();
java
authorsMap1.put("FR", frenchAuthors); - E. Map<String, ? extends List<String>> authorsMap2 = new HashMap<String, ArrayList<String>> (); java authorsMap2.put("FR", frenchAuthors);
Answer: A,B,C
Explanation:
* Option A (Map<String, ArrayList<String>> authorsMap1 = new HashMap<>();)
* #Compilation Fails
* frenchAuthors is declared as List<String>,notArrayList<String>.
* The correct way to declare a Map that allows storing List<String> is to use List<String> as the generic type,notArrayList<String>.
* Fix:
java
Map<String, List<String>> authorsMap1 = new HashMap<>();
authorsMap1.put("FR", frenchAuthors);
* Reason:The type ArrayList<String> is more specific than List<String>, and this would cause a type mismatcherror.
* Option B (Map<String, ? extends List<String>> authorsMap2 = new HashMap<String, ArrayList<String>>();)
* #Compilation Fails
* ? extends List<String>makes the map read-onlyfor adding new elements.
* The line authorsMap2.put("FR", frenchAuthors); causes acompilation errorbecause wildcard (?
extends List<String>) prevents modifying the map.
* Fix:Remove the wildcard:
java
Map<String, List<String>> authorsMap2 = new HashMap<>();
authorsMap2.put("FR", frenchAuthors);
* Option C (var authorsMap3 = new HashMap<>();)
* Compiles Successfully
* The var keyword allows the compiler to infer the type.
* However,the inferred type is HashMap<Object, Object>, which may cause issues when retrieving values.
* Option D (Map<String, List<String>> authorsMap4 = new HashMap<String, ArrayList<String>
>();)
* Compiles Successfully
* Valid declaration:HashMap<K, V> can be assigned to Map<K, V>.
* Using new HashMap<String, ArrayList<String>>() with Map<String, List<String>> isallowed due to polymorphism.
* Correct syntax:
java
Map<String, List<String>> authorsMap4 = new HashMap<String, ArrayList<String>>(); authorsMap4.put("FR", frenchAuthors);
* Option E (Map<String, List<String>> authorsMap5 = new HashMap<String, List<String>>();)
* Compiles Successfully
* HashMap<String, List<String>> isa valid instantiation.
* Correct usage:
java
Map<String, List<String>> authorsMap5 = new HashMap<>();
authorsMap5.put("FR", frenchAuthors);
Thus, the correct answers are:C, D, E
References:
* Java SE 21 - Generics and Type Inference
* Java SE 21 - var Keyword
NEW QUESTION # 36
......
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