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Oracle Java SE 21 Developer Professional Sample Questions (Q28-Q33):
NEW QUESTION # 28
Given:
java
public class Test {
class A {
}
static class B {
}
public static void main(String[] args) {
// Insert here
}
}
Which three of the following are valid statements when inserted into the given program?
- A. B b = new Test.B();
- B. B b = new Test().new B();
- C. A a = new A();
- D. A a = new Test().new A();
- E. B b = new B();
- F. A a = new Test.A();
Answer: A,D,E
Explanation:
In the provided code, we have two inner classes within the Test class:
* Class A:
* An inner (non-static) class.
* Instances of A are associated with an instance of the enclosing Test class.
* Class B:
* A static nested class.
* Instances of B are not associated with any instance of the enclosing Test class and can be instantiated without an instance of Test.
Evaluation of Statements:
A: A a = new A();
* Invalid.Since A is a non-static inner class, it requires an instance of the enclosing class Test to be instantiated. Attempting to instantiate A without an instance of Test will result in a compilation error.
B: B b = new Test.B();
* Valid.B is a static nested class and can be instantiated without an instance of Test. This syntax is correct.
C: A a = new Test.A();
* Invalid.Even though A is referenced through Test, it is a non-static inner class and requires an instance of Test for instantiation. This will result in a compilation error.
D: B b = new Test().new B();
* Invalid.While this syntax is used for instantiating non-static inner classes, B is a static nested class and does not require an instance of Test. This will result in a compilation error.
E: B b = new B();
* Valid.Since B is a static nested class, it can be instantiated directly without referencing the enclosing class.
F: A a = new Test().new A();
* Valid.This is the correct syntax for instantiating a non-static inner class. An instance of Test is created, and then an instance of A is created associated with that Test instance.
Therefore, the valid statements are B, E, and F.
NEW QUESTION # 29
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?
- A. None of the suggestions.
- B. var concurrentHashMap = new ConcurrentHashMap(42, 10);
- C. var concurrentHashMap = new ConcurrentHashMap(42, 0.88f, 10);
- D. var concurrentHashMap = new ConcurrentHashMap(42);
- E. var concurrentHashMap = new ConcurrentHashMap();
Answer: C
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 # 30
Given:
java
List<String> frenchAuthors = new ArrayList<>();
frenchAuthors.add("Victor Hugo");
frenchAuthors.add("Gustave Flaubert");
Which compiles?
- A. Map<String, ? extends List<String>> authorsMap2 = new HashMap<String, ArrayList<String>> (); java authorsMap2.put("FR", frenchAuthors);
- B. Map<String, List<String>> authorsMap5 = new HashMap<String, List<String>>(); java authorsMap5.put("FR", frenchAuthors);
- C. var authorsMap3 = new HashMap<>();
java
authorsMap3.put("FR", frenchAuthors); - D. Map<String, ArrayList<String>> authorsMap1 = new HashMap<>();
java
authorsMap1.put("FR", frenchAuthors); - E. Map<String, List<String>> authorsMap4 = new HashMap<String, ArrayList<String>>(); java authorsMap4.put("FR", frenchAuthors);
Answer: B,C,E
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 # 31
Given:
java
Object myVar = 0;
String print = switch (myVar) {
case int i -> "integer";
case long l -> "long";
case String s -> "string";
default -> "";
};
System.out.println(print);
What is printed?
- A. It throws an exception at runtime.
- B. string
- C. integer
- D. Compilation fails.
- E. long
- F. nothing
Answer: D
Explanation:
* Why does the compilation fail?
* TheJava switch statement does not support primitive type pattern matchingin switch expressions as of Java 21.
* The case pattern case int i -> "integer"; isinvalidbecausepattern matching with primitive types (like int or long) is not yet supported in switch statements.
* The error occurs at case int i -> "integer";, leading to acompilation failure.
* Correcting the Code
* Since myVar is of type Object,autoboxing converts 0 into an Integer.
* To make the code compile, we should use Integer instead of int:
java
Object myVar = 0;
String print = switch (myVar) {
case Integer i -> "integer";
case Long l -> "long";
case String s -> "string";
default -> "";
};
System.out.println(print);
* Output:
bash
integer
Thus, the correct answer is:Compilation fails.
References:
* Java SE 21 - Pattern Matching for switch
* Java SE 21 - switch Expressions
NEW QUESTION # 32
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. static
- B. Compilation fails
- C. default
- D. nothing
Answer: A
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 # 33
......
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