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Oracle Java SE 21 Developer Professional Sample Questions (Q45-Q50):

NEW QUESTION # 45
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. stringBuilder1
B. stringBuilder3
C. stringBuilder2
D. stringBuilder4
E. None of them

Answer: D

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 # 46
Which three of the following are correct about the Java module system?

A. We must add a module descriptor to make an application developed using a Java version prior to SE9 run on Java 11.
B. The unnamed module exports all of its packages.
C. If a package is defined in both a named module and the unnamed module, then the package in the unnamed module is ignored.
D. The unnamed module can only access packages defined in the unnamed module.
E. If a request is made to load a type whose package is not defined in any known module, then the module system will attempt to load it from the classpath.
F. Code in an explicitly named module can access types in the unnamed module.

Answer: B,C,E

Explanation:
The Java Platform Module System (JPMS), introduced in Java 9, modularizes the Java platform and applications. Understanding the behavior of named and unnamed modules is crucial.
* B. The unnamed module exports all of its packages.
Correct. The unnamed module, which includes all code on the classpath, exports all of its packages. This means that any code can access the public types in these packages. However, the unnamed module cannot be explicitly required by named modules.
* C. If a package is defined in both a named module and the unnamed module, then the package in the unnamed module is ignored.
Correct. In cases where a package is present in both a named module and the unnamed module, the version in the named module takes precedence. The package in the unnamed module is ignored to maintain module integrity and avoid conflicts.
* F. If a request is made to load a type whose package is not defined in any known module, then the module system will attempt to load it from the classpath.
Correct. When the module system cannot find a requested type in any known module, it defaults to searching the classpath (i.e., the unnamed module) to locate the type.
Incorrect Options:
* A. Code in an explicitly named module can access types in the unnamed module.
Incorrect. Named modules cannot access types in the unnamed module. The unnamed module can read from named modules, but the reverse is not allowed to ensure strong encapsulation.
* D. We must add a module descriptor to make an application developed using a Java version prior to SE9 run on Java 11.
Incorrect. Adding a module descriptor (module-info.java) is not mandatory for applications developed before Java 9 to run on Java 11. Such applications can run in the unnamed module without modification.
* E. The unnamed module can only access packages defined in the unnamed module.
Incorrect. The unnamed module can access all packages exported by all named modules, in addition to its own packages.

NEW QUESTION # 47
Which methods compile?

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

Answer: B,C

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 # 48
Given:
java
StringBuffer us = new StringBuffer("US");
StringBuffer uk = new StringBuffer("UK");
Stream<StringBuffer> stream = Stream.of(us, uk);
String output = stream.collect(Collectors.joining("-", "=", ""));
System.out.println(output);
What is the given code fragment's output?

A. An exception is thrown.
B. -US=UK
C. =US-UK
D. US-UK
E. US=UK
F. Compilation fails.

Answer: C

Explanation:
In this code, two StringBuffer objects, us and uk, are created with the values "US" and "UK", respectively. A stream is then created from these objects using Stream.of(us, uk).
The collect method is used with Collectors.joining("-", "=", ""). The joining collector concatenates the elements of the stream into a single String with the following parameters:
* Delimiter ("-"):Inserted between each element.
* Prefix ("="):Inserted at the beginning of the result.
* Suffix (""):Inserted at the end of the result.
Therefore, the elements "US" and "UK" are concatenated with "-" between them, resulting in "US-UK". The prefix "=" is added at the beginning, resulting in the final output =US-UK.

NEW QUESTION # 49
Given:
java
Optional o1 = Optional.empty();
Optional o2 = Optional.of(1);
Optional o3 = Stream.of(o1, o2)
.filter(Optional::isPresent)
.findAny()
.flatMap(o -> o);
System.out.println(o3.orElse(2));
What is the given code fragment's output?

A. 0
B. Optional.empty
C. Compilation fails
D. An exception is thrown
E. 1
F. 2
G. Optional[1]

Answer: E

Explanation:
In this code, two Optional objects are created:
* o1 is an empty Optional.
* o2 is an Optional containing the integer 1.
A stream is created from o1 and o2. The filter method retains only the Optional instances that are present (i.e., non-empty). This results in a stream containing only o2.
The findAny method returns an Optional describing some element of the stream, or an empty Optional if the stream is empty. Since the stream contains o2, findAny returns Optional[Optional[1]].
The flatMap method is then used to flatten this nested Optional. It applies the provided mapping function (o -
> o) to the value, resulting in Optional[1].
Finally, o3.orElse(2) returns the value contained in o3 if it is present; otherwise, it returns 2. Since o3 contains
1, the output is 1.

NEW QUESTION # 50
......

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