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Oracle Java SE 21 Developer Professional Sample Questions (Q65-Q70):
NEW QUESTION # 65
Given:
java
int post = 5;
int pre = 5;
int postResult = post++ + 10;
int preResult = ++pre + 10;
System.out.println("postResult: " + postResult +
", preResult: " + preResult +
", Final value of post: " + post +
", Final value of pre: " + pre);
What is printed?
- A. postResult: 15, preResult: 16, Final value of post: 5, Final value of pre: 6
- B. postResult: 16, preResult: 15, Final value of post: 6, Final value of pre: 5
- C. postResult: 16, preResult: 16, Final value of post: 6, Final value of pre: 6
- D. postResult: 15, preResult: 16, Final value of post: 6, Final value of pre: 6
Answer: D
Explanation:
* Understanding post++ (Post-increment)
* post++uses the value first, then increments it.
* postResult = post++ + 10;
* post starts as 5.
* post++ returns 5, then post is incremented to 6.
* postResult = 5 + 10 = 15.
* Final value of post after this line is 6.
* Understanding ++pre (Pre-increment)
* ++preincrements the value first, then uses it.
* preResult = ++pre + 10;
* pre starts as 5.
* ++pre increments pre to 6, then returns 6.
* preResult = 6 + 10 = 16.
* Final value of pre after this line is 6.
Thus, the final output is:
yaml
postResult: 15, preResult: 16, Final value of post: 6, Final value of pre: 6 References:
* Java SE 21 - Operators and Expressions
* Java SE 21 - Arithmetic Operators
NEW QUESTION # 66
Which methods compile?
- A. ```java public List<? extends IOException> getListExtends() { return new ArrayList<Exception>(); } csharp
- B. ```java public List<? super IOException> getListSuper() { return new ArrayList<Exception>(); } csharp
- C. ```java
public List<? extends IOException> getListExtends() {
return new ArrayList<FileNotFoundException>();
}
- D. ```java
public List<? super IOException> getListSuper() {
return new ArrayList<FileNotFoundException>();
}
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 # 67
What is the output of the following snippet? (Assume the file exists)
java
Path path = Paths.get("C:homejoefoo");
System.out.println(path.getName(0));
- A. Compilation error
- B. C
- C. home
- D. C:
- E. IllegalArgumentException
Answer: C
Explanation:
In Java's java.nio.file package, the Path class represents a file path in a file system. The Paths.get(String first, String... more) method is used to create a Path instance by converting a path string or URI.
In the provided code snippet, the Path object path is created with the string "C:homejoefoo". This represents an absolute path on a Windows system.
The getName(int index) method of the Path class returns a name element of the path as a Path object. The index is zero-based, where index 0 corresponds to the first element in the path's name sequence. It's important to note that the root component (e.g., "C:" on Windows) is not considered a name element and is not included in this sequence.
Therefore, for the path "C:homejoefoo":
* Root Component:"C:"
* Name Elements:
* Index 0: "home"
* Index 1: "joe"
* Index 2: "foo"
When path.getName(0) is called, it returns the first name element, which is "home". Thus, the output of the System.out.println statement is home.
NEW QUESTION # 68
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 n1.
- B. Compilation fails at line n2.
- C. markdown
Inner class:
------------
Outer field
- D. Nothing
- E. An exception is thrown at runtime.
Answer: A
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 # 69
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. default
- B. Compilation fails
- C. static
- D. nothing
Answer: C
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 # 70
......
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