Oracle 1z0-830 Exam With Confidence Using Practice Dumps

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.

Error in for Loop Initialization

The initialization part of a for loopcannot declare multiple variables with different types in a single statement.

Error:

java

for (int i = 0, int j = 3; i < j; i++) {

Fix:Declare variables separately:

java

for (int i = 0, j = 3; i < j; i++) {

lyrics.lines() in Java 21

The lines() method of String returns aStream<String>, splitting the string by line breaks.

Calling .toList() on a streamconverts it to a list.

Valid Code After Fixing the Loop:

java

var lyrics = """

Quand il me prend dans ses bras

Qu'il me parle tout bas

Je vois la vie en rose

""";

for (int i = 0, j = 3; i < j; i++) {

System.out.println(lyrics.lines()

toList()

get(i));

}

Expected Output After Fixing:

vbnet

Quand il me prend dans ses bras

Qu'il me parle tout bas

Je vois la vie en rose

Thus, the correct answer is:Compilation fails.

References:

Java SE 21 - String.lines()

Java SE 21 - for Statement Rules

In Java generics, wildcards are used to relax the type constraints of generic types. The extends wildcard () denotes an upper bounded wildcard, allowing any type that is a subclass of Type. Conversely, the super wildcard () denotes a lower bounded wildcard, allowing any type that is a superclass of Type.

Option A:

java

public List getListSuper() {

return new ArrayList();

}

Here, List represents a list that can hold IOException objects and objects of its supertypes. Since Exception is a superclass of IOException, ArrayList is compatible with List. Therefore, this method compiles successfully.

Option B:

java

public List getListExtends() {

return new ArrayList();

}

In this case, List represents a list that can hold objects of IOException and its subclasses. Since FileNotFoundException is a subclass of IOException, ArrayList is compatible with List. Thus, this method compiles successfully.

Option C:

java

public List getListExtends() {

return new ArrayList();

}

Here, List expects a list of IOException or its subclasses. However, Exception is a superclass of IOException, not a subclass. Therefore, ArrayList is not compatible with List, and this method will not compile.

Option D:

java

public List getListSuper() {

return new ArrayList();

}

In this scenario, List expects a list that can hold IOException objects and objects of its supertypes. Since FileNotFoundException is a subclass of IOException, ArrayList is not compatible with List, 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.