Free Practice Questions for Oracle 1z0-830 Exam

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.

Understanding the do-while Loop

The do-while loopexecutes at least oncebefore checking the condition.

The condition i++ > 0 increments iafterchecking.

Step-by-Step Execution

Iteration 1:

i = 0

Prints: "Chanel"

i++ updates i to 1

Condition 1 > 0is true, so the loop exits.

Why Doesn't the Loop Continue?

Since i starts at 0, the conditioni++ > 0 is false after the first iteration.

The loopexits immediately after printing "Chanel".

Final Output

nginx

Chanel

Thus, the correct answer is:Chanel

References:

Java SE 21 - do-while Loop

Java SE 21 - Post-Increment Behavior

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

In this code, the strip() method is used to remove leading and trailing whitespace from strings. The strip() method, introduced in Java 11, is Unicode-aware and removes all leading and trailing characters that are considered whitespace according to the Unicode standard.

docs.oracle.com

Analysis of Each Statement:

First Statement:

java

String s = " ";

System.out.print("[" + s.strip());

The string s contains four spaces.

Applying s.strip() removes all leading and trailing spaces, resulting in an empty string.

The output is "[" followed by the empty string, so the printed result is "[".

Second Statement:

java

s = " hello ";

System.out.print("," + s.strip());

The string s is now " hello ".

Applying s.strip() removes all leading and trailing spaces, resulting in "hello".

The output is "," followed by "hello", so the printed result is ",hello".

Third Statement:

java

s = "h i ";

System.out.print("," + s.strip() + "]");

The string s is now "h i ".

Applying s.strip() removes the trailing spaces, resulting in "h i".

The output is "," followed by "h i" and then "]", so the printed result is ",h i]".

Combined Output:

Combining all parts, the final output is:

css

[,hello,h i]

Understanding Variable Assignments

java

Integer frenchRevolution = 1789;

Object o1 = new String("1789");

Object o2 = frenchRevolution;

frenchRevolution = null;

frenchRevolution is an Integer with value1789.

o1 is aString with value "1789".

o2 storesa reference to frenchRevolution, which is an Integer (1789).

frenchRevolution = null;only nullifies the reference, but o2 still holds the Integer 1789.

Calling toString() on o2

java

Object o3 = o2.toString();

o2 refers to an Integer (1789).

Integer.toString() returns theString representation "1789".

o3 is assigned "1789" (String).

Evaluating o1.equals(o3)

java

System.out.println(o1.equals(o3));

o1.equals(o3) isequivalent to:

java

"1789".equals("1789")

Since both areequal strings, the output is:

arduino

true

Thus, the correct answer is:true

References:

Java SE 21 - Integer.toString()

Java SE 21 - String.equals()