Authentic Oracle Certification Exam 1z0-830 Questions Answers

Understanding Multi-line Strings in Java (""" Text Blocks)

Java 13 introducedtext blocks ("""), allowing multi-line stringswithout needing explicit \n for new lines.

In a text block,each line is preserved as it appears in the source code.

Analyzing the Options

Option A: \ (Backslash Continuation)

The backslash (\) at the end of a lineprevents a new line from being added, meaning:

nginx

red green blue

Incorrect.

Option B: \s (Whitespace Escape)

\s represents asingle space,not a new line.

The output would be:

nginx

red green blue

Incorrect.

Option C: \t (Tab Escape)

\t inserts atab, not a new line.

The output would be:

nginx

red green blue

Incorrect.

Option D: Correct Text Block

java

String colors = """

red

green

blue

""";

Thispreserves the new lines, producing:

nginx

red

green

blue

Correct.

Thus, the correct answer is:"String colors = """ red green blue """."

References:

Java SE 21 - Text Blocks

Java SE 21 - String Formatting

Pattern Matching in switch

The switch expression introduced inJava 21supportspattern matchingfor different types.

However,a switch expression must be exhaustive, meaningit must cover all possible cases or provide a default case.

Why does compilation fail?

input is an Object, and the switch expression attempts to pattern-match it to String, Double, and Integer.

If input had been of another type (e.g., Float or Long), there would beno matching case, leading to anon-exhaustive switch.

Javarequires a default caseto ensure all possible inputs are covered.

Corrected Code (Adding a default Case)

java

Object input = 42;

String result = switch (input) {

case String s -> "It's a string with value: " + s;

case Double d -> "It's a double with value: " + d;

case Integer i -> "It's an integer with value: " + i;

default -> "Unknown type";

};

System.out.println(result);

With this change, the codecompiles and runs successfully.

Output:

vbnet

It's an integer with value: 42

Thus, the correct answer is:Compilation failsdue to a missing default case.

References:

Java SE 21 - Pattern Matching for switch

Java SE 21 - switch Expressions

(BufferedWriter writer = new BufferedWriter("file.txt") is incorrect.)

Theincorrect statementisoption Bbecause BufferedWriterdoes nothave a constructor that accepts a String (file name) directly. The correct way to use BufferedWriter is to wrap it around a FileWriter, like this:

java

try (BufferedWriter writer = new BufferedWriter(new FileWriter("file.txt"))) {

writer.write("Hello");

}

Evaluation of Other Options:

Option A (Files.write)→ Correct

Uses Files.write() to write bytes to a file.

Efficient and concise method for writing small text files.

Option C (FileOutputStream)→ Correct

Uses a FileOutputStream to write raw bytes to a file.

Works for both text and binary data.

Option D (PrintWriter)→ Correct

Uses PrintWriter for formatted text output.

Option F (FileWriter)→ Correct

Uses FileWriter to write text data.

Option E (None of the suggestions)→ Incorrect becauseoption Bis incorrect.