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Oracle Java SE 21 Developer Professional Sample Questions (Q20-Q25):

NEW QUESTION # 20
Given:
java
String colors = "red " +
"green " +
"blue ";
Which text block can replace the above code?

A. java
String colors = """
red s
greens
blue s
""";
B. java
String colors = """
red
green
blue
""";
C. java
String colors = """
red
green
blue
""";
D. java
String colors = """
red
green
blue
""";
E. None of the propositions

Answer: B

Explanation:
* Understanding Multi-line Strings in Java (""" Text Blocks)
* Java 13 introducedtext blocks ("""), allowing multi-line stringswithout needing explicit 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: (Tab Escape)
* 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

NEW QUESTION # 21
Given:
java
var _ = 3;
var $ = 7;
System.out.println(_ + $);
What is printed?

A. _$
B. 0
C. It throws an exception.
D. Compilation fails.

Answer: D

Explanation:
* The var keyword and identifier rules:
* The var keyword is used for local variable type inference introduced inJava 10.
* However,Java does not allow _ (underscore) as an identifiersinceJava 9.
* If we try to use _ as a variable name, the compiler will throw an error:
pgsql
error: as of release 9, '_' is a keyword, and may not be used as an identifier
* The $ symbol as an identifier:
* The $ characteris a valid identifierin Java.
* However, since _ is not allowed, the codefails to compile before even reaching $.
Thus,the correct answer is "Compilation fails."
References:
* Java SE 21 - var Local Variable Type Inference
* Java SE 9 - Restrictions on _ Identifier

NEW QUESTION # 22
Given:
java
String textBlock = """
j
a
v s
a
""";
System.out.println(textBlock.length());
What is the output?

A. 0
B. 1
C. 2
D. 3

Answer: B

Explanation:
In this code, a text block is defined using the """ syntax introduced in Java 13. Text blocks allow for multiline string literals, preserving the format as written in the code.
Text Block Analysis:
The text block is defined as:
java
String textBlock = """
j
a
contentReference[oaicite:0]{index=0}

NEW QUESTION # 23
Which of the following suggestions compile?(Choose two.)

A. java
sealed class Figure permits Rectangle {}
public class Rectangle extends Figure {
float length, width;
}
B. java
public sealed class Figure
permits Circle, Rectangle {}
final class Circle extends Figure {
float radius;
}
non-sealed class Rectangle extends Figure {
float length, width;
}
C. java
sealed class Figure permits Rectangle {}
final class Rectangle extends Figure {
float length, width;
}
D. java
public sealed class Figure
permits Circle, Rectangle {}
final sealed class Circle extends Figure {
float radius;
}
non-sealed class Rectangle extends Figure {
float length, width;
}

Answer: B,C

Explanation:
Option A (sealed class Figure permits Rectangle {} and final class Rectangle extends Figure {}) - Valid
* Why it compiles?
* Figure issealed, meaning itmust explicitly declareits subclasses.
* Rectangle ispermittedto extend Figure and isdeclared final, meaning itcannot be extended further.
* This followsvalid sealed class rules.
Option B (sealed class Figure permits Rectangle {} and public class Rectangle extends Figure {}) -# Invalid
* Why it fails?
* Rectangle extends Figure, but it doesnot specify if it is sealed, final, or non-sealed.
* Fix:The correct declaration must be one of the following:
java
final class Rectangle extends Figure {} // OR
sealed class Rectangle permits OtherClass {} // OR
non-sealed class Rectangle extends Figure {}
Option C (final sealed class Circle extends Figure {}) -#Invalid
* Why it fails?
* A class cannot be both final and sealedat the same time.
* sealed meansit must have permitted subclasses, but final meansit cannot be extended.
* Fix:Change final sealed to just final:
java
final class Circle extends Figure {}
Option D (public sealed class Figure permits Circle, Rectangle {} with final class Circle and non-sealed class Rectangle) - Valid
* Why it compiles?
* Figure issealed, meaning it mustdeclare its permitted subclasses(Circle and Rectangle).
* Circle is declaredfinal, so itcannot have subclasses.
* Rectangle is declarednon-sealed, meaningit can be subclassedfreely.
* This correctly followsJava's sealed class rules.
Thus, the correct answers are:A, D
References:
* Java SE 21 - Sealed Classes
* Java SE 21 - Class Modifiers

NEW QUESTION # 24
Given:
java
StringBuilder result = Stream.of("a", "b")
.collect(
() -> new StringBuilder("c"),
StringBuilder::append,
(a, b) -> b.append(a)
);
System.out.println(result);
What is the output of the given code fragment?

A. bca
B. bac
C. cacb
D. cba
E. abc
F. cbca
G. acb

Answer: D

Explanation:
In this code, a Stream containing the elements "a" and "b" is processed using the collect method. The collect method is a terminal operation that performs a mutable reduction on the elements of the stream using a Collector. In this case, custom implementations for the supplier, accumulator, and combiner are provided.
Components of the collect Method:
* Supplier:
* () -> new StringBuilder("c")
* This supplier creates a new StringBuilder initialized with the string "c".
* Accumulator:
* StringBuilder::append
* This accumulator appends each element of the stream to the StringBuilder.
* Combiner:
* (a, b) -> b.append(a)
* This combiner is used in parallel stream operations to merge two StringBuilder instances. It appends the contents of a to b.
Execution Flow:
* Stream Elements:"a", "b"
* Initial StringBuilder:"c"
* Accumulation:
* The first element "a" is appended to "c", resulting in "ca".
* The second element "b" is appended to "ca", resulting in "cab".
* Combiner:
* In this sequential stream, the combiner is not utilized. The combiner is primarily used in parallel streams to merge partial results.
Final Result:
The StringBuilder contains "cab". Therefore, the output of the program is:
nginx
cab

NEW QUESTION # 25
......

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