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Oracle Java SE 21 Developer Professional Sample Questions (Q10-Q15):
NEW QUESTION # 10
Which of the following can be the body of a lambda expression?
- A. None of the above
- B. A statement block
- C. An expression and a statement
- D. Two expressions
- E. Two statements
Answer: B
Explanation:
In Java, a lambda expression can have two forms for its body:
* Single Expression:A concise form where the body consists of a single expression. The result of this expression is implicitly returned.
Example:
java
(a, b) -> a + b
In this example, (a, b) are the parameters, and a + b is the single expression that adds them together.
* Statement Block:A more detailed form where the body consists of a block of statements enclosed in braces {}. Within this block, you can have multiple statements, and if a return value is expected, you must explicitly use the return statement.
Example:
java
(a, b) -> {
int sum = a + b;
System.out.println("Sum is: " + sum);
return sum;
}
In this example, the lambda body is a statement block that performs multiple actions: it calculates the sum, prints it, and then returns the sum.
Given the options:
* A. Two statements:While a lambda body can contain multiple statements, they must be enclosed within a statement block {}. Simply having two statements without braces is not valid syntax for a lambda expression.
* B. An expression and a statement:Similar to option A, if a lambda body contains more than one element (be it expressions or statements), they need to be enclosed in a statement block.
* C. A statement block:This is correct. A lambda expression can have a body that is a statement block, allowing multiple statements enclosed in braces.
* D. None of the above:This is incorrect since option C is valid.
* E. Two expressions:As with options A and B, multiple expressions must be enclosed in a statement block to form a valid lambda body.
Therefore, the correct answer is C: A statement block.
NEW QUESTION # 11
Given:
java
var ceo = new HashMap<>();
ceo.put("Sundar Pichai", "Google");
ceo.put("Tim Cook", "Apple");
ceo.put("Mark Zuckerberg", "Meta");
ceo.put("Andy Jassy", "Amazon");
Does the code compile?
- A. False
- B. True
Answer: A
Explanation:
In this code, a HashMap is instantiated using the var keyword:
java
var ceo = new HashMap<>();
The diamond operator <> is used without explicit type arguments. While the diamond operatorallows the compiler to infer types in many cases, when using var, the compiler requires explicit type information to infer the variable's type.
Therefore, the code will not compile because the compiler cannot infer the type of the HashMap when both var and the diamond operator are used without explicit type parameters.
To fix this issue, provide explicit type parameters when creating the HashMap:
java
var ceo = new HashMap<String, String>();
Alternatively, you can specify the variable type explicitly:
java
Map<String, String>
contentReference[oaicite:0]{index=0}
NEW QUESTION # 12
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. cbca
- B. abc
- C. bca
- D. acb
- E. cacb
- F. cba
- G. bac
Answer: F
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 # 13
Given:
java
sealed class Vehicle permits Car, Bike {
}
non-sealed class Car extends Vehicle {
}
final class Bike extends Vehicle {
}
public class SealedClassTest {
public static void main(String[] args) {
Class<?> vehicleClass = Vehicle.class;
Class<?> carClass = Car.class;
Class<?> bikeClass = Bike.class;
System.out.print("Is Vehicle sealed? " + vehicleClass.isSealed() +
"; Is Car sealed? " + carClass.isSealed() +
"; Is Bike sealed? " + bikeClass.isSealed());
}
}
What is printed?
- A. Is Vehicle sealed? true; Is Car sealed? false; Is Bike sealed? false
- B. Is Vehicle sealed? true; Is Car sealed? true; Is Bike sealed? true
- C. Is Vehicle sealed? false; Is Car sealed? false; Is Bike sealed? false
- D. Is Vehicle sealed? false; Is Car sealed? true; Is Bike sealed? true
Answer: A
Explanation:
* Understanding Sealed Classes in Java
* Asealed classrestricts which other classes can extend it.
* A sealed classmust explicitly declare its permitted subclassesusing the permits keyword.
* Subclasses can be declared as:
* sealed(restricts further extension).
* non-sealed(removes the restriction, allowing unrestricted subclassing).
* final(prevents further subclassing).
* Analyzing the Given Code
* Vehicle is declared as sealed with permits Car, Bike, meaning only Car and Bike can extend it.
* Car is declared as non-sealed, which means itis no longer sealedand can have subclasses.
* Bike is declared as final, meaningit cannot be subclassed.
* Using isSealed() Method
* vehicleClass.isSealed() #truebecause Vehicle is explicitly marked as sealed.
* carClass.isSealed() #falsebecause Car is marked non-sealed.
* bikeClass.isSealed() #falsebecause Bike is final, and a final class isnot considered sealed.
* Final Output
csharp
Is Vehicle sealed? true; Is Car sealed? false; Is Bike sealed? false
Thus, the correct answer is:"Is Vehicle sealed? true; Is Car sealed? false; Is Bike sealed? false" References:
* Java SE 21 - Sealed Classes
* Java SE 21 - isSealed() Method
NEW QUESTION # 14
Given:
java
CopyOnWriteArrayList<String> list = new CopyOnWriteArrayList<>();
list.add("A");
list.add("B");
list.add("C");
// Writing in one thread
new Thread(() -> {
list.add("D");
System.out.println("Element added: D");
}).start();
// Reading in another thread
new Thread(() -> {
for (String element : list) {
System.out.println("Read element: " + element);
}
}).start();
What is printed?
- A. Compilation fails.
- B. It prints all elements, including changes made during iteration.
- C. It prints all elements, but changes made during iteration may not be visible.
- D. It throws an exception.
Answer: C
Explanation:
* Understanding CopyOnWriteArrayList
* CopyOnWriteArrayList is a thread-safe variant of ArrayList whereall mutative operations (add, set, remove, etc.) create a new copy of the underlying array.
* This meansiterations will not reflect modifications made after the iterator was created.
* Instead of modifying the existing array, a new copy is created for modifications, ensuring that readers always see a consistent snapshot.
* Thread Execution Behavior
* Thread 1 (Writer Thread)adds "D" to the list.
* Thread 2 (Reader Thread)iterates over the list.
* The reader thread gets a snapshot of the listbefore"D" is added.
* The output may look like:
mathematica
Read element: A
Read element: B
Read element: C
Element added: D
* "D" may not appear in the output of the reader threadbecause the iteration occurs on a snapshot before the modification.
* Why doesn't it print all elements including changes?
* Since CopyOnWriteArrayList doesnot allow changes to be visible during iteration, the reader threadwill not see "D"if it started iterating before "D" was added.
Thus, the correct answer is:"It prints all elements, but changes made during iteration may not be visible." References:
* Java SE 21 - CopyOnWriteArrayList
NEW QUESTION # 15
......
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