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Oracle Java SE 21 Developer Professional Sample Questions (Q84-Q89):
NEW QUESTION # 84
What do the following print?
java
public class DefaultAndStaticMethods {
public static void main(String[] args) {
WithStaticMethod.print();
}
}
interface WithDefaultMethod {
default void print() {
System.out.print("default");
}
}
interface WithStaticMethod extends WithDefaultMethod {
static void print() {
System.out.print("static");
}
}
- A. Compilation fails
- B. default
- C. nothing
- D. static
Answer: D
Explanation:
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.
ย
NEW QUESTION # 85
Which three of the following are correct about the Java module system?
- A. The unnamed module can only access packages defined in the unnamed module.
- B. If a request is made to load a type whose package is not defined in any known module, then the module system will attempt to load it from the classpath.
- C. The unnamed module exports all of its packages.
- D. Code in an explicitly named module can access types in the unnamed module.
- E. We must add a module descriptor to make an application developed using a Java version prior to SE9 run on Java 11.
- F. If a package is defined in both a named module and the unnamed module, then the package in the unnamed module is ignored.
Answer: B,C,F
Explanation:
The Java Platform Module System (JPMS), introduced in Java 9, modularizes the Java platform and applications. Understanding the behavior of named and unnamed modules is crucial.
* B. The unnamed module exports all of its packages.
Correct. The unnamed module, which includes all code on the classpath, exports all of its packages. This means that any code can access the public types in these packages. However, the unnamed module cannot be explicitly required by named modules.
* C. If a package is defined in both a named module and the unnamed module, then the package in the unnamed module is ignored.
Correct. In cases where a package is present in both a named module and the unnamed module, the version in the named module takes precedence. The package in the unnamed module is ignored to maintain module integrity and avoid conflicts.
* F. If a request is made to load a type whose package is not defined in any known module, then the module system will attempt to load it from the classpath.
Correct. When the module system cannot find a requested type in any known module, it defaults to searching the classpath (i.e., the unnamed module) to locate the type.
Incorrect Options:
* A. Code in an explicitly named module can access types in the unnamed module.
Incorrect. Named modules cannot access types in the unnamed module. The unnamed module can read from named modules, but the reverse is not allowed to ensure strong encapsulation.
* D. We must add a module descriptor to make an application developed using a Java version prior to SE9 run on Java 11.
Incorrect. Adding a module descriptor (module-info.java) is not mandatory for applications developed before Java 9 to run on Java 11. Such applications can run in the unnamed module without modification.
* E. The unnamed module can only access packages defined in the unnamed module.
Incorrect. The unnamed module can access all packages exported by all named modules, in addition to its own packages.
ย
NEW QUESTION # 86
Given:
java
interface Calculable {
long calculate(int i);
}
public class Test {
public static void main(String[] args) {
Calculable c1 = i -> i + 1; // Line 1
Calculable c2 = i -> Long.valueOf(i); // Line 2
Calculable c3 = i -> { throw new ArithmeticException(); }; // Line 3
}
}
Which lines fail to compile?
- A. The program successfully compiles
- B. Line 2 and line 3
- C. Line 3 only
- D. Line 1 and line 2
- E. Line 1 only
- F. Line 1 and line 3
- G. Line 2 only
Answer: A
Explanation:
In this code, the Calculable interface defines a single abstract method calculate that takes an int parameter and returns a long. The main method contains three lambda expressions assigned to variables c1, c2, and c3 of type Calculable.
* Line 1:Calculable c1 = i -> i + 1;
This lambda expression takes an integer i and returns the result of i + 1. Since the expression i + 1 results in an int, and Java allows implicit widening conversion from int to long, this line compiles successfully.
* Line 2:Calculable c2 = i -> Long.valueOf(i);
Here, the lambda expression takes an integer i and returns the result of Long.valueOf(i). The Long.valueOf (int i) method returns a Long object. However, Java allows unboxing of the Long object to a long primitive type when necessary. Therefore, this line compiles successfully.
* Line 3:Calculable c3 = i -> { throw new ArithmeticException(); };
This lambda expression takes an integer i and throws an ArithmeticException. Since the method calculate has a return type of long, and throwing an exception is a valid way to exit the method without returning a value, this line compiles successfully.
Since all three lines adhere to the method signature defined in the Calculable interface and there are no type mismatches or syntax errors, the program compiles successfully.
ย
NEW QUESTION # 87
Given:
java
List<String> l1 = new ArrayList<>(List.of("a", "b"));
List<String> l2 = new ArrayList<>(Collections.singletonList("c"));
Collections.copy(l1, l2);
l2.set(0, "d");
System.out.println(l1);
What is the output of the given code fragment?
- A. [d]
- B. [d, b]
- C. [a, b]
- D. [c, b]
- E. An IndexOutOfBoundsException is thrown
- F. An UnsupportedOperationException is thrown
Answer: D
Explanation:
In this code, two lists l1 and l2 are created and initialized as follows:
* l1 Initialization:
* Created using List.of("a", "b"), which returns an immutable list containing the elements "a" and
"b".
* Wrapped with new ArrayList<>(...) to create a mutable ArrayList containing the same elements.
* l2 Initialization:
* Created using Collections.singletonList("c"), which returns an immutable list containing the single element "c".
* Wrapped with new ArrayList<>(...) to create a mutable ArrayList containing the same element.
State of Lists Before Collections.copy:
* l1: ["a", "b"]
* l2: ["c"]
Collections.copy(l1, l2):
The Collections.copy method copies elements from the source list (l2) into the destination list (l1). The destination list must have at least as many elements as the source list; otherwise, an IndexOutOfBoundsException is thrown.
In this case, l1 has two elements, and l2 has one element, so the copy operation is valid. After copying, the first element of l1 is replaced with the first element of l2:
* l1 after copy: ["c", "b"]
l2.set(0, "d"):
This line sets the first element of l2 to "d".
* l2 after set: ["d"]
Final State of Lists:
* l1: ["c", "b"]
* l2: ["d"]
The System.out.println(l1); statement outputs the current state of l1, which is ["c", "b"]. Therefore, the correct answer is C: [c, b].
ย
NEW QUESTION # 88
Given:
java
record WithInstanceField(String foo, int bar) {
double fuz;
}
record WithStaticField(String foo, int bar) {
static double wiz;
}
record ExtendingClass(String foo) extends Exception {}
record ImplementingInterface(String foo) implements Cloneable {}
Which records compile? (Select 2)
- A. ImplementingInterface
- B. WithInstanceField
- C. ExtendingClass
- D. WithStaticField
Answer: A,D
Explanation:
In Java, records are a special kind of class designed to act as transparent carriers for immutabledata. They automatically provide implementations for equals(), hashCode(), and toString(), and their fields are final and private by default.
* Option A: ExtendingClass
* Analysis: Records in Java implicitly extend java.lang.Record and cannot extend any other class because Java does not support multiple inheritance. Attempting to extend another class, such as Exception, will result in a compilation error.
* Conclusion: Does not compile.
* Option B: WithInstanceField
* Analysis: Records do not allow the declaration of instance fields outside of their components.
The declaration of double fuz; is not permitted and will cause a compilation error.
* Conclusion: Does not compile.
* Option C: ImplementingInterface
* Analysis: Records can implement interfaces. In this case, ImplementingInterface implements Cloneable, which is valid.
* Conclusion: Compiles successfully.
ย
NEW QUESTION # 89
......
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