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最新的 Java SE 1z1-830 免費考試真題 (Q21-Q26):

問題 #21
Given:
java
import java.io.*;
class A implements Serializable {
int number = 1;
}
class B implements Serializable {
int number = 2;
}
public class Test {
public static void main(String[] args) throws Exception {
File file = new File("o.ser");
A a = new A();
var oos = new ObjectOutputStream(new FileOutputStream(file));
oos.writeObject(a);
oos.close();
var ois = new ObjectInputStream(new FileInputStream(file));
B b = (B) ois.readObject();
ois.close();
System.out.println(b.number);
}
}
What is the given program's output?

A. 0
B. Compilation fails
C. ClassCastException
D. 1
E. NotSerializableException

答案：C

解題說明：
In this program, we have two classes, A and B, both implementing the Serializable interface, and a Test class with the main method.
Program Flow:
* Serialization:
* An instance of class A is created and assigned to the variable a.
* An ObjectOutputStream is created to write to the file "o.ser".
* The object a is serialized and written to the file.
* The ObjectOutputStream is closed.
* Deserialization:
* An ObjectInputStream is created to read from the file "o.ser".
* The program attempts to read an object from the file and cast it to an instance of class B.
* The ObjectInputStream is closed.
Analysis:
* Serialization Process:
* The object a is an instance of class A and is serialized into the file "o.ser".
* Deserialization Process:
* When deserializing, the program reads the object from the file and attempts to cast it to class B.
* However, the object in the file is of type A, not B.
* Since A and B are distinct classes with no inheritance relationship, casting an A instance to B is invalid.
Exception Details:
* Attempting to cast an object of type A to type B results in a ClassCastException.
* The exception message would be similar to:
pgsql
Exception in thread "main" java.lang.ClassCastException: class A cannot be cast to class B Conclusion:
The program compiles successfully but throws a ClassCastException at runtime when it attempts to cast the deserialized object to class B.

問題 #22
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. Line 2 only
B. The program successfully compiles
C. Line 3 only
D. Line 1 only
E. Line 2 and line 3
F. Line 1 and line 3
G. Line 1 and line 2

答案：B

解題說明：
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.

問題 #23
What does the following code print?
java
import java.util.stream.Stream;
public class StreamReduce {
public static void main(String[] args) {
Stream<String> stream = Stream.of("J", "a", "v", "a");
System.out.print(stream.reduce(String::concat));
}
}

A. null
B. Compilation fails
C. Optional[Java]
D. Java

答案：C

解題說明：
In this code, a Stream of String elements is created containing the characters "J", "a", "v", and "a". The reduce method is then used with String::concat as the accumulator function.
The reduce method with a single BinaryOperator parameter performs a reduction on the elements of the stream, using an associative accumulation function, and returns an Optional describing the reduced value, if any. In this case, it concatenates the strings in the stream.
Since the stream contains elements, the reduction operation concatenates them to form the string "Java". The result is wrapped in an Optional, resulting in Optional[Java]. The print statement outputs this Optional object, displaying Optional[Java].

問題 #24
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

答案：D

解題說明：
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.

問題 #25
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
public sealed class Figure
permits Circle, Rectangle {}
final sealed class Circle extends Figure {
float radius;
}
non-sealed class Rectangle extends Figure {
float length, width;
}
D. java
sealed class Figure permits Rectangle {}
final class Rectangle extends Figure {
float length, width;
}

答案：B,D

解題說明：
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

問題 #26
......

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