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1z0-830 Free Practice, Latest 1z0-830 Mock Test

1z0-830 Free Practice, Latest 1z0-830 Mock Test

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Oracle Java SE 21 Developer Professional Sample Questions (Q80-Q85):

NEW QUESTION # 80
Given:
java
try (FileOutputStream fos = new FileOutputStream("t.tmp");
ObjectOutputStream oos = new ObjectOutputStream(fos)) {
fos.write("Today");
fos.writeObject("Today");
oos.write("Today");
oos.writeObject("Today");
} catch (Exception ex) {
// handle exception
}
Which statement compiles?

  • A. oos.write("Today");
  • B. fos.write("Today");
  • C. fos.writeObject("Today");
  • D. oos.writeObject("Today");

Answer: D

Explanation:
In Java, FileOutputStream and ObjectOutputStream are used for writing data to files, but they have different purposes and methods. Let's analyze each statement:
* fos.write("Today");
The FileOutputStream class is designed to write raw byte streams to files. The write method in FileOutputStream expects a parameter of type int or byte[]. Since "Today" is a String, passing it directly to fos.
write("Today"); will cause a compilation error because there is no write method in FileOutputStream that accepts a String parameter.
* fos.writeObject("Today");
The FileOutputStream class does not have a method named writeObject. The writeObject method is specific to ObjectOutputStream. Therefore, attempting to call fos.writeObject("Today"); will result in a compilation error.
* oos.write("Today");
The ObjectOutputStream class is used to write objects to an output stream. However, it does not have a write method that accepts a String parameter. The available write methods in ObjectOutputStream are for writing primitive data types and objects. Therefore, oos.write("Today"); will cause a compilation error.
* oos.writeObject("Today");
The ObjectOutputStream class provides the writeObject method, which is used to serialize objects and write them to the output stream. Since String implements the Serializable interface, "Today" can be serialized.
Therefore, oos.writeObject("Today"); is valid and compiles successfully.
In summary, the only statement that compiles without errors is oos.writeObject("Today");.
References:
* Java SE 21 & JDK 21 - ObjectOutputStream
* Java SE 21 & JDK 21 - FileOutputStream


NEW QUESTION # 81
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. WithInstanceField
  • B. ExtendingClass
  • C. WithStaticField
  • D. ImplementingInterface

Answer: C,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.


NEW QUESTION # 82
Which three of the following are correct about the Java module system?

  • A. 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.
  • B. The unnamed module can only access packages defined in the unnamed module.
  • C. If a package is defined in both a named module and the unnamed module, then the package in the unnamed module is ignored.
  • D. We must add a module descriptor to make an application developed using a Java version prior to SE9 run on Java 11.
  • E. The unnamed module exports all of its packages.
  • F. Code in an explicitly named module can access types in the unnamed module.

Answer: A,C,E

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.


NEW QUESTION # 83
What do the following print?
java
import java.time.Duration;
public class DividedDuration {
public static void main(String[] args) {
var day = Duration.ofDays(2);
System.out.print(day.dividedBy(8));
}
}

  • A. PT0D
  • B. Compilation fails
  • C. PT0H
  • D. PT6H
  • E. It throws an exception

Answer: D

Explanation:
In this code, a Duration object day is created representing a duration of 2 days using the Duration.ofDays(2) method. The dividedBy(long divisor) method is then called on this Duration object with the argument 8.
The dividedBy(long divisor) method returns a copy of the original Duration divided by the specified value. In this case, dividing 2 days by 8 results in a duration of 0.25 days. In the ISO-8601 duration format used by Java's Duration class, this is represented as PT6H, which stands for a period of 6 hours.
Therefore, the output of the System.out.print statement is PT6H.


NEW QUESTION # 84
Given:
java
public class SpecialAddition extends Addition implements Special {
public static void main(String[] args) {
System.out.println(new SpecialAddition().add());
}
int add() {
return --foo + bar--;
}
}
class Addition {
int foo = 1;
}
interface Special {
int bar = 1;
}
What is printed?

  • A. 0
  • B. Compilation fails.
  • C. 1
  • D. 2
  • E. It throws an exception at runtime.

Answer: B

Explanation:
1. Why does the compilation fail?
* The interface Special contains bar as int bar = 1;.
* In Java, all interface fields are implicitly public, static, and final.
* This means that bar is a constant (final variable).
* The method add() contains bar--, which attempts to modify bar.
* Since bar is final, it cannot be modified, causing acompilation error.
2. Correcting the Code
To make the code compile, bar must not be final. One way to fix this:
java
class SpecialImpl implements Special {
int bar = 1;
}
Or modify the add() method:
java
int add() {
return --foo + bar; // No modification of bar
}
Thus, the correct answer is:Compilation fails.
References:
* Java SE 21 - Interfaces
* Java SE 21 - Final Variables


NEW QUESTION # 85
......

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