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| Section | Objectives |
|---|---|
| Java Language Fundamentals | - Java syntax and language structure
|
| Core APIs | - Java standard library usage
|
| Object-Oriented Programming | - Core OOP principles
|
| Exception Handling and Debugging | - Error handling mechanisms
|
| Input/Output and File Handling | - NIO and file operations
|
| Advanced Java Features (Java SE 21) | - Modern language features
|
| Concurrency and Multithreading | - Thread management
|
| Database Connectivity (JDBC) | - Database interaction
|
You can avail all the above-mentioned characteristics of the desktop software in this web-based Oracle 1z0-830 practice test. While you appear in the Oracle 1z0-830 real examination, you will feel the same environment you faced during our Oracle 1z0-830 practice test.
NEW QUESTION # 39
Given:
java
void verifyNotNull(Object input) {
boolean enabled = false;
assert enabled = true;
assert enabled;
System.out.println(input.toString());
assert input != null;
}
When does the given method throw a NullPointerException?
Answer: D
Explanation:
In the verifyNotNull method, the following operations are performed:
* Assertion to Enable Assertions:
java
boolean enabled = false;
assert enabled = true;
assert enabled;
* The variable enabled is initially set to false.
* The first assertion assert enabled = true; assigns true to enabled if assertions are enabled. If assertions are disabled, this assignment does not occur.
* The second assertion assert enabled; checks if enabled is true. If assertions are enabled and the previous assignment occurred, this assertion passes. If assertions are disabled, this assertion is ignored.
* Dereferencing the input Object:
java
System.out.println(input.toString());
* This line attempts to call the toString() method on the input object. If input is null, this will throw a NullPointerException.
* Assertion to Check input for null:
java
assert input != null;
* This assertion checks that input is not null. If input is null and assertions are enabled, this assertion will fail, throwing an AssertionError. If assertions are disabled, this assertion is ignored.
Analysis:
* If Assertions Are Enabled:
* The enabled variable is set to true by the first assertion, and the second assertion passes.
* If input is null, calling input.toString() will throw a NullPointerException before the final assertion is reached.
* If input is not null, input.toString() executes without issue, and the final assertion assert input != null; passes.
* If Assertions Are Disabled:
* The enabled variable remains false, but the assertions are ignored, so this has no effect.
* If input is null, calling input.toString() will throw a NullPointerException.
* If input is not null, input.toString() executes without issue.
Conclusion:
A NullPointerException is thrown if input is null, regardless of whether assertions are enabled or disabled.
Therefore, the correct answer is:
C: Only if assertions are disabled and the input argument is null
NEW QUESTION # 40
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?
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 # 41
Given a properties file on the classpath named Person.properties with the content:
ini
name=James
And:
java
public class Person extends ListResourceBundle {
protected Object[][] getContents() {
return new Object[][]{
{"name", "Jeanne"}
};
}
}
And:
java
public class Test {
public static void main(String[] args) {
ResourceBundle bundle = ResourceBundle.getBundle("Person");
String name = bundle.getString("name");
System.out.println(name);
}
}
What is the given program's output?
Answer: B
Explanation:
In this scenario, we have a Person class that extends ListResourceBundle and a properties file named Person.
properties. Both define a resource with the key "name" but with different values:
* Person class (ListResourceBundle):Defines the key "name" with the value "Jeanne".
* Person.properties file:Defines the key "name" with the value "James".
When the ResourceBundle.getBundle("Person") method is called, the Java runtime searches for a resource bundle with the base name "Person". The search order is as follows:
* Class-Based Resource Bundle:The runtime first looks for a class named Person (i.e., Person.class).
* Properties File Resource Bundle:If the class is not found, it then looks for a properties file named Person.properties.
In this case, since the Person class is present and accessible, the runtime will load the Person class as the resource bundle. Therefore, the getBundle method returns an instance of the Person class.
Subsequently, when bundle.getString("name") is called, it retrieves the value associated with the key "name" from the Person class, which is "Jeanne".
Thus, the output of the program is:
nginx
Jeanne
NEW QUESTION # 42
Given:
java
public class Test {
static int count;
synchronized Test() {
count++;
}
public static void main(String[] args) throws InterruptedException {
Runnable task = Test::new;
Thread t1 = new Thread(task);
Thread t2 = new Thread(task);
t1.start();
t2.start();
t1.join();
t2.join();
System.out.println(count);
}
}
What is the given program's output?
Answer: C
Explanation:
In this code, the Test class has a static integer field count and a constructor that is declared with the synchronized modifier. In Java, the synchronized modifier can be applied to methods to control access to critical sections, but it cannot be applied directly to constructors. Attempting to declare a constructor as synchronized will result in a compilation error.
Compilation Error Details:
The Java Language Specification does not permit the use of the synchronized modifier on constructors.
Therefore, the compiler will produce an error indicating that the synchronized modifier is not allowed in this context.
Correct Usage:
If you need to synchronize the initialization of instances, you can use a synchronized block within the constructor:
java
public class Test {
static int count;
Test() {
synchronized (Test.class) {
count++;
}
}
public static void main(String[] args) throws InterruptedException {
Runnable task = Test::new;
Thread t1 = new Thread(task);
Thread t2 = new Thread(task);
t1.start();
t2.start();
t1.join();
t2.join();
System.out.println(count);
}
}
In this corrected version, the synchronized block within the constructor ensures that the increment operation on count is thread-safe.
Conclusion:
The original program will fail to compile due to the illegal use of the synchronized modifier on the constructor. Therefore, the correct answer is E: Compilation fails.
NEW QUESTION # 43
Given:
java
String colors = "red\n" +
"green\n" +
"blue\n";
Which text block can replace the above code?
Answer: B
Explanation:
* Understanding Multi-line Strings in Java (""" Text Blocks)
* Java 13 introducedtext blocks ("""), allowing multi-line stringswithout needing explicit \n 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: \t (Tab Escape)
* \t 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 # 44
......
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