1Z0-829 Practice Test 29 — 15 Questions

Question 1

Refer to the exhibit. A Java application is deployed in /opt/app/lib/ and attempts to perform the following operations:
1) Read the file /data/config/settings.xml
2) Write to the file /logs/app.log
3) Read the file /data/config/subdir/extra.conf
Which statement is true?

Accepted Answer

Operations 1 and 2 are allowed; operation 3 is denied.. The Java application is deployed in /opt/app/lib/, so its code base is /opt/app/lib/. The security policy grants read permission for /data/config/*, which matches /data/config/settings.xml (operation 1) but does not match /data/config/subdir/extra.conf because the wildcard * does not descend into subdirectories (operation 3 is denied). Write permission is granted for /logs/*, so /logs/app.log (operation 2) is allowed. Thus, operations 1 and 2 are allowed, and operation 3 is denied.

Answer

All three operations are allowed because the code base matches.

Answer

Only operation 1 is allowed; operations 2 and 3 are denied.

Answer

Operations 1 and 3 are allowed; operation 2 is denied.

Question 2

A financial application processes a large log file containing millions of timestamp entries in two different formats: "2024-01-15T10:30:00Z" (ISO 8601) and "01/15/2024 10:30:00 AM" (US locale). The current implementation uses a single DateTimeFormatter with a pattern that throws an exception for the other format. The team wants to parse both formats efficiently without using try-catch for each line. Performance is critical, and the code must be thread-safe. Which approach should be used?

Accepted Answer

Use DateTimeFormatterBuilder.appendOptional to combine both formatters into one thread-safe formatter.. `DateTimeFormatterBuilder.appendOptional` allows combining multiple formatters into a single thread-safe `DateTimeFormatter`. This approach parses each input against the optional formatters in order, succeeding on the first match without exceptions, which is ideal for performance-critical, thread-safe log processing.

Answer

Use a single DateTimeFormatter and catch DateTimeParseException for the failing format, then retry with a second formatter.

Answer

Use DateTimeFormatter.ofPattern("yyyy-MM-dd'T'HH:mm:ss'Z'" or "MM/dd/yyyy hh:mm:ss a") with a union pattern.

Answer

Use a SimpleDateFormat with a try-catch block for each format inside a synchronized method.

Question 3

What is the output of the code above?

Accepted Answer

18. The code creates a stream of lists containing numbers 1 through 9, flatMaps to a single stream of integers, filters for numbers divisible by 3 (leaving 3, 6, 9), and reduces them by summing. The sum is 3+6+9=18, making option A correct.

Answer

15

Answer

9

Answer

27

Question 4

Refer to the exhibit. What is the output?

Accepted Answer

-15. The code uses `reduce(0, (a, b) -> a - b)` on a stream of integers. This operation starts with an identity value of 0 and applies the subtraction lambda cumulatively: 0 - 1 = -1, then -1 - 2 = -3, then -3 - 3 = -6, then -6 - 4 = -10, then -10 - 5 = -15. Therefore, the output is -15, making option C correct.

Answer

1

Answer

15

Answer

0

Question 5

A company uses serialization to transfer objects between microservices. To prevent deserialization attacks, they want to restrict which classes can be deserialized. Which approach should be used in Java 17?

Accepted Answer

Use ObjectInputFilter with a custom filter. Java 17 provides the `ObjectInputFilter` API (introduced in Java 9) as the standard mechanism to restrict which classes can be deserialized. By setting a custom filter via `ObjectInputStream.setObjectInputFilter()` or a system-wide `jdk.serialFilter` property, you can whitelist or blacklist classes based on patterns, effectively preventing deserialization attacks without modifying the serialization stream itself.

Answer

Override resolveClass in ObjectInputStream

Answer

Use SecurityManager with a policy file

Answer

Implement Externalizable and control fields

Question 6

Which TWO statements about the break and continue statements in Java are correct?

Accepted Answer

The break statement without a label always terminates the innermost enclosing loop or switch.. The break statement without a label, when executed inside a loop or switch, immediately terminates the innermost enclosing loop or switch construct. This is defined by the Java Language Specification (JLS §14.15), which states that an unlabeled break statement transfers control to the immediately enclosing statement of the break target.

Answer

The continue statement can be used inside a switch statement to skip the rest of the switch body.

Answer

The break statement can be used only inside loops and switch statements.

Answer

The continue statement without a label skips the current iteration and proceeds to the next iteration without reevaluating the loop condition.

Question 7

A developer needs to filter a list of transactions where the amount is greater than 100 and collect the results into a new list. Which approach is best practice for readability and performance?

Accepted Answer

transactions.stream().filter(t -> t.getAmount() > 100).collect(Collectors.toList()). It uses the Stream API's `filter` method to declaratively select transactions with an amount greater than 100, and `collect(Collectors.toList())` to gather results into a new list. This approach is both readable and performant, leveraging lazy evaluation and internal iteration for efficient processing.

Answer

List result = new ArrayList<>(); for(Transaction t : transactions) { if(t.getAmount() > 100) result.add(t); }

Answer

transactions.stream().filter(t -> t.getAmount() > 100).toArray()

Answer

transactions.removeIf(t -> t.getAmount() <= 100)

Question 8

Given: Object obj = "Hello"; String result = switch(obj) { case String s -> "String of length " + s.length(); case Integer i -> "Integer"; default -> "Unknown"; }; What is result?

Accepted Answer

"String of length 5". The switch expression uses pattern matching with type patterns. The runtime type of obj is String, so the first case matches, binding the value to s and executing the expression "String of length " + s.length(). Since s is "Hello", s.length() returns 5, so result is "String of length 5". Option C is correct.

Answer

"Unknown"

Answer

Compilation error

Answer

"Integer"

Question 9

A developer wrote a method that uses a for-each loop to iterate over a list of strings and remove elements that match a certain condition. However, the method throws a ConcurrentModificationException at runtime. What is the most likely cause?

Accepted Answer

The loop modifies the list by calling remove() on the list itself.. A for-each loop internally uses an Iterator to traverse the list. If the list is structurally modified (e.g., by calling remove() directly on the list) during iteration, the iterator's expected modCount will differ from the actual modCount, triggering a ConcurrentModificationException. This is a fail-fast behavior of the ArrayList and other Collection implementations.

Answer

The loop uses a for-each loop with an array instead of a list.

Answer

The loop uses an iterator that is not properly initialized.

Answer

The list is unmodifiable.

Question 10

What is the result of attempting to compile and run the following code?
```java
import java.util.*;
List<String> list = Arrays.asList("A", "B");
list.removeIf(s -> s.startsWith("A"));
System.out.println(list);
```

Accepted Answer

Runtime exception. The code compiles successfully because `removeIf` is a default method in `Collection` and the predicate is a valid lambda. However, `Arrays.asList` returns a fixed-size list backed by the array, and `removeIf` attempts to modify the list structurally, which throws `UnsupportedOperationException` at runtime. Therefore, the correct answer is E: Runtime exception.

Answer

[A, B]

Answer

[A]

Answer

[B]

Answer

Compilation fails

Question 11

What is the output of the above code?

Accepted Answer

[a, c]. The code uses `List.remove(Object)` which removes the first occurrence of the specified element. After removing "b" from the list [a, b, c], the list becomes [a, c]. Option B is correct because the output is [a, c].

Answer

[a, b]

Answer

[a, b, c]

Answer

[b, c]

Answer

Compilation fails

Question 12

A company has a service layer with a method performOperation() that originally declared throws BaseException (a checked exception). Several client methods call performOperation() and catch BaseException, wrapping it in a RuntimeException for propagation. After a system upgrade, the implementation of performOperation() now explicitly throws two new checked exceptions: SubException1 and SubException2, both of which extend BaseException. The client methods must continue to compile without modifying their catch signatures, and they must still handle the new exceptions appropriately. What is the best way to modify the client methods?

Accepted Answer

No change needed because the catch for BaseException catches all subclasses.. In Java, a catch block for a superclass exception type (BaseException) will catch all subclasses (SubException1 and SubException2) due to polymorphism in exception handling. Since the client methods already catch BaseException and wrap it in a RuntimeException, no changes are needed to the catch signatures; the new exceptions are automatically handled. This satisfies the requirement that client methods continue to compile without modifying their catch signatures while still handling the new exceptions appropriately.

Answer

Use a multi-catch block to catch SubException1 and SubException2 and rethrow as RuntimeException.

Answer

Modify the client method signature to declare throws SubException1, SubException2.

Answer

Add separate catch blocks for SubException1 and SubException2 before the existing BaseException catch.

Question 13

You are developing a Java application that processes sensitive user data. The application runs on a server with strict security policies. You need to read configuration properties from a file located at /etc/app/config.properties. The application uses a SecurityManager. During testing, you get a security exception: java.security.AccessControlException: access denied (java.io.FilePermission /etc/app/config.properties read). You have already added a file permission grant in the policy file for the application codebase. However, the exception persists. What is the most likely cause?

Accepted Answer

A library or caller on the execution stack does not have the required FilePermission.. Even if the application's own codebase has the required FilePermission, any library or caller on the execution stack that does not have the permission will cause an AccessControlException. The SecurityManager performs stack inspection: every class on the call stack must have the required permission. If a library or framework method invoked by your code lacks the grant, the check fails.

Answer

The application codebase URL in the policy file is incorrect.

Answer

The file path is relative and needs to be resolved against user.dir.

Answer

The policy file is not being loaded because the JVM was not started with -Djava.security.policy.

Question 14

Which THREE are valid ways to package a Java 17 application for distribution? (Choose three.)

Accepted Answer

JMOD file. A is correct because a JMOD file is a native packaging format introduced in Java 9 (JEP 261) specifically for distributing Java modules. It can include compiled classes, native code, configuration files, and even other JMOD files, making it a valid distribution format for Java 17 applications, especially when using the Java Module System.

Answer

ZIP file containing compiled classes

Answer

Shell script that compiles and runs the application

Question 15

A Java application needs to refer to a file using the path "data/input.txt". To ensure platform independence (correct file separator), which approach is recommended?

Accepted Answer

Paths.get("data", "input.txt"). `Paths.get("data", "input.txt")` uses the default file system's path separator, ensuring platform independence. This method internally calls `FileSystems.getDefault().getPath()` to construct the path with the correct separator for the operating system (e.g., '/' on Linux/macOS, '\' on Windows).

Answer

FileSystems.getDefault().getPath("data", "input.txt")

Answer

new File("data/input.txt").toPath().normalize()

Answer

new File("data/input.txt").toPath()