Updated Jun-2024 1Z0-829 Free Exam Files Downloaded Instantly [Q15-Q39]

The code fragment compares four pairs of strings using the equals() and intern() methods. The equals() method compares the content of two strings, while the intern() method returns a canonical representation of a string, which means that it returns a reference to an existing string with the same content in the string pool. The string pool is a memory area where strings are stored and reused to save space and improve performance. The results of the comparisons are as follows:
s1.equals(s2): This returns true because both s1 and s2 have the same content, “Hello Java 17”.
s1 == s2: This returns false because s1 and s2 are different objects with different references, even though they have the same content. The == operator compares the references of two objects, not their content.
s1.intern() == s2.intern(): This returns true because both s1.intern() and s2.intern() return a reference to the same string object in the string pool, which has the content “Hello Java 17”. The intern() method ensures that there is only one copy of each distinct string value in the string pool.
“Hello Java 17” == s2: This returns false because “Hello Java 17” is a string literal, which is automatically interned and stored in the string pool, while s2 is a string object created with the new operator, which is not interned by default and stored in the heap. Therefore, they have different references and are not equal using the == operator.

The code fragment is trying to read an object from a file using the ObjectInputStream class. This class throws an IOException and a ClassNotFoundException. To handle these exceptions, the main method signature should declare that it throws these exceptions. Otherwise, the code will not compile. If the main method throws these exceptions, the code will print Wagon : 200, which is the result of calling the toString method of the LuxuryCar object that was written to the file. Reference: ObjectInputStream (Java SE 17 & JDK 17) – Oracle, ObjectOutputStream (Java SE 17 & JDK 17) – Oracle

The code fragment is trying to sort a list of books using the Collections.sort() method. The correct answer is D, because the compareTo() method is not the correct way to compare two objects in a Comparator. The compare() method is the correct way to compare two objects in a Comparator and return an int value that indicates their order1. The compareTo() method is used to implement the Comparable interface, which defines the natural order of objects of a class2. The other options are incorrect because they either do not change the type of the list, which is already mutable, or they do not use the correct syntax for sorting a stream, which requires a terminal operation such as collect()3. Reference: Comparator (Java SE 17 & JDK 17), Comparable (Java SE 17 & JDK 17), Stream (Java SE 17 & JDK 17)

Explanation
The code given is an interface and a class that implements the interface. The interface has three methods, m1(), m2(), and m3(). The class has one method, m1(). The only two method invocations that will execute are D and E.
D is a call to the m2() method in the class, and E is a call to the m3() method in the interface. References:
https://education.oracle.com/products/trackp_OCPJSE17, 3, 4, 5

Explanation
The answer is B because the code uses the writeObject and readObject methods of the ObjectOutputStream and ObjectInputStream classes to serialize and deserialize the Game object. These methods use the default serialization mechanism, which writes and reads the state of the object’s fields, including the inherited ones. Therefore, the title field of the Software class is also serialized and deserialized along with the players field of the Game class. The toString method of the Game class calls the toString method of the Software class using super.toString(), which returns the value of the title field.
Hence, when the deserialized object is printed, it shows “Software Game Software Game Chess 2”.
References:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
Serialization and Deserialization in Java with Example

Explanation
The code fragment is comparing the values of a, b, and c using the < and > operators. The first comparison, d, is checking if a is less than b and greater than c. Since a is equal to 2, b is equal to -2, and c is equal to -4, this comparison will evaluate to true. The second comparison, e, is checking if a is greater than b and a is greater than c. Since a is equal to 2, b is equal to -2, and c is equal to -4, this comparison will evaluate to false.
Therefore, the result will be true 1 false 2. References: Operators (The Java™ Tutorials > Learning the Java Language – Oracle

The answer is B because a bottom-up migration is a strategy for modularizing an existing application by moving its dependencies to the module path one by one, starting from the lowest-level libraries and ending with the application itself. This way, each module can declare its dependencies on other modules using the module-info.java file, and benefit from the features of the Java Platform Module System (JPMS), such as reliable configuration, strong encapsulation, and service loading.
Option A is incorrect because a top-down migration is a strategy for modularizing an existing application by moving it to the module path first, along with its dependencies as automatic modules. Automatic modules are non-modular JAR files that are treated as modules with some limitations, such as not having a module descriptor or a fixed name. A top-down migration allows the application to use the module path without requiring all of its dependencies to be modularized first.
Option C is incorrect because a top-down migration does not require any specific order of migrating modules, as long as the application is moved first and its dependencies are moved as automatic modules. A bottom-up migration, on the other hand, requires the required modules to migrate before the modules that depend on them.
Option D is incorrect because unnamed modules are not automatic modules in any migration strategy. Unnamed modules are modules that do not have a name or a module descriptor, such as classes loaded from the class path or dynamically generated classes. Unnamed modules have unrestricted access to all other modules, but they cannot be accessed by named modules, except through reflection with reduced security checks. Reference:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
Migrating to Modules (How and When) – JavaDeploy
Java 9 Modularity: Patterns and Practices for Developing Maintainable Applications

Explanation
The answer is B because the code fragment contains a logical error that causes a MissingResourceException at runtime. The code fragment tries to load a resource bundle with the base name “Captions.properties” and the locale “en_US”. However, there is no such resource bundle available in the classpath. The available resource bundles are:
Captions.properties
Captions_en.properties
Captions_US.properties
Captions_en_US.properties
The ResourceBundle class follows a fallback mechanism to find the best matching resource bundle for a given locale. It first tries to find the resource bundle with the exact locale, then it tries to find the resource bundle with the same language and script, then it tries to find the resource bundle with the same language, and finally it tries to find the default resource bundle with no locale. If none of these resource bundles are found, it throws a MissingResourceException.
In this case, the code fragment is looking for a resource bundle with the base name “Captions.properties” and the locale “en_US”. The ResourceBundle class will try to find the following resource bundles in order:
Captions.properties_en_US
Captions.properties_en
Captions.properties
However, none of these resource bundles exist in the classpath. Therefore, the ResourceBundle class will throw a MissingResourceException.
To fix this error, the code fragment should use the correct base name of the resource bundle family, which is
“Captions” without the “.properties” extension. For example:
ResourceBundle captions = ResourceBundle.getBundle(“Captions”, currentLocale); This will load the appropriate resource bundle for the current locale, which is “Captions_en_US.properties” in this case. References:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
ResourceBundle (Java Platform SE 8 )
About the ResourceBundle Class (The Java™ Tutorials > Internationalization)

The code fragment is comparing the values of a, b, and c using the < and > operators. The first comparison, d, is checking if a is less than b and greater than c. Since a is equal to 2, b is equal to -2, and c is equal to -4, this comparison will evaluate to true. The second comparison, e, is checking if a is greater than b and a is greater than c. Since a is equal to 2, b is equal to -2, and c is equal to -4, this comparison will evaluate to false. Therefore, the result will be true 1 false 2. Reference: Operators (The Java™ Tutorials > Learning the Java Language – Oracle

Explanation
The code snippet is an example of Java SE 17 code. The code is checking if the object is an instance of class C and if it is, it will print “mC”. If it is not an instance of class C, it will print “mA”. In this case, the object is not an instance of class C, so the output will be “mA”. References: Pattern Matching for instanceof – Oracle Help Center

The code fragment in Option C invokes all callable objects in the workers set by using the ExecutorService’s invokeAll() method. This method takes a collection of Callable objects and returns a list of Future objects representing the results of the tasks. The other options are incorrect because they either use the wrong method (invokeAny() or submit()) or have syntax errors (missing parentheses or semicolons). Reference: AbstractExecutorService (Java SE 17 & JDK 17) – Oracle

The answer is C because the code demonstrates the concept of method overloading and type conversion in Java. Method overloading allows different methods to have the same name but different parameters. Type conversion allows values of one data type to be assigned to another data type, either automatically or explicitly. In the code, the class Test has four methods named sum, each with different parameter types: int, float, and double. The main method creates an instance of Test and calls the sum method with different arguments. The compiler will choose the most specific method that matches the arguments, based on the following rules:
If there is an exact match between the argument types and the parameter types, that method is chosen.
If there is no exact match, but there is a method with compatible parameter types, that method is chosen. Compatible types are those that can be converted from one to another automatically, such as int to long or float to double.
If there is more than one method with compatible parameter types, the most specific method is chosen. The most specific method is the one whose parameter types are closest to the argument types in terms of size or precision.
In the code, the following method calls are made:
test.sum(10, 10.5) -> This matches the sum(int a, float b) method exactly, so it is chosen. The result is 20.5, which is converted to int and printed as 20 (B).
test.sum(10) -> This does not match any method exactly, but it matches the sum(double a) method with compatible types, as int can be converted to double automatically. The result is 10.0, which is printed as 10 (A).
test.sum(10.5, 10) -> This does not match any method exactly, but it matches two methods with compatible types: sum(float a, float b) and sum(double a, double b). The latter is more specific, as double is closer to the argument types than float. The result is 20.5, which is printed as 20 (D).
Therefore, the output is B A D. Reference:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
Method Overloading in Java
Type conversion in Java with Examples
Java Method Overloading with automatic type conversions

Explanation
The answer is B because the code fragment contains a syntax error that prevents it from compiling. The code fragment tries to catch a StockException in line 10, but the catch block does not have a parameter of type StockException. The catch block should have a parameter of type StockException, such as:
catch (StockException e) { // handle the exception }
This is required by the Java syntax for the catch clause, which must have a parameter that is a subclass of Throwable. Without a parameter, the catch block is invalid and causes a compilation error.
Option A is incorrect because the program does not throw a StockException, as it does not compile.
Option C is incorrect because the program does not throw an OutofStockException, as it does not compile.
Option D is incorrect because the program does not throw a ClassCastException, as it does not compile.
References:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
The try-with-resources Statement (The Java™ Tutorials > Essential Classes > Exceptions) The catch Blocks (The Java™ Tutorials > Essential Classes > Exceptions)

Explanation
The answer is D because the code fragment reads the content of the in.txt file and writes it to the out.txt file.
The content of the in.txt file is “23456789”. The code fragment uses a char array buffer of size 8 to read the content of the in.txt file. The while loop reads the content of the in.txt file and writes it to the out.txt file until the end of the file is reached. Therefore, the content of the out.txt file will be “0123456789”.

Explanation
A module path is a sequence of directories that contain modules or JAR files. A named module is a module that has a name and a module descriptor (module-info.class) that declares its dependencies and exports. An automatic module is a module that does not have a module descriptor, but is derived from the name and contents of a JAR file. Both named and automatic modules can be placed on the module path, and they can be resolved by the Java runtime. An unnamed module is a special module that contains all the classes that are not in any other module, such as those on the class path. An unnamed module is not on the module path, but it can read all other modules.

Explanation
The tryLock () method of the Lock interface is a non-blocking attempt to acquire a lock. It returns true if the lock is available and acquired by the current thread, and false otherwise. It does not wait for the lock to be released by another thread. This is different from the lock () method, which blocks the current thread until the lock is acquired, and does not return any value. References:
https://docs.oracle.com/en/java/javase/17/docs/api/java.base/java/util/concurrent/locks/Lock.html#tryLock(), 3, 4

The answer is C because the code fragment contains a syntax error in line 7, where the method display is declared without any parameter type. This causes a compilation error, as Java requires the parameter type to be specified for each method parameter. To fix this error, the parameter type should be added before the parameter name, such as string design. This will enable the code to compile and run without any errors. Reference:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
Java Methods

The tryLock () method of the Lock interface is a non-blocking attempt to acquire a lock. It returns true if the lock is available and acquired by the current thread, and false otherwise. It does not wait for the lock to be released by another thread. This is different from the lock () method, which blocks the current thread until the lock is acquired, and does not return any value. Reference: https://docs.oracle.com/en/java/javase/17/docs/api/java.base/java/util/concurrent/locks/Lock.html#tryLock(), 3, 4, 5

Explanation
The code fragment is trying to read an object from a file using the ObjectInputStream class. This class throws an IOException and a ClassNotFoundException. To handle these exceptions, the main method signature should declare that it throws these exceptions. Otherwise, the code will not compile. If the main method throws these exceptions, the code will print Wagon : 200, which is the result of calling the toString method of the LuxuryCar object that was written to the file. References: ObjectInputStream (Java SE 17 & JDK 17) – Oracle, ObjectOutputStream (Java SE 17 & JDK 17) – Oracle

Explanation
The code fragment compares four pairs of strings using the equals() and intern() methods. The equals() method compares the content of two strings, while the intern() method returns a canonical representation of a string, which means that it returns a reference to an existing string with the same content in the string pool. The string pool is a memory area where strings are stored and reused to save space and improve performance. The results of the comparisons are as follows:
s1.equals(s2): This returns true because both s1 and s2 have the same content, “Hello Java 17”.
s1 == s2: This returns false because s1 and s2 are different objects with different references, even though they have the same content. The == operator compares the references of two objects, not their content.
s1.intern() == s2.intern(): This returns true because both s1.intern() and s2.intern() return a reference to the same string object in the string pool, which has the content “Hello Java 17”. The intern() method ensures that there is only one copy of each distinct string value in the string pool.
“Hello Java 17” == s2: This returns false because “Hello Java 17” is a string literal, which is automatically interned and stored in the string pool, while s2 is a string object created with the new operator, which is not interned by default and stored in the heap. Therefore, they have different references and are not equal using the == operator.
References: String (Java SE 17 & JDK 17) – Oracle

Explanation
The answer is C because the code fragment contains a syntax error in line 7, where the method display is declared without any parameter type. This causes a compilation error, as Java requires the parameter type to be specified for each method parameter. To fix this error, the parameter type should be added before the parameter name, such as string design. This will enable the code to compile and run without any errors.
References:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
Java Methods

The code fragment uses the Collections.binarySearch method to search for the string “e3” in the list. The first search returns the index of the element, which is 2. The second search returns the index of the element, which is 0. The third search returns the index of the element, which is -4. The final result is 2. Reference: Collections (Java SE 17 & JDK 17) – Oracle

The answer is D because the code fragment uses the Stream API to create two streams, s1 and s2, and then concatenates them using the concat() method. The resulting stream is then processed in parallel using the parallel() method, and the distinct() method is used to remove duplicate elements. Finally, the forEach() method is used to print the elements of the resulting stream to the console. Since the order of elements in a parallel stream is unpredictable, the output could be any of the options given, but option D is the most likely. Reference:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
Parallelizing Streams

Explanation
The code creates two threads, t1 and t2, and starts them. The threads will print their names and the value of the Atomic Integer object, x, which is initially set to 1. The threads will then increment the value of x and print their names and the new value of x. Since the threads are started at the same time, the output will be in random order. However, the final output will always be t1 : 1 : t2: 1 : t1 : 2 : t2: 2: References: AtomicInteger (Java SE
17 & JDK 17) – Oracle

The code snippet is an example of Java SE 17 code. The code is checking if the object is an instance of class C and if it is, it will print “mC”. If it is not an instance of class C, it will print “mA”. In this case, the object is not an instance of class C, so the output will be “mA”. Reference: Pattern Matching for instanceof – Oracle Help Center