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Title: [Mar-2024] Oracle 1z1-829 Actual Questions and Braindumps [Q25-Q42]

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 [Mar-2024] Oracle 1z1-829 Actual Questions and Braindumps
Pass 1z1-829 Exam with Updated 1z1-829 Exam Dumps PDF 2024


To take the Oracle 1Z0-829 exam, candidates must have a good understanding of Java SE 17, including its syntax, core libraries, and key features. They should also be familiar with software development practices, such as agile methodologies, and have experience with popular development tools such as IntelliJ IDEA and Eclipse.

Oracle 1Z0-829 certification exam is designed for Java developers who are looking to validate their skills in Java SE 17 development. 1z1-829 exam is a credible way to demonstrate your proficiency in Java SE 17 programming and gain a competitive advantage in the job market. It is an intermediate-level exam that requires candidates to have a solid understanding of Java syntax, object-oriented programming concepts, and Java SE 17 API.
 


NO.25 Given:Which action enables the code to compile?
 Replace 15 with item.display (”Flower”);
 Replace 2 with static string name;
 Replace 7 with public void display (string design) {
 Replace 3 with private static void display () {
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 DeveloperJava SE 17 DeveloperOCP Oracle Certified Professional Java SE 17 Developer Study GuideJava MethodsNO.26 Given the code fragment:What is the result?
 Range 1Range 2Range 3
 Range1Note a valid rank.
 Range 1Range 2Range 3Range 1Not a valida rank
 Range 1
ExplanationThe code fragment is using the switch statement with the new Java 17 syntax. The switch statement checks the value of the variable rank and executes the corresponding case statement. In this case, the value of rank is 4, so the first case statement is executed, printing “Range1”. The second and third case statements are also executed, printing “Range2” and “Range3”. The default case statement is also executed, printing “Not a valid rank”. References: Java Language Changes – Oracle Help CenterNO.27 Given the code fragments:Which is true?
 The program prints t1 : 1: t2 : 1: t1 : t2 : 2 : in random order.
 The program prints t1 : 1 : t2: 1 : t1 : 2 : t2: 2:
 The program prints t1 : 1: t2 : 1: t1 : 1 : t2 : 1 : indefinitely
 The program prints an exception
ExplanationThe 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 SE17 & JDK 17) – OracleNO.28 Given the code fragment:What is the result?
 CatDog
 A NullPointerException is thrownCatCat
 DogDog
 Catnull
The answer is E because the code fragment creates a new Pet object with the name “Dog” and assigns it to the variable p. Then, it assigns p to p1. Next, it changes the name of p1 to “Cat”. Then, it assigns p1 to p. Finally, it sets p to null and prints the name of p and p1. The output will be “Cat” and “null” because p is set to null and p1 still points to the Pet object with the name “Cat”.NO.29 Given:What is the result?
 Software Game Chess 0
 Software Game Software Game Chese 2
 Software game write error
 Software Game Software Game chess 0
 Software Game Chess 2
 Software Game read error
ExplanationThe 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 DeveloperJava SE 17 DeveloperOCP Oracle Certified Professional Java SE 17 Developer Study GuideSerialization and Deserialization in Java with ExampleNO.30 Given the code fragment:Which code line n1, obtains the java.io.Console object?
 
 
 
 
 
ExplanationThe code fragment is trying to obtain the java.io.Console object, which is a class that provides methods to access the character-based console device, if any, associated with the current Java virtual machine. The correct way to obtain the Console object is to call the static method Console console() in the java.lang.System class.This method returns the unique Console object associated with the current Java virtual machine, if any.Therefore, option A is correct, as it calls System.console() and assigns it to a Console variable. References:https://docs.oracle.com/javase/17/docs/api/java.base/java/io/Console.htmlhttps://docs.oracle.com/javase/17/docs/api/java.base/java/lang/System.html#console()https://education.oracle.com/products/trackp_OCPJSE17https://mylearn.oracle.com/ou/learning-path/java-se-17-developer/99487NO.31 Which two code fragments compile?
 
 
 
 
 
The two code fragments that compile are B and E. These are the only ones that use the correct syntax for declaring and initializing a var variable. The var keyword is a reserved type name that allows the compiler to infer the type of the variable based on the initializer expression. However, the var variable must have an initializer, and the initializer must not be null or a lambda expression. Therefore, option A is invalid because it does not have an initializer, option C is invalid because it has a null initializer, and option D is invalid because it has a lambda expression as an initializer. Option B is valid because it has a String initializer, and option E is valid because it has an int initializer.https://docs.oracle.com/en/java/javase/17/language/local-variable-type-inference.htmlNO.32 Given the code fragment:Which code fragment invokes all callable objects in the workers set?
 
 
 
 
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) – OracleNO.33 Given:What is the result?
 flipsflips
 Compilation fails
 flipsruns
 runsflips
 runsruns
The code fragment will fail to compile because the play method in the Dog class is declared as private, which means that it cannot be accessed from outside the class. The main method is trying to call the play method on a Dog object, which is not allowed. Therefore, the code fragment will produce a compilation error.NO.34 Given the code fragment:
 True:false:true:true
 True:true:false:false
 True:false:true:false
 True:false:false:false
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.NO.35 Given the code fragment:abstract sealed interface SInt permits Story, Art { default String getTitle() { return “Book Title” ; }}Which set of class definitions compiles?
 Interace story extends STnt {}Interface Art extends SInt {}
 Public interface story extends sInd {}Public interface Art extends SInt {}
 Sealed interface Storty extends sInt {}Non-sealed class Art implements Sint {}
 Non-sealed interface story extends SInt {}Class Art implements Sint {}
 Non-sealed interface story extends SInt {}Non-sealed interaface Art extends Sint {}
ExplanationThe answer is C because the code fragment given is an abstract sealed interface SInt that permits Story and Art. The correct answer is option C, which is a sealed interface Story that extends SInt and a non-sealed class Art that implements SInt. This is because a sealed interface can only be extended by the classes or interfaces that it permits, and a non-sealed class can implement a sealed interface.Option A is incorrect because interface is misspelled as interace, and Story and Art should be capitalized as they are the names of the permitted classes or interfaces.Option B is incorrect because public is misspelled as public, and sInd should be SInt as it is the name of the sealed interface.Option D is incorrect because a non-sealed interface cannot extend a sealed interface, as it would violate the restriction of permitted subtypes.Option E is incorrect because both Story and Art cannot be non-sealed interfaces, as they would also violate the restriction of permitted subtypes.References:Oracle Certified Professional: Java SE 17 DeveloperJava SE 17 DeveloperOCP Oracle Certified Professional Java SE 17 Developer Study GuideSealed Classes and Interfaces in Java 15 | BaeldungSealed Class in Java – JavatpointNO.36 Given:and the code fragment:
 300.00CellPhone,ToyCar,Motor,Fan
 100.00CellPhone,ToyCar,Motor,Fan
 100.00 CellPhone,ToyCar
 300.00CellPhone.ToyCar
The code fragment is using the Stream API to perform a reduction operation on a list of ElectricProduct objects. The reduction operation consists of three parts: an identity value, an accumulator function, and a combiner function. The identity value is the initial value of the result, which is 0.0 in this case. The accumulator function is a BiFunction that takes two arguments: the current result and the current element of the stream, and returns a new result. In this case, the accumulator function is (a,b) -> a + b.getPrice (), which means that it adds the price of each element to the current result. The combiner function is a BinaryOperator that takes two partial results and combines them into one. In this case, the combiner function is (a,b) -> a + b, which means that it adds the two partial results together.The code fragment then applies a filter operation on the stream, which returns a new stream that contains only the elements that match the given predicate. The predicate is p -> p.getPrice () > 10, which means that it selects only the elements that have a price greater than 10. The code fragment then applies a map operation on the filtered stream, which returns a new stream that contains the results of applying the given function to each element. The function is p -> p.getName (), which means that it returns the name of each element.The code fragment then calls the collect method on the mapped stream, which performs a mutable reduction operation on the elements of the stream using a Collector. The Collector is Collectors.joining (“,”), which means that it concatenates the elements of the stream into a single String, separated by commas.The code fragment then prints out the result of the reduction operation and the result of the collect operation, separated by a new line. The result of the reduction operation is 300.00, which is the sum of the prices of all ElectricProduct objects that have a price greater than 10. The result of the collect operation is CellPhone,ToyCar,Motor,Fan, which is the concatenation of the names of all ElectricProduct objects that have a price greater than 10.Therefore, the output of the code fragment is:300.00 CellPhone,ToyCar,Motor,FanNO.37 Given the code fragment:Which code fragment returns different values?
 int sum = listOfNumbers. parallelStream () reduce (5, Integer:: sum) ;
 int sum = listOfNumbers. Stream () reduce (5, (a, b) -> a + b) ;
 int sum = listOfNumbers. Stream () reduce ( Integer:: sum) ; +5;
 int sum = listOfNumbers. parallelStream () reduce ({m, n) -> m +n) orElse (5) +5;
 int sum = listOfNumbers. Stream () reduce (0, Integer:: sum) + 5
ExplanationThe answer is C because the code fragment uses a different syntax and logic for the reduce operation than the other options. The reduce method in option C takes a single parameter, which is a BinaryOperator that combines two elements of the stream into one. The method returns an Optional, which may or may not contain a value depending on whether the stream is empty or not. The code fragment then adds 5 to the result of the reduce method, regardless of whether it is present or not. This may cause an exception if the Optional is empty, or produce a different value than the other options if the Optional is not empty.The other options use a different syntax and logic for the reduce operation. They all take two parameters, which are an identity value and a BinaryOperator that combines an element of the stream with an accumulator.The method returns the final accumulator value, which is equal to the identity value if the stream is empty, or the result of applying the BinaryOperator to all elements of the stream otherwise. The code fragments then add5 to the result of the reduce method, which will always produce a valid value.For example, suppose listOfNumbers contains [1, 2, 3]. Then, option A will perform the following steps:Initialize accumulator to identity value 5Apply BinaryOperator Integer::sum to accumulator and first element: 5 + 1 = 6 Update accumulator to 6 Apply BinaryOperator Integer::sum to accumulator and second element: 6 + 2 = 8 Update accumulator to 8 Apply BinaryOperator Integer::sum to accumulator and third element: 8 + 3 = 11 Update accumulator to 11 Return final accumulator value 11 Add 5 to final accumulator value: 11 + 5 = 16 Option B will perform the same steps as option A, except using a lambda expression instead of a method reference for the BinaryOperator. Option D will perform the same steps as option A, except using parallelStream instead of stream, which may change the order of applying the BinaryOperator but not the final result. Option E will perform the same steps as option A, except using identity value 0 instead of 5.Option C, however, will perform the following steps:Apply BinaryOperator Integer::sum to first and second element: 1 + 2 = 3 Apply BinaryOperator Integer::sum to previous result and third element: 3 + 3 = 6 Return Optional containing final result value 6 Add 5 to Optional value: Optional.of(6) + 5 = Optional.of(11) As you can see, option C produces a different value than the other options, and also uses a different syntax and logic for the reduce operation. References Oracle Certified Professional: Java SE 17 Developer Java SE 17 Developer OCP Oracle Certified Professional Java SE 17 Developer Study Guide Guide to Stream.reduce()NO.38 Which statement is true?
 IllegalStateException is thrown if a thread in waiting state is moved back to runnable.
 thread in waiting state consumes CPU cycles.
 A thread in waiting state must handle InterrupedException.
 After the timed wait expires, the waited thread moves to the terminated state.
A thread in waiting state is waiting for another thread to perform a particular action, such as calling notify() or notifyAll() on a shared object, or terminating a joined thread. A thread in waiting state can be interrupted by another thread, which will cause the waiting thread to throw an InterruptedException and return to the runnable state. Therefore, a thread in waiting state must handle InterruptedException, either by catching it or declaring it in the throws clause. Reference: Thread.State (Java SE 17 & JDK 17), [Thread (Java SE 17 & JDK 17)]NO.39 Given the code fragment:Which code fragment returns different values?
 int sum = listOfNumbers. parallelStream () reduce (5, Integer:: sum) ;
 int sum = listOfNumbers. Stream () reduce (5, (a, b) -> a + b) ;
 int sum = listOfNumbers. Stream () reduce ( Integer:: sum) ; +5;
 int sum = listOfNumbers. parallelStream () reduce ({m, n) -> m +n) orElse (5) +5;
 int sum = listOfNumbers. Stream () reduce (0, Integer:: sum) + 5
The answer is C because the code fragment uses a different syntax and logic for the reduce operation than the other options. The reduce method in option C takes a single parameter, which is a BinaryOperator that combines two elements of the stream into one. The method returns an Optional, which may or may not contain a value depending on whether the stream is empty or not. The code fragment then adds 5 to the result of the reduce method, regardless of whether it is present or not. This may cause an exception if the Optional is empty, or produce a different value than the other options if the Optional is not empty.The other options use a different syntax and logic for the reduce operation. They all take two parameters, which are an identity value and a BinaryOperator that combines an element of the stream with an accumulator. The method returns the final accumulator value, which is equal to the identity value if the stream is empty, or the result of applying the BinaryOperator to all elements of the stream otherwise. The code fragments then add 5 to the result of the reduce method, which will always produce a valid value.For example, suppose listOfNumbers contains [1, 2, 3]. Then, option A will perform the following steps:Initialize accumulator to identity value 5Apply BinaryOperator Integer::sum to accumulator and first element: 5 + 1 = 6 Update accumulator to 6 Apply BinaryOperator Integer::sum to accumulator and second element: 6 + 2 = 8 Update accumulator to 8 Apply BinaryOperator Integer::sum to accumulator and third element: 8 + 3 = 11 Update accumulator to 11 Return final accumulator value 11 Add 5 to final accumulator value: 11 + 5 = 16 Option B will perform the same steps as option A, except using a lambda expression instead of a method reference for the BinaryOperator. Option D will perform the same steps as option A, except using parallelStream instead of stream, which may change the order of applying the BinaryOperator but not the final result. Option E will perform the same steps as option A, except using identity value 0 instead of 5.Option C, however, will perform the following steps:Apply BinaryOperator Integer::sum to first and second element: 1 + 2 = 3 Apply BinaryOperator Integer::sum to previous result and third element: 3 + 3 = 6 Return Optional containing final result value 6 Add 5 to Optional value: Optional.of(6) + 5 = Optional.of(11) As you can see, option C produces a different value than the other options, and also uses a different syntax and logic for the reduce operation. Reference:Oracle Certified Professional: Java SE 17 DeveloperJava SE 17 DeveloperOCP Oracle Certified Professional Java SE 17 Developer Study GuideGuide to Stream.reduce()NO.40 Given the code fragment:What is the result:
 ADEABCB // the order of element is unpredictable
 ABCE
 ABCDE // the order of elements is unpredictable
 ABBCDE // the order of elements is unpredictable
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 DeveloperJava SE 17 DeveloperOCP Oracle Certified Professional Java SE 17 Developer Study GuideParallelizing StreamsNO.41 Which statement is true?
 The tryLock () method returns a boolean indicator immediately regardless if it has or has not managed to acquire the lock.
 The tryLock () method returns a boolean indicator immediately if it has managed to acquire the lock, otherwise it waits for the lock acquisition.
 The lock () method returns a boolean indicator immediately if it has managed to acquire the lock, otherwise it waits for the lock acquisition.
 The Lock () method returns a boolean indicator immediately regardless if it has or has not managed to acquire the lock
ExplanationThe 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, 4NO.42 Given the code fragment:Which action sorts the book list?
 At Line n2, replace books,sort() with books.stream().sort(0.
 At line n1, convert books type to mutable ArrayList type.
 At Line n1, convert type to mutable array type.
 At Line n2, replace compareTo () with compare ().
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) Loading …






	
	

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