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Oracle 1Z0-829 is an exam that is designed for Java programmers who have at least two years of programming experience. 1Z0-829 exam allows candidates to demonstrate their skills and knowledge to design and develop modern Java applications using Java SE 17. 1Z0-829 exam covers various topics such as modern Java features, design patterns, error handling, concurrency, and security, among others.
To prepare for the Oracle 1Z0-829 certification exam, candidates are recommended to have a solid understanding of Java SE 17 concepts and programming techniques. There are various training resources available, including official Oracle training courses, study guides, practice exams, and online forums, which can help you to prepare for the exam.
Oracle 1Z0-829 (Java SE 17 Developer) Certification Exam is designed to test the candidate's knowledge and skills in developing applications using Java SE 17. It is a globally recognized certification that demonstrates the candidate's expertise in Java programming language, including advanced features such as lambda expressions, modules, and streams. Java SE 17 Developer certification exam is intended for developers who have a strong understanding of Java programming and are looking to validate their skills and knowledge.
NEW QUESTION # 14
Given the code fragment:
What is the result?
- A. True true false NewYear
- B. False true true optional (Newyear)
- C. 010 optional (Newyear)
- D. 0110
Answer: B
Explanation:
The code fragment is using the stream methods allMatch, anyMatch, noneMatch, and findFirst on a list of strings called specialDays. These methods are used to perform matching operations on the elements of a stream, such as checking if all, any, or none of the elements satisfy a given predicate, or finding the first element that matches a predicate1. The predicate in this case is that the string equals "Labour" or "Halloween". The output will be:
False: because not all of the elements in specialDays are equal to "Labour" or "Halloween".
true: because at least one of the elements in specialDays is equal to "Labour" or "Halloween".
true: because none of the elements in specialDays are equal to both "Labour" and "Halloween".
Optional[NewYear]: because the first element in specialDays that matches the predicate is "NewYear", and the findFirst method returns an Optional object that may or may not contain a non-null value2.
NEW QUESTION # 15
Given:
What is the result
- A. Marketing
Finance
Technical - B. Marketing
- C. UnDefined
- D. Marketing
Undefined
Answer: C
Explanation:
The code fragment is using the switch statement with the new Java 17 syntax. The switch statement checks the value of the variable desig and executes the corresponding case statement. In this case, the value of desig is "CTO", which does not match any of the case labels. Therefore, the default case statement is executed, which prints "Undefined". The other case statements are not executed, because there is no fall through in the new syntax. Therefore, the output of the code fragment is:
Undefined
NEW QUESTION # 16
Given:
What is the result?
- A. Software Game Chess 0
- B. Software Game Software Game Chese 2
- C. Software game write error
- D. Software Game Software Game chess 0
- E. Software Game read error
- F. Software Game Chess 2
Answer: B
Explanation:
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
NEW QUESTION # 17
Given:
Which two modifications enable the code to print Open Close?
- A.
- B.
- C.
- D.
- E.
Answer: A,D
Explanation:
The code given is a try-with-resources statement that declares a resource of type AutoCloseable. The resource is an anonymous class that implements the AutoCloseable interface and overrides the close() method. The code also has a print() method that prints the value of the variable s. The code is supposed to print "Open Close", but it does not compile because of two errors.
The first error is at line n1, where the anonymous class is missing a semicolon at the end of its declaration. This causes a syntax error and prevents the code from compiling. To fix this error, option B adds a semicolon after the closing curly brace of the anonymous class.
The second error is at line n2, where the print() method is called without an object reference. This causes a compilation error because the print() method is not static and cannot be invoked without an object. To fix this error, option E adds an object reference to the print() method by using the variable t.
Therefore, options B and E are correct and enable the code to print "Open Close".
NEW QUESTION # 18
Which statement is true about modules?
- A. Automatic and unnamed modules are on the module path.
- B. Automatic and named modules are on the module path.
- C. Only unnamed modules are on the module path.
- D. Only automatic modules are on the module path.
- E. Only named modules are on the module path.
Answer: B
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.
NEW QUESTION # 19
Given:
What is the result
- A. Marketing
Finance
Technical - B. Marketing
- C. UnDefined
- D. Marketing
Undefined
Answer: C
Explanation:
Explanation
The code fragment is using the switch statement with the new Java 17 syntax. The switch statement checks the value of the variable desig and executes the corresponding case statement. In this case, the value of desig is
"CTO", which does not match any of the case labels. Therefore, the default case statement is executed, which prints "Undefined". The other case statements are not executed, because there is no fall through in the new syntax. Therefore, the output of the code fragment is:
Undefined
NEW QUESTION # 20
Given the code fragment:
What is the result:
- A. ABCDE // the order of elements is unpredictable
- B. ABCE
- C. ABBCDE // the order of elements is unpredictable
- D. ADEABCB // the order of element is unpredictable
Answer: C
Explanation:
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
NEW QUESTION # 21
Given:
and the code fragment:
- A. 300.00
CellPhone,ToyCar,Motor,Fan - B. 100.00 CellPhone,ToyCar
- C. 300.00
CellPhone.ToyCar - D. 100.00
CellPhone,ToyCar,Motor,Fan
Answer: A
Explanation:
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,Fan
NEW QUESTION # 22
Given the code fragments:
Which is true?
- A. The program prints t1 : 1: t2 : 1: t1 : t2 : 2 : in random order.
- B. The program prints t1 : 1 : t2: 1 : t1 : 2 : t2: 2:
- C. The program prints t1 : 1: t2 : 1: t1 : 1 : t2 : 1 : indefinitely
- D. The program prints an exception
Answer: B
Explanation:
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
NEW QUESTION # 23
Given:
What is the result?
- A. 0
- B. 1
- C. A runtime exception is thrown.
- D. 2
- E. 3
- F. 4
- G. Compilation fails.
- H. Nothing is printed because of an indefinite loop.
Answer: G
Explanation:
Explanation
The code will not compile because the variable 'x' is declared as final and then it is being modified in the switch statement. This is not allowed in Java. A final variable is a variable whose value cannot be changed once it is initialized1. The switch statement tries to assign different values to 'x' depending on the value of 'y', which violates the final modifier. The compiler will report an error: The final local variable x cannot be assigned. It must be blank and not using a compound assignment. References: The final Keyword (The Java™ Tutorials > Learning the Java Language > Classes and Objects)
NEW QUESTION # 24
Given:
What is the result?
- A. B A C
- B. B A D
- C. D A D
- D. D D D
Answer: B
Explanation:
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
NEW QUESTION # 25
Which statement is true?
- A. The Lock () method returns a boolean indicator immediately regardless if it has or has not managed to acquire the lock
- B. The tryLock () method returns a boolean indicator immediately regardless if it has or has not managed to acquire the lock.
- C. The tryLock () method returns a boolean indicator immediately if it has managed to acquire the lock, otherwise it waits for the lock acquisition.
- D. The lock () method returns a boolean indicator immediately if it has managed to acquire the lock, otherwise it waits for the lock acquisition.
Answer: B
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. Reference: https://docs.oracle.com/en/java/javase/17/docs/api/java.base/java/util/concurrent/locks/Lock.html#tryLock(), 3, 4, 5
NEW QUESTION # 26
Given the code fragment:
What is the result?
- A. Range 1
Range 2
Range 3
Range 1
Not a valida rank - B. Range1
Note a valid rank. - C. Range 1
Range 2
Range 3 - D. Range 1
Answer: A
Explanation:
The 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". Reference: Java Language Changes - Oracle Help Center
NEW QUESTION # 27
Given:
What is the result?
- A. Software Game Chess 0
- B. Software Game Software Game Chese 2
- C. Software game write error
- D. Software Game Software Game chess 0
- E. Software Game read error
- F. Software Game Chess 2
Answer: B
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". Reference:
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
NEW QUESTION # 28
Given:
Which two method invocation execute?
- A. IFace.m2();
- B. Ifnce.m3 ();
- C. IFace myclassobj = new Myc ();
myclassObj.m3 (); - D. new MyC() .m2 ();
- E. IFace .,4():
- F. iFace mucloassObj = new Myc ();
myClassObj.m4();
Answer: D,E
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. Reference: https://education.oracle.com/products/trackp_OCPJSE17, 3, 4, 5
NEW QUESTION # 29
Given:
- A. A NumberFormatException is thrown
- B. Compilation fails
- C. Hello
- D. there
Answer: B
Explanation:
Explanation
The code fragment will fail to compile because the parseInt method of the Integer class is a static method, which means that it can be invoked without creating an object of the class. However, the code is trying to invoke the parseInt method on an object of type Integer, which is not allowed. The correct way to invoke the parseInt method is by using the class name, such as Integer.parseInt (s). Therefore, the code fragment will produce a compilation error. References: Integer (Java SE 17 & JDK 17) - Oracle
NEW QUESTION # 30
Given the product class:
And the shop class:
What is the result?
- A. Compilation fails
- B. Cookie 3.99 2.99
- C. Cookie 0.0 0.0
- D. Cookie 2.99 2.99
- E. Cookie 0.0 2.99
- F. An exception is produced at runtime
Answer: A
Explanation:
Explanation
The code fragment will fail to compile because the readObject method in the Product class is missing the
@Override annotation. The readObject method is a special method that is used to customize the deserialization process of an object. It must be declared as private, have no return type, and take a single parameter of type ObjectInputStream. It must also be annotated with @Override to indicate that it overrides the default behavior of the ObjectInputStream class. Without the @Override annotation, the compiler will treat the readObject method as a normal method and not as a deserialization hook. Therefore, the code fragment will produce a compilation error. References: Object Serialization - Oracle, [ObjectInputStream (Java SE 17 & JDK 17) - Oracle]
NEW QUESTION # 31
Given the code fragment:
What is the result?
- A. true 1
false 2 - B. false 1
false 2 - C. falase 0
true 1 - D. false 1
ture 2
Answer: C
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. Reference: Operators (The Java™ Tutorials > Learning the Java Language - Oracle
NEW QUESTION # 32
Given the code fragment:
What is the result?
- A. 5000$60M6D
- B. $SIM6D
- C. PT5SPTIMP6D
- D. PT5000PT60MP6D
Answer: D
Explanation:
Explanation
The code fragment is creating a Duration object with a value of 5000 milliseconds, then printing it. Then, it is creating another Duration object with a value of 60 seconds, then printing it. Finally, it is creating a Period object with a value of 6 days, then printing it. The output will be "PT5000PT60MP6D". References:
https://docs.oracle.com/javase/8/docs/api/java/time/Duration.html,
https://docs.oracle.com/javase/8/docs/api/java/time/Period.html
NEW QUESTION # 33
Given the code fragments:
Which is true?
- A. The program prints t1 : 1: t2 : 1: t1 : t2 : 2 : in random order.
- B. The program prints t1 : 1 : t2: 1 : t1 : 2 : t2: 2:
- C. The program prints t1 : 1: t2 : 1: t1 : 1 : t2 : 1 : indefinitely
- D. The program prints an exception
Answer: B
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: Reference: AtomicInteger (Java SE 17 & JDK 17) - Oracle
NEW QUESTION # 34
Given the code fragment:
What is the result?
- A. True true false NewYear
- B. False true true optional (Newyear)
- C. 010 optional (Newyear)
- D. 0110
Answer: B
Explanation:
Explanation
The code fragment is using the stream methods allMatch, anyMatch, noneMatch, and findFirst on a list of strings called specialDays. These methods are used to perform matching operations on the elements of a stream, such as checking if all, any, or none of the elements satisfy a given predicate, or finding the first element that matches a predicate1. The predicate in this case is that the string equals "Labour" or
"Halloween". The output will be:
False: because not all of the elements in specialDays are equal to "Labour" or "Halloween".
true: because at least one of the elements in specialDays is equal to "Labour" or "Halloween".
true: because none of the elements in specialDays are equal to both "Labour" and "Halloween".
Optional[NewYear]: because the first element in specialDays that matches the predicate is "NewYear", and the findFirst method returns an Optional object that may or may not contain a non-null value2.
References: Stream (Java SE 17 & JDK 17), Optional (Java SE 17 & JDK 17)
NEW QUESTION # 35
Given the code fragment:
Which code fragment invokes all callable objects in the workers set?
- A.
- B.
- C.
- D.
Answer: D
Explanation:
Explanation
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). References: AbstractExecutorService (Java SE 17 & JDK 17) - Oracle
NEW QUESTION # 36
Given:
What is the result?
- A. 100
100
1000 - B. 1001
100
1000 - C. 101
101
1000 - D. 1001
1001
1000
Answer: B
Explanation:
The code fragment is using the bitwise operators & (AND), | (OR), and ^ (XOR) to perform operations on the binary representations of the integer values. The & operator returns a 1 in each bit position where both operands have a 1, the | operator returns a 1 in each bit position where either operand has a 1, and the ^ operator returns a 1 in each bit position where only one operand has a 1. The binary representations of the integer values are as follows:
1000 = 1111101000
100 = 1100100
101 = 1100101
The code fragment performs the following operations:
x = x ^ y; // x becomes 1111010101, which is 1001 in decimal
y = x ^ y; // y becomes 1100100, which is 100 in decimal
x = x ^ y; // x becomes 1100101, which is 101 in decimal
The code fragment then prints out the values of x, y, and z, which are 1001, 100, and 1000 respectively. Therefore, option D is correct.
NEW QUESTION # 37
......
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