Pass 1Z0-829 Exam Latest Practice Questions Updated on Mar 30, 2024
Oracle 1Z0-829 Study Guide Archives
Passing the Oracle 1Z0-829 exam requires extensive knowledge of Java programming concepts, as well as practical experience in working with Java SE 17. 1Z0-829 exam is designed to test the developer's ability to solve real-world problems and develop efficient and scalable applications using Java SE 17. Java SE 17 Developer certification can help developers showcase their expertise and stand out in a competitive job market.
NEW QUESTION # 16
Given:
What is the result?
- A. (Bicycle, car, motorcycle, truck)
- B. (3:bicycle, 0:car, 0motercycle, 5:truck)
- C. Bicycle-1, car=3, motorcycle=1, truck=2)
- D. Bicycle =7, car=7, motorcycle=7, truck=7)
- E. Compilation fails.
Answer: E
Explanation:
The answer is E because the code fragment contains several syntax errors that prevent it from compiling. Some of the errors are:
The enum declaration is missing a semicolon after the list of constants.
The enum constants are not capitalized, which violates the Java naming convention for enums.
The switch expression is missing parentheses around the variable name.
The case labels are missing colons after the enum constants.
The default label is missing a break statement, which causes a fall-through to the next case.
The println statement is missing a closing parenthesis and a semicolon.
A possible corrected version of the code fragment is:
enum Vehicle { BICYCLE, CAR, MOTORCYCLE, TRUCK; } public class Test { public static void main(String[] args) { Vehicle v = Vehicle.BICYCLE; switch (v) { case BICYCLE: System.out.print("1"); break; case CAR: System.out.print("3"); break; case MOTORCYCLE: System.out.print("1"); break; case TRUCK: System.out.print("2"); break; default: System.out.print("0"); break; } System.out.println(); } } This would print 1 as the output. Reference:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
Enum Types
The switch Statement
NEW QUESTION # 17
Given:
What is the result?
- A. D D D
- B. B A D
- C. D A D
- D. B A C
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 # 18
Given:
Which action enables the code to compile?
- A. Replace 15 with item.display (''Flower'');
- B. Replace 2 with static string name;
- C. Replace 7 with public void display (string design) {
- D. Replace 3 with private static void display () {
Answer: C
Explanation:
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
NEW QUESTION # 19
Given:
Which statement is true?
- A. The program fails to compile.
- B. The program throws StockException.
- C. The program throws outofStockException.
- D. The program throws ClassCastException
Answer: A
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. Reference:
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)
NEW QUESTION # 20
Given:
What is the result?
- A. 100
100
1000 - B. 1001
100
1000 - C. 101
101
1000 - D. 1001
1001
1000
Answer: B
Explanation:
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 # 21
Given:
Which two modifications enable the code to print Open Close?
- A.
- B.
- C.
- D.
- E.
Answer: B,C
Explanation:
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 # 22
Given:
What is the result?
- A. (Bicycle, car, motorcycle, truck)
- B. (3:bicycle, 0:car, 0motercycle, 5:truck)
- C. Bicycle-1, car=3, motorcycle=1, truck=2)
- D. Bicycle =7, car=7, motorcycle=7, truck=7)
- E. Compilation fails.
Answer: E
Explanation:
Explanation
The answer is E because the code fragment contains several syntax errors that prevent it from compiling.
Some of the errors are:
The enum declaration is missing a semicolon after the list of constants.
The enum constants are not capitalized, which violates the Java naming convention for enums.
The switch expression is missing parentheses around the variable name.
The case labels are missing colons after the enum constants.
The default label is missing a break statement, which causes a fall-through to the next case.
The println statement is missing a closing parenthesis and a semicolon.
A possible corrected version of the code fragment is:
enum Vehicle { BICYCLE, CAR, MOTORCYCLE, TRUCK; } public class Test { public static void main(String[] args) { Vehicle v = Vehicle.BICYCLE; switch (v) { case BICYCLE: System.out.print("1"); break; case CAR: System.out.print("3"); break; case MOTORCYCLE: System.out.print("1"); break; case TRUCK: System.out.print("2"); break; default: System.out.print("0"); break; } System.out.println(); } } This would print 1 as the output. References:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
Enum Types
The switch Statement
NEW QUESTION # 23
Given:
Which two modifications enable the code to print Open Close?
- A.
- B.
- C.
- D.
- E.
Answer: B,C
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 # 24
Given the content of the in. tart file:
23456789
and the code fragment:
What is the content of the out .txt file?
- A. 01234567801234
- B. 012345678901234
- C. 0123456789
- D. 012345678
- E. 0123456789234567
- F. 01234567
Answer: C
Explanation:
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".
NEW QUESTION # 25
Given the code fragment:
Which code fragment returns different values?
- A. int sum = listOfNumbers. parallelStream () reduce ({m, n) -> m +n) orElse (5) +5;
- B. int sum = listOfNumbers. Stream () reduce ( Integer:: sum) ; +5;
- C. int sum = listOfNumbers. Stream () reduce (0, Integer:: sum) + 5
- D. int sum = listOfNumbers. parallelStream () reduce (5, Integer:: sum) ;
- E. int sum = listOfNumbers. Stream () reduce (5, (a, b) -> a + b) ;
Answer: B
Explanation:
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 5
Apply 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 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
Guide to Stream.reduce()
NEW QUESTION # 26
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 automatic modules are on the module path.
- D. Only unnamed modules are on the module path.
- E. Only named modules are on the module path.
Answer: B
Explanation:
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 # 27
Given:
What is the result?
- A. Compilation fails
- B. runsflips
- C. flipsruns
- D. runsruns
- E. flipsflips
Answer: A
Explanation:
Explanation
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.
NEW QUESTION # 28
Assuming that the data, txt file exists and has the following content:
Text1
Text2
Text3
Given the code fragment:
What is the result?
- A. text1-text2-text3
A java.lang.indexoutofBoundsException is thrown. - B. text1-text2-text3
text1
text2
text3 - C. text1-
text2-
text3-
text3 - D. text1-text2-text3
text3
Answer: D
Explanation:
The answer is D because the code fragment reads the file "data.txt" and collects all the lines in the file into a single string, separated by hyphens. Then, it prints the resulting string. Next, it attempts to read the fourth line in the file (index 3) and print it. However, since the file only has three lines, an IndexOutOfBoundsException is thrown. Reference:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
Read contents of a file using Files class in Java
NEW QUESTION # 29
Given:
What is the result
- A. UnDefined
- B. Marketing
Finance
Technical - C. Marketing
- D. Marketing
Undefined
Answer: A
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 # 30
Daylight Saving Time (DST) is the practice of advancing clocks at the start of spring by one hour and adjusting them backward by one hour in autumn.
Considering that in 2021, DST in Chicago (Illinois) ended on November 7th at 2 AM, and given the fragment:
What is the output?
- A. false
true - B. False
false - C. true
false - D. true
true
Answer: C
Explanation:
The answer is A because the code fragment uses the ZoneId and ZonedDateTime classes to create two date-time objects with the same local date-time but different zone offsets. The ZoneId class represents a time-zone ID, such as America/Chicago, and the ZonedDateTime class represents a date-time with a time-zone in the ISO-8601 calendar system. The code fragment creates two ZonedDateTime objects with the same local date-time of 2021-11-07T01:30, but different zone IDs of America/Chicago and UTC. The code fragment then compares the two objects using the equals and isEqual methods.
The equals method compares the state of two objects for equality. In this case, it compares the local date-time, zone offset, and zone ID of the two ZonedDateTime objects. Since the zone offsets and zone IDs are different, the equals method returns false.
The isEqual method compares the instant of two temporal objects for equality. In this case, it compares the instant of the two ZonedDateTime objects, which is derived from the local date-time and zone offset. Since DST in Chicago ended on November 7th at 2 AM in 2021, the local date-time of 2021-11-07T01:30 in America/Chicago corresponds to the same instant as 2021-11-07T06:30 in UTC. Therefore, the isEqual method returns true.
Hence, the output is true false. Reference:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
ZoneId (Java Platform SE 8 )
ZonedDateTime (Java Platform SE 8 )
Time Zone & Clock Changes in Chicago, Illinois, USA
Daylight Saving Time Changes 2023 in Chicago, USA
NEW QUESTION # 31
Given:
What is the result?
- A. Mb
- B. mA
- C. MA
- D. Mb
- E. Mb
MC
Answer: B
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". Reference: Pattern Matching for instanceof - Oracle Help Center
NEW QUESTION # 32
Given the code fragment:
Which two statements at Line nl independently enable you to print 1250?
- A. Integer res = 250 + s:
- B. Integer res= 250 + s2;
- C. Integer res = 250;
- D. Integer res = 250 + integer.parseint (s)
- E. Integer res = 250 + integer (s2):
- F. Integer res = 250 + integer . valueof (s);
Answer: D,F
Explanation:
Res = + s2;
Explanation:
The code fragment is creating a string variable "s" with the value "10_00" and an integer variable "s2" with the value 10. The string "s" is using an underscore as a separator for readability, which is allowed in Java SE 171. The question is asking for two statements that can add 250 to the numeric value of "s" and assign it to an integer variable "res". The correct answers are A and E because they use the methods parseInt and valueOf of the Integer class to convert the string "s" to an integer. Both methods interpret the string as a signed decimal integer and return the equivalent int or Integer value23. The other options are incorrect because they either use invalid syntax, such as B and C, or they do not convert the string "s" to an integer, such as D and F. Reference: Binary Literals (The Java™ Tutorials > Learning the Java Language > Numbers and Strings), Integer (Java SE 17 & JDK 17), Integer (Java SE 17 & JDK 17)
NEW QUESTION # 33
Given:
Which two should the module-info file include for it to represent the service provider interface?
- A. Provides.com.transport.vehicle.cars.Car with com.transport.vehicle.cars. impt, CatImpI;
- B. Exports com.transport.vehicle;
- C. Provides.com.transport.vehicle.cars.Car impl,CarImp1 to com.transport.vehicle.cars. Cars
- D. Requires cm.transport.vehicle,cars:
- E. exports com.transport.vehicle.cars.Car;
- F. Exports com.transport.vehicle.cars;
- G. Requires cm.transport.vehicle,cars:
Answer: A,E
Explanation:
The answer is B and E because the module-info file should include a provides directive and an exports directive to represent the service provider interface. The provides directive declares that the module provides an implementation of a service interface, which is com.transport.vehicle.cars.Car in this case. The with clause specifies the fully qualified name of the service provider class, which is com.transport.vehicle.cars.impl.CarImpl in this case. The exports directive declares that the module exports a package, which is com.transport.vehicle.cars in this case, to make it available to other modules. The package contains the service interface that other modules can use.
Option A is incorrect because requires is not the correct keyword to declare a service provider interface. Requires declares that the module depends on another module, which is not the case here.
Option C is incorrect because it has a typo in the module name. It should be com.transport.vehicle.cars, not cm.transport.vehicle.cars.
Option D is incorrect because it has a typo in the keyword provides. It should be provides, not Provides. It also has a typo in the service interface name. It should be com.transport.vehicle.cars.Car, not com.transport.vehicle.cars.Car impl. It also has an unnecessary to clause, which is used to limit the accessibility of an exported package to specific modules.
Option F is incorrect because it exports the wrong package. It should export com.transport.vehicle.cars, not com.transport.vehicle.cars.impl. The impl package contains the service provider class, which should not be exposed to other modules.
Option G is incorrect because it exports the wrong package. It should export com.transport.vehicle.cars, not com.transport.vehicle. The vehicle package does not contain the service interface or the service provider class. Reference:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
Java Modules - Service Interface Module - GeeksforGeeks
Java Service Provider Interface | Baeldung
NEW QUESTION # 34
Given:
What is the result?
- A. Software Game Chess 0
- B. Software Game Software Game Chese 2
- C. Software Game Software Game chess 0
- D. Software Game Chess 2
- E. Software game write error
- F. Software Game read error
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 # 35
Given the code fragment:
What is the result?
- A. false 1
ture 2 - B. false 1
false 2 - C. falase 0
true 1 - D. true 1
false 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 # 36
......
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