Most Recent Oracle 1Z0-829 Exam Dumps

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''.

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

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The 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.

Oracle Certified Professional: Java SE 17 Developer

Java SE 17 Developer

OCP Oracle Certified Professional Java SE 17 Developer Study Guide

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Sealed Class in Java - Javatpoint

The code fragment is creating a string variable ''a'' with the value ''Hello! Java''. Then, it is printing the index of ''Java'' in ''a''. Next, it is replacing ''Hello!'' with ''Welcome!'' in ''a''. Then, it is printing the index of ''Java'' in ''a''. Finally, it is creating a new StringBuilder object ''b'' with the value of ''a'' and printing the index of ''Java'' in ''b''. The output will be 8109 because the index of ''Java'' in ''a'' is 8, the index of ''Java'' in ''a'' after replacing ''Hello!'' with ''Welcome!'' is 10, and the index of ''Java'' in ''b'' is 9.Reference:Oracle Java SE 17 Developer source and documents: [String (Java SE 17 & JDK 17)], [StringBuilder (Java SE 17 & JDK 17)]

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.Reference:The final Keyword (The Java Tutorials > Learning the Java Language > Classes and Objects)