1Z0-829 VALID TEST VOUCHER & 1Z0-829 NEW REAL TEST

1Z0-829 Valid Test Voucher & 1Z0-829 New Real Test

1Z0-829 Valid Test Voucher & 1Z0-829 New Real Test

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Oracle Java SE 17 Developer Sample Questions (Q48-Q53):

NEW QUESTION # 48
Given the code fragment:

Which code fragment returns different values?

  • A. int sum = listOfNumbers. Stream () reduce ( Integer:: sum) ; +5;
  • B. int sum = listOfNumbers. Stream () reduce (5, (a, b) -> a + b) ;
  • C. int sum = listOfNumbers. Stream () reduce (0, Integer:: sum) + 5
  • D. int sum = listOfNumbers. parallelStream () reduce ({m, n) -> m +n) orElse (5) +5;
  • E. int sum = listOfNumbers. parallelStream () reduce (5, Integer:: sum) ;

Answer: A

Explanation:
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. 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()


NEW QUESTION # 49
Given the code fragments:

Which is true?

  • A. The program prints t1 : 1 : t2: 1 : t1 : 2 : t2: 2:
  • B. The program prints t1 : 1: t2 : 1: t1 : t2 : 2 : in random order.
  • C. The program prints t1 : 1: t2 : 1: t1 : 1 : t2 : 1 : indefinitely
  • D. The program prints an exception

Answer: A

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 # 50
Given the code fragment:

abstract sealed interface SInt permits Story, Art { default String getTitle() { return "Book Title" ; }
}
Which set of class definitions compiles?

  • A. Interace story extends STnt {}
    Interface Art extends SInt {}
  • B. Non-sealed interface story extends SInt {}
    Non-sealed interaface Art extends Sint {}
  • C. Public interface story extends sInd {}
    Public interface Art extends SInt {}
  • D. Sealed interface Storty extends sInt {}
    Non-sealed class Art implements Sint {}
  • E. Non-sealed interface story extends SInt {}
    Class Art implements Sint {}

Answer: D

Explanation:
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.
Reference:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
Sealed Classes and Interfaces in Java 15 | Baeldung
Sealed Class in Java - Javatpoint


NEW QUESTION # 51
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 # 52
Given:

What is the result?

  • A. Compilation fails
  • B. 1 Snowy
  • C. 0 CLOUDY
  • D. 0 Snowy
  • E. 1 RAINY

Answer: D

Explanation:
The code is defining an enum class called Forecast with three values: SUNNY, CLOUDY, and RAINY. The toString() method is overridden to always return "SNOWY". In the main method, the ordinal value of SUNNY is printed, which is 0, followed by the value of CLOUDY converted to uppercase, which is "CLOUDY". However, since the toString() method of Forecast returns "SNOWY" regardless of the actual value, the output will be "0 SNOWY". Reference: Enum (Java SE 17 & JDK 17), Enum.EnumDesc (Java SE 17 & JDK 17)


NEW QUESTION # 53
......

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