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Oracle Java SE 21 Developer Professional Sample Questions (Q54-Q59):
NEW QUESTION # 54
Which two of the following aren't the correct ways to create a Stream?
- A. Stream stream = Stream.generate(() -> "a");
- B. Stream<String> stream = Stream.builder().add("a").build();
- C. Stream stream = Stream.empty();
- D. Stream stream = Stream.ofNullable("a");
- E. Stream stream = Stream.of("a");
- F. Stream stream = new Stream();
- G. Stream stream = Stream.of();
Answer: B,F
Explanation:
In Java, the Stream API provides several methods to create streams. However, not all approaches are valid.
NEW QUESTION # 55
Given:
java
Object myVar = 0;
String print = switch (myVar) {
case int i -> "integer";
case long l -> "long";
case String s -> "string";
default -> "";
};
System.out.println(print);
What is printed?
- A. nothing
- B. It throws an exception at runtime.
- C. string
- D. long
- E. integer
- F. Compilation fails.
Answer: F
Explanation:
* Why does the compilation fail?
* TheJava switch statement does not support primitive type pattern matchingin switch expressions as of Java 21.
* The case pattern case int i -> "integer"; isinvalidbecausepattern matching with primitive types (like int or long) is not yet supported in switch statements.
* The error occurs at case int i -> "integer";, leading to acompilation failure.
* Correcting the Code
* Since myVar is of type Object,autoboxing converts 0 into an Integer.
* To make the code compile, we should use Integer instead of int:
java
Object myVar = 0;
String print = switch (myVar) {
case Integer i -> "integer";
case Long l -> "long";
case String s -> "string";
default -> "";
};
System.out.println(print);
* Output:
bash
integer
Thus, the correct answer is:Compilation fails.
References:
* Java SE 21 - Pattern Matching for switch
* Java SE 21 - switch Expressions
NEW QUESTION # 56
Given a properties file on the classpath named Person.properties with the content:
ini
name=James
And:
java
public class Person extends ListResourceBundle {
protected Object[][] getContents() {
return new Object[][]{
{"name", "Jeanne"}
};
}
}
And:
java
public class Test {
public static void main(String[] args) {
ResourceBundle bundle = ResourceBundle.getBundle("Person");
String name = bundle.getString("name");
System.out.println(name);
}
}
What is the given program's output?
- A. MissingResourceException
- B. Jeanne
- C. James
- D. JeanneJames
- E. Compilation fails
- F. JamesJeanne
Answer: B
Explanation:
In this scenario, we have a Person class that extends ListResourceBundle and a properties file named Person.
properties. Both define a resource with the key "name" but with different values:
* Person class (ListResourceBundle):Defines the key "name" with the value "Jeanne".
* Person.properties file:Defines the key "name" with the value "James".
When the ResourceBundle.getBundle("Person") method is called, the Java runtime searches for a resource bundle with the base name "Person". The search order is as follows:
* Class-Based Resource Bundle:The runtime first looks for a class named Person (i.e., Person.class).
* Properties File Resource Bundle:If the class is not found, it then looks for a properties file named Person.properties.
In this case, since the Person class is present and accessible, the runtime will load the Person class as the resource bundle. Therefore, the getBundle method returns an instance of the Person class.
Subsequently, when bundle.getString("name") is called, it retrieves the value associated with the key "name" from the Person class, which is "Jeanne".
Thus, the output of the program is:
nginx
Jeanne
NEW QUESTION # 57
Which of the following statements oflocal variables declared with varareinvalid?(Choose 4)
- A. var b = 2, c = 3.0;
- B. var a = 1;(Valid: var correctly infers int)
- C. var d[] = new int[4];
- D. var h = (g = 7);
- E. var f = { 6 };
- F. var e;
Answer: A,C,E,F
Explanation:
1. Valid Use Cases of var
* var is alocal variable type inferencefeature.
* The compilerinfers the type from the assigned value.
* Example of valid use:
java
var a = 10; // Type inferred as int
var str = "Hello"; // Type inferred as String
2. Analyzing the Given Statements
Statement
Valid/Invalid
Reason
var a = 1;
Valid
Type inferred as int.
var b = 2, c = 3.0;
#Invalid
var doesnot allow multiple declarationsin one statement.
var d[] = new int[4];
#Invalid
Array brackets []are not allowedwith var.
var e;
#Invalid
varrequires an initializer(cannot be declared without assignment).
var f = { 6 };
#Invalid
{ 6 } is anarray initializer, which must have an explicit type.
var h = (g = 7);
Valid
g is assigned 7, and h gets its value.
Thus, the correct answers are:B, C, D, E
References:
* Java SE 21 - Local Variable Type Inference (var)
* Java SE 21 - var Restrictions
NEW QUESTION # 58
Given:
java
Optional o1 = Optional.empty();
Optional o2 = Optional.of(1);
Optional o3 = Stream.of(o1, o2)
.filter(Optional::isPresent)
.findAny()
.flatMap(o -> o);
System.out.println(o3.orElse(2));
What is the given code fragment's output?
- A. 0
- B. An exception is thrown
- C. 1
- D. 2
- E. Optional.empty
- F. Compilation fails
- G. Optional[1]
Answer: C
Explanation:
In this code, two Optional objects are created:
* o1 is an empty Optional.
* o2 is an Optional containing the integer 1.
A stream is created from o1 and o2. The filter method retains only the Optional instances that are present (i.e., non-empty). This results in a stream containing only o2.
The findAny method returns an Optional describing some element of the stream, or an empty Optional if the stream is empty. Since the stream contains o2, findAny returns Optional[Optional[1]].
The flatMap method is then used to flatten this nested Optional. It applies the provided mapping function (o -
> o) to the value, resulting in Optional[1].
Finally, o3.orElse(2) returns the value contained in o3 if it is present; otherwise, it returns 2. Since o3 contains
1, the output is 1.
NEW QUESTION # 59
......
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