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Oracle Java SE 21 Developer Professional Sample Questions (Q47-Q52):

NEW QUESTION # 47
Which of the following doesnotexist?

A. They all exist.
B. BooleanSupplier
C. BiSupplier<T, U, R>
D. DoubleSupplier
E. LongSupplier
F. Supplier<T>

Answer: C

Explanation:
1. Understanding Supplier Functional Interfaces
* The Supplier<T> interface is part of java.util.function and provides valueswithout taking any arguments.
* Java also provides primitive specializations of Supplier<T>:
* BooleanSupplier# Returns a boolean. Exists
* DoubleSupplier# Returns a double. Exists
* LongSupplier# Returns a long. Exists
* Supplier<T># Returns a generic T. Exists
2. What about BiSupplier<T, U, R>?
* There is no BiSupplier<T, U, R> in Java.
* In Java, suppliers donot take arguments, so abi-supplierdoes not exist.
* If you need a function thattakes two arguments and returns a value, use BiFunction<T, U, R>.
Thus, the correct answer is:BiSupplier<T, U, R> does not exist.
References:
* Java SE 21 - Supplier<T>
* Java SE 21 - Functional Interfaces

NEW QUESTION # 48
Given:
java
final Stream<String> strings =
Files.readAllLines(Paths.get("orders.csv"));
strings.skip(1)
.limit(2)
.forEach(System.out::println);
And that the orders.csv file contains:
mathematica
OrderID,Customer,Product,Quantity,Price
1,Kylian Mbappe,Keyboard,2,25.50
2,Teddy Riner,Mouse,1,15.99
3,Sebastien Loeb,Monitor,1,199.99
4,Antoine Griezmann,Headset,3,45.00
What is printed?

A. arduino
2,Teddy Riner,Mouse,1,15.99
3,Sebastien Loeb,Monitor,1,199.99
B. An exception is thrown at runtime.
C. Compilation fails.
D. arduino
1,Kylian Mbappe,Keyboard,2,25.50
2,Teddy Riner,Mouse,1,15.99
3,Sebastien Loeb,Monitor,1,199.99
4,Antoine Griezmann,Headset,3,45.00
E. arduino
1,Kylian Mbappe,Keyboard,2,25.50
2,Teddy Riner,Mouse,1,15.99

Answer: B,C

Explanation:
1. Why Does Compilation Fail?
* The error is in this line:
java
final Stream<String> strings = Files.readAllLines(Paths.get("orders.csv"));
* Files.readAllLines(Paths.get("orders.csv")) returns a List<String>,not a Stream<String>.
* A List<String> cannot be assigned to a Stream<String>.
2. Correcting the Code
* The correct way to create a stream from the file:
java
Stream<String> strings = Files.lines(Paths.get("orders.csv"));
* This correctly creates a Stream<String> from the file.
3. Expected Output After Fixing
java
Files.lines(Paths.get("orders.csv"))
skip(1) // Skips the header row
limit(2) // Limits to first two data rows
forEach(System.out::println);
* Output:
arduino
1,Kylian Mbappe,Keyboard,2,25.50
2,Teddy Riner,Mouse,1,15.99
Thus, the correct answer is:Compilation fails.
References:
* Java SE 21 - Files.readAllLines
* Java SE 21 - Files.lines

NEW QUESTION # 49
Given:
java
List<String> frenchAuthors = new ArrayList<>();
frenchAuthors.add("Victor Hugo");
frenchAuthors.add("Gustave Flaubert");
Which compiles?

A. var authorsMap3 = new HashMap<>();
java
authorsMap3.put("FR", frenchAuthors);
B. Map<String, List<String>> authorsMap5 = new HashMap<String, List<String>>(); java authorsMap5.put("FR", frenchAuthors);
C. Map<String, ArrayList<String>> authorsMap1 = new HashMap<>();
java
authorsMap1.put("FR", frenchAuthors);
D. Map<String, List<String>> authorsMap4 = new HashMap<String, ArrayList<String>>(); java authorsMap4.put("FR", frenchAuthors);
E. Map<String, ? extends List<String>> authorsMap2 = new HashMap<String, ArrayList<String>> (); java authorsMap2.put("FR", frenchAuthors);

Answer: A,B,D

Explanation:
* Option A (Map<String, ArrayList<String>> authorsMap1 = new HashMap<>();)
* #Compilation Fails
* frenchAuthors is declared as List<String>,notArrayList<String>.
* The correct way to declare a Map that allows storing List<String> is to use List<String> as the generic type,notArrayList<String>.
* Fix:
java
Map<String, List<String>> authorsMap1 = new HashMap<>();
authorsMap1.put("FR", frenchAuthors);
* Reason:The type ArrayList<String> is more specific than List<String>, and this would cause a type mismatcherror.
* Option B (Map<String, ? extends List<String>> authorsMap2 = new HashMap<String, ArrayList<String>>();)
* #Compilation Fails
* ? extends List<String>makes the map read-onlyfor adding new elements.
* The line authorsMap2.put("FR", frenchAuthors); causes acompilation errorbecause wildcard (?
extends List<String>) prevents modifying the map.
* Fix:Remove the wildcard:
java
Map<String, List<String>> authorsMap2 = new HashMap<>();
authorsMap2.put("FR", frenchAuthors);
* Option C (var authorsMap3 = new HashMap<>();)
* Compiles Successfully
* The var keyword allows the compiler to infer the type.
* However,the inferred type is HashMap<Object, Object>, which may cause issues when retrieving values.
* Option D (Map<String, List<String>> authorsMap4 = new HashMap<String, ArrayList<String>
>();)
* Compiles Successfully
* Valid declaration:HashMap<K, V> can be assigned to Map<K, V>.
* Using new HashMap<String, ArrayList<String>>() with Map<String, List<String>> isallowed due to polymorphism.
* Correct syntax:
java
Map<String, List<String>> authorsMap4 = new HashMap<String, ArrayList<String>>(); authorsMap4.put("FR", frenchAuthors);
* Option E (Map<String, List<String>> authorsMap5 = new HashMap<String, List<String>>();)
* Compiles Successfully
* HashMap<String, List<String>> isa valid instantiation.
* Correct usage:
java
Map<String, List<String>> authorsMap5 = new HashMap<>();
authorsMap5.put("FR", frenchAuthors);
Thus, the correct answers are:C, D, E
References:
* Java SE 21 - Generics and Type Inference
* Java SE 21 - var Keyword

NEW QUESTION # 50
Given:
java
Optional<String> optionalName = Optional.ofNullable(null);
String bread = optionalName.orElse("Baguette");
System.out.print("bread:" + bread);
String dish = optionalName.orElseGet(() -> "Frog legs");
System.out.print(", dish:" + dish);
try {
String cheese = optionalName.orElseThrow(() -> new Exception());
System.out.println(", cheese:" + cheese);
} catch (Exception exc) {
System.out.println(", no cheese.");
}
What is printed?

A. bread:Baguette, dish:Frog legs, cheese.
B. bread:Baguette, dish:Frog legs, no cheese.
C. bread:bread, dish:dish, cheese.
D. Compilation fails.

Answer: B

Explanation:
Understanding Optional.ofNullable(null)
* Optional.ofNullable(null); creates an empty Optional (i.e., it contains no value).
* Optional.of(null); would throw a NullPointerException, but ofNullable(null); safely creates an empty Optional.
Execution of orElse, orElseGet, and orElseThrow
* orElse("Baguette")
* Since optionalName is empty, "Baguette" is returned.
* bread = "Baguette"
* Output:"bread:Baguette"
* orElseGet(() -> "Frog legs")
* Since optionalName is empty, "Frog legs" is returned from the lambda expression.
* dish = "Frog legs"
* Output:", dish:Frog legs"
* orElseThrow(() -> new Exception())
* Since optionalName is empty, an exception is thrown.
* The catch block catches this exception and prints ", no cheese.".
Thus, the final output is:
makefile
bread:Baguette, dish:Frog legs, no cheese.
References:
* Java SE 21 & JDK 21 - Optional
* Java SE 21 - Functional Interfaces

NEW QUESTION # 51
Given:
java
Map<String, Integer> map = Map.of("b", 1, "a", 3, "c", 2);
TreeMap<String, Integer> treeMap = new TreeMap<>(map);
System.out.println(treeMap);
What is the output of the given code fragment?

A. {a=3, b=1, c=2}
B. {b=1, a=3, c=2}
C. {b=1, c=2, a=3}
D. {c=1, b=2, a=3}
E. Compilation fails
F. {c=2, a=3, b=1}
G. {a=1, b=2, c=3}

Answer: A

Explanation:
In this code, a Map named map is created using Map.of with the following key-value pairs:
* "b": 1
* "a": 3
* "c": 2
The Map.of method returns an immutable map containing these mappings.
Next, a TreeMap named treeMap is instantiated by passing the map to its constructor:
java
TreeMap<String, Integer> treeMap = new TreeMap<>(map);
The TreeMap constructor with a Map parameter creates a new tree map containing the same mappings as the given map, ordered according to the natural ordering of its keys. In Java, the natural ordering for String keys is lexicographical order.
Therefore, the TreeMap will store the entries in the following order:
* "a": 3
* "b": 1
* "c": 2
When System.out.println(treeMap); is executed, it outputs the TreeMap in its natural order, resulting in:
r
{a=3, b=1, c=2}
Thus, the correct answer is option F: {a=3, b=1, c=2}.

NEW QUESTION # 52
......

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