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Oracle Java SE 21 Developer Professional Sample Questions (Q46-Q51):

NEW QUESTION # 46
Given:
java
try (FileOutputStream fos = new FileOutputStream("t.tmp");
ObjectOutputStream oos = new ObjectOutputStream(fos)) {
fos.write("Today");
fos.writeObject("Today");
oos.write("Today");
oos.writeObject("Today");
} catch (Exception ex) {
// handle exception
}
Which statement compiles?

A. oos.write("Today");
B. fos.write("Today");
C. fos.writeObject("Today");
D. oos.writeObject("Today");

Answer: D

Explanation:
In Java, FileOutputStream and ObjectOutputStream are used for writing data to files, but they have different purposes and methods. Let's analyze each statement:
* fos.write("Today");
The FileOutputStream class is designed to write raw byte streams to files. The write method in FileOutputStream expects a parameter of type int or byte[]. Since "Today" is a String, passing it directly to fos.
write("Today"); will cause a compilation error because there is no write method in FileOutputStream that accepts a String parameter.
* fos.writeObject("Today");
The FileOutputStream class does not have a method named writeObject. The writeObject method is specific to ObjectOutputStream. Therefore, attempting to call fos.writeObject("Today"); will result in a compilation error.
* oos.write("Today");
The ObjectOutputStream class is used to write objects to an output stream. However, it does not have a write method that accepts a String parameter. The available write methods in ObjectOutputStream are for writing primitive data types and objects. Therefore, oos.write("Today"); will cause a compilation error.
* oos.writeObject("Today");
The ObjectOutputStream class provides the writeObject method, which is used to serialize objects and write them to the output stream. Since String implements the Serializable interface, "Today" can be serialized.
Therefore, oos.writeObject("Today"); is valid and compiles successfully.
In summary, the only statement that compiles without errors is oos.writeObject("Today");.
References:
* Java SE 21 & JDK 21 - ObjectOutputStream
* Java SE 21 & JDK 21 - FileOutputStream

NEW QUESTION # 47
Given:
java
var counter = 0;
do {
System.out.print(counter + " ");
} while (++counter < 3);
What is printed?

A. Compilation fails.
B. 1 2 3 4
C. 1 2 3
D. 0 1 2 3
E. An exception is thrown.
F. 0 1 2

Answer: F

Explanation:
* Understanding do-while Execution
* A do-while loopexecutes at least oncebefore checking the condition.
* ++counter < 3 increments counterbeforeevaluating the condition.
* Step-by-Step Execution
* Iteration 1:counter = 0, print "0", then ++counter becomes 1, condition 1 < 3 istrue.
* Iteration 2:counter = 1, print "1", then ++counter becomes 2, condition 2 < 3 istrue.
* Iteration 3:counter = 2, print "2", then ++counter becomes 3, condition 3 < 3 isfalse, so loop exits.
* Final Output
0 1 2
Thus, the correct answer is:0 1 2
References:
* Java SE 21 - Control Flow Statements
* Java SE 21 - do-while Loop

NEW QUESTION # 48
Which methods compile?

A. ```java
public List<? super IOException> getListSuper() {
return new ArrayList<FileNotFoundException>();
}
B. ```java public List<? super IOException> getListSuper() { return new ArrayList<Exception>(); } csharp
C. ```java
public List<? extends IOException> getListExtends() {
return new ArrayList<FileNotFoundException>();
}
D. ```java public List<? extends IOException> getListExtends() { return new ArrayList<Exception>(); } csharp

Answer: B,C

Explanation:
In Java generics, wildcards are used to relax the type constraints of generic types. The extends wildcard (<?
extends Type>) denotes an upper bounded wildcard, allowing any type that is a subclass of Type. Conversely, the super wildcard (<? super Type>) denotes a lower bounded wildcard, allowing any type that is a superclass of Type.
Option A:
java
public List<? super IOException> getListSuper() {
return new ArrayList<Exception>();
}
Here, List<? super IOException> represents a list that can hold IOException objects and objects of its supertypes. Since Exception is a superclass of IOException, ArrayList<Exception> is compatible with List<?
super IOException>. Therefore, this method compiles successfully.
Option B:
java
public List<? extends IOException> getListExtends() {
return new ArrayList<FileNotFoundException>();
}
In this case, List<? extends IOException> represents a list that can hold objects of IOException and its subclasses. Since FileNotFoundException is a subclass of IOException, ArrayList<FileNotFoundException> is compatible with List<? extends IOException>. Thus, this method compiles successfully.
Option C:
java
public List<? extends IOException> getListExtends() {
return new ArrayList<Exception>();
}
Here, List<? extends IOException> expects a list of IOException or its subclasses. However, Exception is a superclass of IOException, not a subclass. Therefore, ArrayList<Exception> is not compatible with List<?
extends IOException>, and this method will not compile.
Option D:
java
public List<? super IOException> getListSuper() {
return new ArrayList<FileNotFoundException>();
}
In this scenario, List<? super IOException> expects a list that can hold IOException objects and objects of its supertypes. Since FileNotFoundException is a subclass of IOException, ArrayList<FileNotFoundException> is not compatible with List<? super IOException>, and this method will not compile.
Therefore, the methods in options A and B compile successfully, while those in options C and D do not.

NEW QUESTION # 49
Given:
java
String textBlock = """
j
a
v s
a
""";
System.out.println(textBlock.length());
What is the output?

A. 0
B. 1
C. 2
D. 3

Answer: A

Explanation:
In this code, a text block is defined using the """ syntax introduced in Java 13. Text blocks allow for multiline string literals, preserving the format as written in the code.
Text Block Analysis:
The text block is defined as:
java
String textBlock = """
j
a
contentReference[oaicite:0]{index=0}

NEW QUESTION # 50
Given:
java
public class ThisCalls {
public ThisCalls() {
this(true);
}
public ThisCalls(boolean flag) {
this();
}
}
Which statement is correct?

A. It does not compile.
B. It throws an exception at runtime.
C. It compiles.

Answer: A

Explanation:
In the provided code, the class ThisCalls has two constructors:
* No-Argument Constructor (ThisCalls()):
* This constructor calls the boolean constructor with this(true);.
* Boolean Constructor (ThisCalls(boolean flag)):
* This constructor attempts to call the no-argument constructor with this();.
This setup creates a circular call between the two constructors:
* The no-argument constructor calls the boolean constructor.
* The boolean constructor calls the no-argument constructor.
Such a circular constructor invocation leads to a compile-time error in Java, specifically "recursiveconstructor invocation." The Java Language Specification (JLS) states:
"It is a compile-time error for a constructor to directly or indirectly invoke itself through a series of one or more explicit constructor invocations involving this." Therefore, the code will not compile due to this recursive constructor invocation.

NEW QUESTION # 51
......

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