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  2. Oracle Certification
  3. 1z1-829 Exam
  4. Oracle.1z1-829.v2024-08-03.q35 Dumps
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Question 16

Given:

What is the result?

Correct Answer: E
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)
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Question 17

Given:

Which action enables the code to compile?

Correct Answer: C
Explanation
The answer is C because the code fragment contains a syntax error in line 7, where the method display is declared without any parameter type. This causes a compilation error, as Java requires the parameter type to be specified for each method parameter. To fix this error, the parameter type should be added before the parameter name, such as string design. This will enable the code to compile and run without any errors.
References:
Oracle Certified Professional: Java SE 17 Developer
Java SE 17 Developer
OCP Oracle Certified Professional Java SE 17 Developer Study Guide
Java Methods
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Question 18

Which statement is true about modules?

Correct Answer: C
A module path is a sequence of directories that contain modules or JAR files. A named module is a module that has a name and a module descriptor (module-info.class) that declares its dependencies and exports. An automatic module is a module that does not have a module descriptor, but is derived from the name and contents of a JAR file. Both named and automatic modules can be placed on the module path, and they can be resolved by the Java runtime. An unnamed module is a special module that contains all the classes that are not in any other module, such as those on the class path. An unnamed module is not on the module path, but it can read all other modules.
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Question 19

Given the content of the in. tart file:
23456789
and the code fragment:

What is the content of the out .txt file?

Correct Answer: D
Explanation
The answer is D because the code fragment reads the content of the in.txt file and writes it to the out.txt file.
The content of the in.txt file is "23456789". The code fragment uses a char array buffer of size 8 to read the content of the in.txt file. The while loop reads the content of the in.txt file and writes it to the out.txt file until the end of the file is reached. Therefore, the content of the out.txt file will be "0123456789".
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Question 20

Given the code fragment:

Correct Answer: D
The code fragment compares four pairs of strings using the equals() and intern() methods. The equals() method compares the content of two strings, while the intern() method returns a canonical representation of a string, which means that it returns a reference to an existing string with the same content in the string pool. The string pool is a memory area where strings are stored and reused to save space and improve performance. The results of the comparisons are as follows:
s1.equals(s2): This returns true because both s1 and s2 have the same content, "Hello Java 17".
s1 == s2: This returns false because s1 and s2 are different objects with different references, even though they have the same content. The == operator compares the references of two objects, not their content.
s1.intern() == s2.intern(): This returns true because both s1.intern() and s2.intern() return a reference to the same string object in the string pool, which has the content "Hello Java 17". The intern() method ensures that there is only one copy of each distinct string value in the string pool.
"Hello Java 17" == s2: This returns false because "Hello Java 17" is a string literal, which is automatically interned and stored in the string pool, while s2 is a string object created with the new operator, which is not interned by default and stored in the heap. Therefore, they have different references and are not equal using the == operator.
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