Java SE 8 Programmer I practice questions

1. Question 1 of 25

   Given:

   

   What is the result?

   • A200.0 : 100.0
   • B400.0 : 200.0
   • C400.0 : 100.0
   • DCompilation fails.

   Show answer

   Correct answer: C

   How others answered

   • C89%

   Explanation

   You're wrong. I took the test last week and some of this questions were on the test. Remember, test questions are not the same for everyone. They keep changing questions every time you take the test, but concept is the same. Don't memorize these questions, but try to understand them. Only then you'll pass. Good luck!

   Adapted from community discussion by Harid · 2020-08-24 · 8 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

   View community discussion4

   Candidate context

   Question 1 discussion

   These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

   1. H

      Harid

      2020-08-24

      8 upvotes

      You're wrong. I took the test last week and some of this questions were on the test. Remember, test questions are not the same for everyone. They keep changing questions every time you take the test, but concept is the same. Don't memorize these questions, but try to understand them. Only then you'll pass. Good luck!

   2. D

      DriftKing

      2023-08-28

      Selected C3 upvotes

      Tested. Answer is C: 400.0 : 100.0 class Product{ double price; } public class Test { public void updatePrice(Product product, double price){ price = price*2; product.price = product.price + price; } public static void main(String args[]) { Product prt = new Product(); prt.price = 200; double newPrice = 100; Test t = new Test(); t.updatePrice(prt, newPrice); System.out.println(prt.price+ " : "+newPrice); } }

   3. D

      dumma

      2019-12-14

      4 upvotes

      No, because they are accessing variables using instance. You would get error if you try to access method directly instead of instantiation

   4. H

      hackGh

      2020-11-20

      3 upvotes

      price= price*2;// 100*2=200 product.price=product.price + price;//prt.price =200+200=400 so prt.price 400

2. Question 2 of 25

   Given the code fragment:

   

   What is the result?

   • AMatch 1
   • BMatch 2
   • CNo Match
   • DA NullPointerException is thrown at runtime.

   Show answer

   Correct answer: A

   How others answered

   • A100%

   Explanation

   sb.toString() => converts it into String. and thereafter, using equals() method on String will do the value comparison. So, answer is A since the values are equal.

   Adapted from community discussion by Ankit1010 · 2023-02-07 · 3 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

   View community discussion4

   Candidate context

   Question 2 discussion

   These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

   1. A

      Ankit1010

      2023-02-07

      3 upvotes

      sb.toString() => converts it into String. and thereafter, using equals() method on String will do the value comparison. So, answer is A since the values are equal.

   2. P

      pedone_valerio

      2024-09-25

      1 upvote

      public static void main(String[] args) { StringBuilder sb = new StringBuilder("Java"); String s = "Java"; if (sb.toString().equals(s.toString())) { System.out.println("Match 1"); } else if (sb.equals(s)) { System.out.println("Match 2"); } else { System.out.println("No Match"); } Answer A

   3. W

      willokans

      2022-11-22

      1 upvote

      Inconvertible type error between String "Java" and StringBuilder "Java". To resolve StringBuilder to String append .toString() method to sb

   4. A

      AhmadTechie

      2023-11-20

      1 upvote

      Answer is A. As sb.toString() => new string object gets created and String equals() method do content check

3. Question 3 of 25

   Select 2

   Which two code fragments cause a compilation error? (Choose two.)

   • Afloat flt = 100.00F;
   • Bfloat flt = (float) 1_11.00;
   • CFloat flt = 100.00;
   • Ddouble y1 = 203.22; float flt = y1;
   • Eint y2 = 100; float flt = (float) y2 ;

   Show answer

   Correct answer: C, D

   How others answered

   • CD100%

   Explanation

   The answer is C and D. A. float flt = 100.00F; // correct B. float flt = (float) 1_11.00; // correct C. Float flt = 100.00; // assigning a double to Float object without explicit cast D. double y1 = 203.22; float flt = y1; // assigning a double to a float without explicit cast E. int y2 = 100; float flt = (float) y2 ; // correct

   Adapted from community discussion by anastacia · 2024-10-04 · 3 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

   View community discussion4

   Candidate context

   Question 3 discussion

   These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

   1. A

      anastacia

      2024-10-04

      3 upvotes

      The answer is C and D. A. float flt = 100.00F; // correct B. float flt = (float) 1_11.00; // correct C. Float flt = 100.00; // assigning a double to Float object without explicit cast D. double y1 = 203.22; float flt = y1; // assigning a double to a float without explicit cast E. int y2 = 100; float flt = (float) y2 ; // correct

   2. A

      akbiyik

      2023-11-30

      1 upvote

      Example 2; import java.util.Arrays; import java.util.Scanner; public class ArrayDefaultValues { int intArray[] = new int[3]; float floatArray[] = new float[3]; byte byteArray[] = new byte[3]; boolean boolArray[] = new boolean[3]; String stringArray[] = new String[3]; public static void main(String args[]){ ArrayDefaultValues obj = new ArrayDefaultValues(); System.out.println(Arrays.toString(obj.intArray)); System.out.println(Arrays.toString(obj.floatArray)); System.out.println(Arrays.toString(obj.byteArray)); System.out.println(Arrays.toString(obj.boolArray)); System.out.println(Arrays.toString(obj.stringArray)); } } Output [0, 0, 0] [0.0, 0.0, 0.0] [0, 0, 0] [false, false, false] [null, null, null]

   3. C

      carloswork

      2023-11-02

      Selected C, D1 upvote

      Answer is CD. To test, uncomment one option at a time in the IDE and check the error. public static void main(String[] args) { //float flt = 100.00F; // A - Ok //float flt = (float) 1_11.00; // B - Ok //Float flt = 100.00; // C - Error - need cast to Float. //double y1 = 203.22; float flt = y1; // D - Error - need cast to float. //int y2 = 100; float flt = (float) y2 ; // F - Ok }

   4. I

      iSnover

      2023-09-30

      1 upvote

      The reason of A and C is that the number 100.00 cannot be a float

4. Question 4 of 25

   Given:

   

   And given the code fragment: Book book1 = new EBook(); book1.readBook(); Which option enables the code to compile?

   

   • AOption A
   • BOption B
   • COption C
   • DOption D

   Show answer

   Correct answer: D

   How others answered

   • D100%

   Explanation

   The answer is letter D, answers A and B are clearly wrong, C would also be right, but what shows that it is wrong is the code fragment that calls the class directly. You cannot instantiate abstract classes directly as it was done in the code snippet, so the letter D is correct, implement the abstracted method.

   Adapted from community discussion by iSnover · 2022-10-26 · 2 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

   View community discussion4

   Candidate context

   Question 4 discussion

   These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

   1. I

      iSnover

      2022-10-26

      Selected D2 upvotes

      The answer is letter D, answers A and B are clearly wrong, C would also be right, but what shows that it is wrong is the code fragment that calls the class directly. You cannot instantiate abstract classes directly as it was done in the code snippet, so the letter D is correct, implement the abstracted method.

   2. D

      Def8

      2022-10-28

      Selected D2 upvotes

      A concrete class needs to implement all the abstract methods.

   3. W

      willokans

      2022-11-22

      2 upvotes

      D - As concrete class will need to implement unimplemented interface method. Java Abstract class can implement interfaces without even providing the implementation of interface methods

   4. V

      Vicky_65

      2023-04-01

      Selected D1 upvote

      D is the correct answer.because we need to give all abstract methods in interface in class which is implementing the interface

5. Question 5 of 25

   Given the code fragments:

   

   What is the result?

   • ACompilation fails only at line n2.
   • BRTool::export Tool::export
   • CTool::export Tool:export
   • DCompilation fails only at line n1.
   • ECompilation fails at both line n1 and line n2.

   Show answer

   Correct answer: D

   How others answered

   • D100%

   Explanation

   D. By default, method declaration in an interface is implicitly Public abstract. Overiding with protected is a more restrictive access modifier.

   Adapted from community discussion by letmein2 · 2020-09-05 · 21 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

   Supporting references

   • Oracle documentation

   View community discussion4

   Candidate context

   Question 5 discussion

   These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

   1. L

      letmein2

      2020-09-05

      21 upvotes

      D. By default, method declaration in an interface is implicitly Public abstract. Overiding with protected is a more restrictive access modifier.

   2. C

      carloswork

      2023-11-03

      Selected D2 upvotes

      Answer is D. "Tool::export" is printed and java.lang.IllegalAccessError error is thrown. Although there is a print on the terminal, this error is classified as a compilation error, according to oracle documentation. https://docs.oracle.com/javase/8/docs/api/java/lang/IllegalAccessError.html To test: interface Exportable { void export(); } class Tool implements Exportable { protected void export () { // line n1 - throwIllegalAccessError System.out.println("Tool::export"); } } class ReportTool extends Tool implements Exportable { public void export () { // line n2 System.out.println("RTool::export"); } public static void main(String[] args) { Tool aTool = new ReportTool(); Tool bTool = new Tool(); callExport(aTool); callExport(bTool); } public static void callExport (Exportable ex) { ex.export(); } }

   3. L

      lnrdgst

      2023-01-30

      4 upvotes

      Guys, in my case, testing here at bluej, I verified that the problem is in the interface, marking the Void with a capital v, in this case, the compilation of line n1 and n2 fails. So what convinced me the most was the answer "E".

   4. A

      akbiyik

      2023-11-30

      Selected D1 upvote

      All abstract, default, and static methods in an interface are implicitly public. void export() has not a package-private access. It is implicitly public. Answer is D. The other typo are not delibarely generated I think.

6. Question 6 of 25

   Given the code fragments:

   

   And:

   

   Which statement is true?

   • AAfter line 11, three objects are eligible for garbage collection.
   • BAfter line 11, two objects are eligible for garbage collection.
   • CAfter line 11, one object is eligible for garbage collection.
   • DAfter line 11, none of the objects are eligible for garbage collection.

   Show answer

   Correct answer: C

   How others answered

   • C60%
   • A20%
   • B20%

   Only 60% agreed on this answer — treat it as contested rather than settled.

   Explanation

   after line 11, nobody referencing to s1, thus makes s1 gc eligible. Even though s2 is null, it is not eligible for gc, because s3 still pointing to it. S3 is not eligible for gc because s1 is pointing to it. Only object that nobody pointing to is S1. After line 11 only one object is eligible for GC = S1.

   Adapted from community discussion by Harid · 2021-02-28 · 15 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

   View community discussion4

   Candidate context

   Question 6 discussion

   These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

   1. H

      Harid

      2021-02-28

      15 upvotes

      after line 11, nobody referencing to s1, thus makes s1 gc eligible. Even though s2 is null, it is not eligible for gc, because s3 still pointing to it. S3 is not eligible for gc because s1 is pointing to it. Only object that nobody pointing to is S1. After line 11 only one object is eligible for GC = S1.

   2. H

      hemassridhar

      2020-07-04

      5 upvotes

      Correct, object created at line num 6 will be eligible for GC.

   3. B

      bshailesh3

      2020-08-08

      3 upvotes

      Answer B is correct, as after assignment of null two references are still pointing. Tested.

   4. J

      JongHwa

      2022-05-21

      2 upvotes

      No. s1, s2, s3 are just variable that has Object's reference value(=address value). they are not Object. you said "nobody referencing to s1". that's incorrect. -> "Nobody is referencing the object that s1 initially had." is correct.

7. Question 7 of 25

   Given:

   

   Which code fragment should you use at line n1 to instantiate the dvd object successfully?

   

   • AOption A
   • BOption B
   • COption C
   • DOption D

   Show answer

   Correct answer: C

   How others answered

   • C100%

   Explanation

   C is the correct answer. --- A is incorrect as there is no default constructor available in CD class. B is incorrect as you cannot have two constructor calls D is incorrect as call to super must be first line of constructor

   Adapted from community discussion by Stewart125 · 2021-10-06 · 7 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

   View community discussion4

   Candidate context

   Question 7 discussion

   These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

   1. S

      Stewart125

      2021-10-06

      7 upvotes

      C is the correct answer. --- A is incorrect as there is no default constructor available in CD class. B is incorrect as you cannot have two constructor calls D is incorrect as call to super must be first line of constructor

   2. L

      levilevi

      2021-08-01

      4 upvotes

      A is not correct because there is no default constructor in CD.

   3. B

      bakhdak

      2023-06-14

      Selected C1 upvote

      Call to super() must be first statement in Derived Class constructor.

   4. A

      alexandrustefanescu

      2022-06-28

      1 upvote

      Answer is C. super(); MUST be first all the time. In the example we have B and C but B is not correct cause I not using the this.r variable so only C is correct

8. Question 8 of 25

   Select 2

   Given the code fragment:

   

   Which two modifications should you make so that the code compiles successfully? (Choose two.)

   • AReplace line 13 with:
   • BReplace line 7 with throw IOException (ג€Exception raisedג€);
   • CReplace line 11 with public static void main(String[]) args) throws Exception {
   • DAt line 14, insert throw new IOException();
   • EReplace line 5 with public void printFileContent() throws IOException {

   Show answer

   Correct answer: C, E

   How others answered

   • CE100%

   Explanation

   The answer is CE, no comments, putting this in the class/method will make it compile normally. In relation to the OCA, in this type of question you have to go with the most obvious.

   Adapted from community discussion by iSnover · 2024-04-08 · 2 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

   View community discussion4

   Candidate context

   Question 8 discussion

   These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

   1. I

      iSnover

      2024-04-08

      Selected C, E2 upvotes

      The answer is CE, no comments, putting this in the class/method will make it compile normally. In relation to the OCA, in this type of question you have to go with the most obvious.

   2. T

      TOPPSI

      2024-07-08

      2 upvotes

      Result: CE. A not. Because in the catche statment Exception was caught first. In the following catch statment ia a IO Exception caught, since this is a subclass of Exception. The compiler recognizes that the code would never get to this place in the code

   3. J

      jackymak

      2024-12-02

      1 upvote

      class X { public void printFileContent() throws IOException { throw new IOException(); } } public class Test { public static void main(String[] args) throws IOException { X xobj = new X(); xobj.printFileContent(); } }

   4. I

      iSnover

      2024-04-10

      1 upvote

      And forgot to comment, Exceptions need to be declared in the constructor declaration.

9. Question 9 of 25

   Given: MainTest.java:

   

   and commands:

   

   What is the result?

   • AString main 1
   • BAn exception is thrown at runtime
   • CString main 1 2 3
   • DString main 123

   Show answer

   Correct answer: C

   How others answered

   • C83%
   • A17%

   Explanation

   The answer is letter C, prints String main 1 2 3. When we pass something in the argument at the time of execution between " " we are saying that everything that is between quotes will occupy a single space in the array that in the case of "1 2 3" as it is on the command line it is at position 0 of the args array.

   Adapted from community discussion by iSnover · 2023-10-08 · 2 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

   View community discussion4

   Candidate context

   Question 9 discussion

   These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

   1. I

      iSnover

      2023-10-08

      Selected C2 upvotes

      The answer is letter C, prints String main 1 2 3. When we pass something in the argument at the time of execution between " " we are saying that everything that is between quotes will occupy a single space in the array that in the case of "1 2 3" as it is on the command line it is at position 0 of the args array.

   2. C

      carloswork

      2023-11-20

      Selected C1 upvote

      Answer is C. This question is similar to question 56. The iSnover explanation below is correct. I was wrong in the comment where I informed you that the answer is C.

   3. J

      javaforlife

      2023-09-21

      2 upvotes

      the correct answer is C if it was java 1 2 3 then it would have been A

   4. C

      carloswork

      2023-11-22

      1 upvote

      I was wrong in the comment where I informed you that the answer is A.

10. Question 10 of 25

    Given this code for the classes MyException and Test:

    

    What is the result?

    • AA
    • BAB
    • CA compile time error occurs at line n1.
    • DB
    • EI

    Show answer

    Correct answer: E

    How others answered

    • E100%

    Explanation

    class MyException extends RuntimeException { }; public class Test { public static void main(String[] args) { try { method1(); } catch (MyException ne) { System.out.print("A"); } } public static void method1() { try { throw 3 > 10 ? new MyException() : new IOException(); } catch (IOException ie) { System.out.println("I"); }catch(Exception re ){ System.out.println("B"); } } }

    Adapted from community discussion by jackymak · 2024-07-24 · 2 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

    View community discussion4

    Candidate context

    Question 10 discussion

    These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

    1. I

       iSnover

       2023-10-08

       Selected E3 upvotes

       The answer is the letter E, in Try-Catch it falls into the false one that prints "I" and it's all right that it passes smoothly in the execution.

    2. J

       jackymak

       2024-07-24

       Selected E2 upvotes

       class MyException extends RuntimeException { }; public class Test { public static void main(String[] args) { try { method1(); } catch (MyException ne) { System.out.print("A"); } } public static void method1() { try { throw 3 > 10 ? new MyException() : new IOException(); } catch (IOException ie) { System.out.println("I"); }catch(Exception re ){ System.out.println("B"); } } }

    3. H

       Hirushi

       2024-02-13

       2 upvotes

       Compile error at first catch block (Exception 'java.io.IOException' is never thrown in the corresponding try block)

    4. Y

       yyfabc

       2024-04-13

       1 upvote

       no compiling error in my side but it does create warning in my code from VSCODE

11. Question 11 of 25

    Given the code fragment from three files:

    

    Which code fragment, when inserted at line 2, enables the code to compile?

    

    • AOption A
    • BOption B
    • COption C
    • DOption D
    • EOption E

    Show answer

    Correct answer: E

    How others answered

    • E100%

    Explanation

    Incorrect question: if class name of public class is USMarket then file can't have name Market.java If we rename public class or file - correct answer "E"

    Adapted from community discussion by Krok · 2024-06-17 · 3 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

    View community discussion4

    Candidate context

    Question 11 discussion

    These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

    1. D

       DanielLeeee

       2021-09-30

       12 upvotes

       import sales.*; import sales.products.*; OR import sales.SalesMan; import sales.products.Product;

    2. K

       Krok

       2024-06-17

       Selected E3 upvotes

       Incorrect question: if class name of public class is USMarket then file can't have name Market.java If we rename public class or file - correct answer "E"

    3. P

       PD1985

       2020-09-28

       3 upvotes

       Answer is E, but there is a typo for last class. file name mentioned as Market.java however the class name is defined as USMarket.java.

    4. A

       andiks

       2021-07-19

       2 upvotes

       Import sales* - Imports all classes in sales but sales itself. Correct answer is C

12. Question 12 of 25

    Select 2

    Given these classes:

    

    and this main method:

    

    Which two options compile when placed at line n1 of the main method? (Choose two.)

    • Adirector.stockOptions = 1_000;
    • Bemployee.salary = 50_000;
    • Cmanager.budget = 1_000_000;
    • Dmanager.stockOption = 500;
    • Eemployee.budget = 200_000;
    • Fdirector.salary = 80_000;

    Show answer

    Correct answer: B, F

    How others answered

    • BF67%
    • DE33%

    Only 67% agreed on this answer — treat it as contested rather than settled.

    Explanation

    BF are correct. 1. The type of the object determines which properties exist within the object in memory. 2. The type of the reference to the object determines which methods and variables are accessible to the Java program. class Employee{ public int salary; } class Manager extends Employee{ public int budget; } public class Director extends Manager{ public int stockOptions; public static void main(String[] args) { Employee employee = new Employee(); Employee manager = new Manager(); Employee director = new Director(); // director.stockOptions = 1_000; // Compilation Error employee.salary = 50_000; // manager.budget = 1_000_000; // Compilation Error // manager.stockOption = 500; // Compilation Error // employee.budget = 200_000; // Compilation Error director.salary = 80_000; } }

    Adapted from community discussion by UAK94 · 2022-10-14 · 3 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

    View community discussion4

    Candidate context

    Question 12 discussion

    These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

    1. U

       UAK94

       2022-10-14

       3 upvotes

       BF are correct. 1. The type of the object determines which properties exist within the object in memory. 2. The type of the reference to the object determines which methods and variables are accessible to the Java program. class Employee{ public int salary; } class Manager extends Employee{ public int budget; } public class Director extends Manager{ public int stockOptions; public static void main(String[] args) { Employee employee = new Employee(); Employee manager = new Manager(); Employee director = new Director(); // director.stockOptions = 1_000; // Compilation Error employee.salary = 50_000; // manager.budget = 1_000_000; // Compilation Error // manager.stockOption = 500; // Compilation Error // employee.budget = 200_000; // Compilation Error director.salary = 80_000; } }

    2. I

       iSnover

       2022-10-08

       Selected B, F2 upvotes

       The answer is BF but this is a tricky question as it involves inheritance and polymorphism. But you can make it less complicated if we eliminate the wrong ones: -> Let's start with the letter B, she is evidently right, because "Employee" is the parent class and the method is hers, so we already have a right alternative. -> We are already sure that the letter B is the first correct one, when looking at the code, we realize that all objects are a variable of type "Employee", that is, they can access anything public of the "Employee" class without problems , and the only answer we have with that is the letter F.

    3. S

       Sisar

       2024-08-27

       Selected B, F1 upvote

       Correct is BF. public class App { public static void main(String[] args) { Employee employee = new Employee(); Employee manager = new Manager(); Employee director = new Director(); director.stockOptions = 1_000; employee.salary = 50_000; manager.budget = 1_000_000; manager.stockOption = 500; employee.budget = 200_000; director.salary = 80_000; } } class Employee { public int salary; } class Manager extends Employee { public int budget; } class Director extends Manager { public int stockOptions; }

    4. A

       alex_au

       2022-10-09

       1 upvote

       The original question asks for the two **wrong** options. Actually manager can access .budget and director can also access .stockOptions.

13. Question 13 of 25

    Given the code fragment:

    

    Which action enables it to print AB?

    • AComment lines 18 to 21.
    • BComment line 20.
    • CComment line 19.
    • DComment line 16.

    Show answer

    Correct answer: C

    How others answered

    • C73%
    • B27%

    Only 73% agreed on this answer — treat it as contested rather than settled.

    Explanation

    Answer is C. List lst = Arrays.asList("A","B", "C","D" ); Iterator itr = lst.iterator(); while(itr.hasNext()) { String e=itr.next(); if (e=="C") { break; } else { // continue; System.out.println(e); } } // AB

    Adapted from community discussion by UAK94 · 2022-10-14 · 6 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

    View community discussion4

    Candidate context

    Question 13 discussion

    These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

    1. U

       UAK94

       2022-10-14

       6 upvotes

       Answer is C. List lst = Arrays.asList("A","B", "C","D" ); Iterator itr = lst.iterator(); while(itr.hasNext()) { String e=itr.next(); if (e=="C") { break; } else { // continue; System.out.println(e); } } // AB

    2. T

       TOPPSI

       2023-01-10

       Selected C2 upvotes

       The question is wich action prits AB. If you comment out line 19 it will print AB

    3. T

       TOPPSI

       2023-01-09

       2 upvotes

       Correct anwser ist B: line 20 is an unrechable statment, because of the countinue Statment

    4. C

       carloswork

       2022-10-26

       Selected C1 upvote

       The correct is, answer C. I'm wrong, The UAK94's source code corresponds to the option C.

14. Question 14 of 25

    Given the definitions of the Bird class and the Peacock class:

    

    and the code fragment:

    

    Which code snippet can be inserted to print Fly.Dance. ?

    • ABird p = new Peacock();
    • BBird b = new Bird(); Peacock p = (Peacock) b;
    • CPeacock b = new Peacock (); Bird p = (Bird) b;
    • DBird b = new Peacock (); Peacock p = (Peacock) b;

    Show answer

    Correct answer: D

    How others answered

    • D83%
    • B17%

    Explanation

    Tested. Answer is D. To test: class Bird { public void fly() { System.out.print("Fly."); } } class Peacock extends Bird { public void dance() { System.out.print("Dance."); } } public class Test { public static void main(String[] args) { Bird b = new Peacock (); Peacock p = (Peacock) b; p.fly(); p.dance(); } }

    Adapted from community discussion by carloswork · 2022-10-26 · 3 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

    View community discussion4

    Candidate context

    Question 14 discussion

    These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

    1. C

       carloswork

       2022-10-26

       Selected D3 upvotes

       Tested. Answer is D. To test: class Bird { public void fly() { System.out.print("Fly."); } } class Peacock extends Bird { public void dance() { System.out.print("Dance."); } } public class Test { public static void main(String[] args) { Bird b = new Peacock (); Peacock p = (Peacock) b; p.fly(); p.dance(); } }

    2. U

       UAK94

       2022-10-14

       3 upvotes

       Answer is D. class Bird{ public void fly() { System.out.println("fly"); } } public class Peacock extends Bird { public void dance() { System.out.println("dance"); } public static void main(String[] args) { Bird b= new Peacock(); Peacock p= (Peacock) b; // Bird b= new Bird(); // Peacock p= (Peacock) b; // ClassCastException p.fly(); p.dance(); } }

    3. A

       alex_au

       2022-10-09

       4 upvotes

       Correct answer is D. For option B there is ClassCastException as class Bird cannot be casted to class Peacock

    4. D

       DriftKing

       2023-09-13

       Selected D2 upvotes

       Tested. Answer is - D. Bird b = new Peacock (); Peacock p = (Peacock) b;

15. Question 15 of 25

    Given the code fragment:

    

    You want the code to print: Sold: 5 items. Stock in Hand: 5 - Purchased: 5 items. Stock in Hand: 10? Which action enables the code to print this?

    • ADeclare the stock variable and the purchase(), sell(), and printStock() methods static.
    • BDeclare the stock variable and the printStock() method static.
    • CDeclare the stock and qty variables and the printStock() method static.
    • DDeclare the stock variable static.

    Show answer

    Correct answer: C

    How others answered

    • C100%

    Explanation

    To test: class StockRoom { private int stock = 10; public void purchase(int qty) { stock += qty; } public void sell (int qty) { stock -= qty; } public void printStock (String action) { System.out.println( action + ":" + qty + " items. Stock in Hand: " + stock); // qty needs to be static variable to compile. } } public class Test { public static void main(String[] args) { StockRoom k1 = new StockRoom(); k1.sell(5); k1.printStock("Sold"); StockRoom k2 = new StockRoom(); k2.purchase(5); k2.printStock("Purchased"); } }

    Adapted from community discussion by carloswork · 2022-10-28 · 2 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

    View community discussion4

    Candidate context

    Question 15 discussion

    These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

    1. C

       carloswork

       2022-10-28

       2 upvotes

       To test: class StockRoom { private int stock = 10; public void purchase(int qty) { stock += qty; } public void sell (int qty) { stock -= qty; } public void printStock (String action) { System.out.println( action + ":" + qty + " items. Stock in Hand: " + stock); // qty needs to be static variable to compile. } } public class Test { public static void main(String[] args) { StockRoom k1 = new StockRoom(); k1.sell(5); k1.printStock("Sold"); StockRoom k2 = new StockRoom(); k2.purchase(5); k2.printStock("Purchased"); } }

    2. I

       iSnover

       2022-10-09

       Selected C1 upvote

       The answer is C, to perform the operation, the variables must be static to refer to a single value and the method must also be static, otherwise, variables that are static cannot be used inside it.

    3. A

       alex_au

       2022-10-10

       2 upvotes

       Methods no need to be static in order to use static variables. However, static methods cannot use instance variables.

    4. A

       af8c3f3

       2024-09-20

       1 upvote

       isn't qty a local variable? and local variables can't be static?

16. Question 16 of 25

    Given:

    

    and the code fragment: S2 sobj = new S2(); sobj.display(10, 100); What is the result?

    • AChild 10 Child 100 Parent 100
    • BParent 10 Child 10 Parent 1000
    • CChild 10 Parent 100 Parent 100
    • DA compile time error occurs.

    Show answer

    Correct answer: A

    How others answered

    • A71%
    • D29%

    Only 71% agreed on this answer — treat it as contested rather than settled.

    Explanation

    Answer is A. class S1 { protected void display(int x) { System.out.println("Parent" + x); } } public class S2 extends S1 { public void display(int x, int y) { this.display(x); display(y); super.display(y); } public void display(int x) { System.out.println("Child" + x); } public static void main(String[] args) { S2 sobj = new S2(); sobj.display(10, 100); // Child10 Child100 Parent100 } }

    Adapted from community discussion by UAK94 · 2022-10-15 · 2 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

    View community discussion4

    Candidate context

    Question 16 discussion

    These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

    1. U

       UAK94

       2022-10-15

       2 upvotes

       Answer is A. class S1 { protected void display(int x) { System.out.println("Parent" + x); } } public class S2 extends S1 { public void display(int x, int y) { this.display(x); display(y); super.display(y); } public void display(int x) { System.out.println("Child" + x); } public static void main(String[] args) { S2 sobj = new S2(); sobj.display(10, 100); // Child10 Child100 Parent100 } }

    2. A

       amit_lad88

       2023-12-19

       3 upvotes

       The correct answer is A. it will print Child10 Child100 Parent100

    3. A

       alex_au

       2022-10-10

       2 upvotes

       JVM will automatically create a default constructor to initialize all fields (which in these cases there are no any attribute fields) so there will be no any compile time error.

    4. M

       MPignaProTech

       2024-10-27

       Selected A1 upvote

       super can be called at any time in a method, but inside a constructor it must be called called first. Good answer is A

17. Question 17 of 25

    Given:

    

    Which statement is true?

    • AThe program executes and prints: 500.0
    • BCommenting line 16 enables the program to print: Thank You! 500.0
    • CCommenting line 13 enables the program to print: Thank You! 500.0
    • DThe program executes and prints: Thank You! 500.0

    Show answer

    Correct answer: B

    How others answered

    • B100%

    Explanation

    Correct, the answer is B. If the question's source code is executed, it will throw the following exception: - Exception in "main" thread java.lang.Error: Unresolved compilation issue: continue cannot be used outside of a loop The continue statement cannot be used outside of a loop, as described in the documentation. When commenting the continue statement line, the code compiles normally and then print "Thank You! 500.0" https://docs.oracle.com/javase/tutorial/java/nutsandbolts/branch.html To test: public static void main(String[] args) { int price = 1000; int qty = 2; String grade = "2"; double discount = 0.0; switch(grade) { case "1": discount = price * 0.1; break; case "2": discount = price * 0.5; //continue; default: System.out.println("Thank You!"); } System.out.println(discount); }

    Adapted from community discussion by carloswork · 2024-10-28 · 2 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

    Supporting references

    • Oracle documentation

    View community discussion2

    Candidate context

    Question 17 discussion

    These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

    1. C

       carloswork

       2024-10-28

       Selected B2 upvotes

       Correct, the answer is B. If the question's source code is executed, it will throw the following exception: - Exception in "main" thread java.lang.Error: Unresolved compilation issue: continue cannot be used outside of a loop The continue statement cannot be used outside of a loop, as described in the documentation. When commenting the continue statement line, the code compiles normally and then print "Thank You! 500.0" https://docs.oracle.com/javase/tutorial/java/nutsandbolts/branch.html To test: public static void main(String[] args) { int price = 1000; int qty = 2; String grade = "2"; double discount = 0.0; switch(grade) { case "1": discount = price * 0.1; break; case "2": discount = price * 0.5; //continue; default: System.out.println("Thank You!"); } System.out.println(discount); }

    2. I

       iSnover

       2024-10-09

       Selected B1 upvote

       The answer is the letter B, notice that on line 16 the command ends with ":" instead of ";" causing the code not to compile. commenting out the line the code compiles normally.

18. Question 18 of 25

    Given the code fragment:

    

    Which modification enables the code fragment to print TrueDone?

    • AReplace line 5 With String opt = "true"; Replace line 7 with case "true":
    • BReplace line 5 with boolean opt = l; Replace line 7 with case 1:
    • CAt line 9, remove the break statement.
    • DRemove the default section.

    Show answer

    Correct answer: A

    How others answered

    • A100%

    Explanation

    A is correct, you cannot have a boolean, double or long in a switch statement thus B is incorrect. C would result in the default statement being printed out

    Adapted from community discussion by Fuego_412_ · 2020-12-14 · 2 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

    View community discussion4

    Candidate context

    Question 18 discussion

    These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

    1. A

       Anton2020

       2021-10-10

       5 upvotes

       There are more options: char, byte, short, Character, Byte, Short, Integer or enum

    2. V

       v323rs

       2020-01-12

       4 upvotes

       Agree, the correct answer "A" Replace line 5 With String opt = "true"; Replace line 7 with case "true":

    3. F

       Fuego_412_

       2020-12-14

       2 upvotes

       A is correct, you cannot have a boolean, double or long in a switch statement thus B is incorrect. C would result in the default statement being printed out

    4. D

       dya45792

       2019-12-12

       1 upvote

       Antwort B ergibt kein sinn , also es muss ein String sein , und bei C und D auslassen oder nicht kommt nur ; DONE raus public static void main(String[] args) { String opt = " true"; switch (opt) { case "true": System.out.print("True"); // break; // default: System.out.print("***"); } System.out.println("Done"); } } also nur Done / oder ?

19. Question 19 of 25

    Given these requirements: ✑ Bus and Boat are Vehicle type classes. ✑ The start() and stop() methods perform common operations across the Vehicle class type. ✑ The ride() method performs a unique operations for each type of Vehicle. Which set of actions meets the requirements with optimized code?

    • A1. Create an abstract class Vehicle by defining start() and stop() methods, and declaring the ride() abstract method. 2. Create Bus and Boat classes by inheriting the Vehicle class and overriding the ride() method.
    • B1. Create an interface Vehicle by defining start() and stop() methods, and declaring the ride() abstract method. 2. Create Bus and Boat classes by implementing the Vehicle class.
    • C1. Create an abstract class Vehicle by declaring stop(), start(), and ride() abstract methods. 2. Create Bus and Boat classes by inheriting the Vehicle class and overriding all the methods.
    • D1. Create an interface Vehicle by defining default stop(), start(), and ride() methods. 2. Create Bus and Boat classes by implementing the Vehicle interface and overriding the ride() method.

    Show answer

    Correct answer: A

    How others answered

    • A71%
    • D29%

    Only 71% agreed on this answer — treat it as contested rather than settled.

    Explanation

    both A and D are correct. in A, we define an abstract class with concrete methods for start() and stop(), and an abstract ride(). in B, we define default start() and stop(), and ride() we keep it abstract. meaning in both approaches we are defining a common operation, and require an overridden version for ride() for the code to work. the only difference is if we want to extend another class other than Vehicle, then we need option D, however with the requirements we have here, both options works fine.

    Adapted from community discussion by charbelishak · 2024-01-18 · 2 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

    View community discussion4

    Candidate context

    Question 19 discussion

    These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

    1. C

       charbelishak

       2024-01-18

       Selected A2 upvotes

       both A and D are correct. in A, we define an abstract class with concrete methods for start() and stop(), and an abstract ride(). in B, we define default start() and stop(), and ride() we keep it abstract. meaning in both approaches we are defining a common operation, and require an overridden version for ride() for the code to work. the only difference is if we want to extend another class other than Vehicle, then we need option D, however with the requirements we have here, both options works fine.

    2. S

       Syyyyyyyy

       2024-03-11

       Selected A2 upvotes

       Apparently it's A. Both A and D can work, but default methods in interfaces are introduced to provide extension to interfaces, not really to use it to provide abstraction.

    3. U

       UAK94

       2022-10-15

       2 upvotes

       interface Vehicle{ default void start() {} default void stop() {} default void ride() {} } class Bus implements Vehicle{ public void ride() {System.out.println("Bus");} } public class Boat implements Vehicle{ public void ride() {System.out.println("Boat");} } Answer is D

    4. Y

       yyfabc

       2023-04-21

       1 upvote

       It feels weird to me ... if i have to solve this problem in a realistic scenario, i think i could create abstract class with implementations of those 2 methods and make ride() abstract, or using default in interface to put fixed logic for the 2 methods and let Bus and Boat implement the interface and override ride().. plz correct me for any mistake, tx!

20. Question 20 of 25

    Select 2

    Given:

    

    Which two code fragments can be inserted at line n1? (Choose two.)

    • AString str = ג€Javaג€;
    • Bfor(int iVal = 0; iVal <=5; iVal++){}
    • CTest() {}
    • Dpackage p1;
    • Eimport java.io.*;

    Show answer

    Correct answer: A, C

    How others answered

    • AC100%

    Explanation

    The correct answer is AC, I will explain: B -> You can't put a for loop outside the main method D -> This one is very wrong, you can only declare the package in the first line of code E -> This one is as wrong as the letter D, you can only import one package before instantiating the class and if you have a package declared, the declaration can only be made between it and the class declaration, anything outside of that is wrong

    Adapted from community discussion by iSnover · 2023-10-09 · 2 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

    View community discussion4

    Candidate context

    Question 20 discussion

    These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

    1. I

       iSnover

       2023-10-09

       Selected A, C2 upvotes

       The correct answer is AC, I will explain: B -> You can't put a for loop outside the main method D -> This one is very wrong, you can only declare the package in the first line of code E -> This one is as wrong as the letter D, you can only import one package before instantiating the class and if you have a package declared, the declaration can only be made between it and the class declaration, anything outside of that is wrong

    2. D

       DriftKing

       2024-08-26

       3 upvotes

       Option A shows ---> String str = ג€Javaג€; Looks like this is a mistake as it should be ---> String str = "Java";

    3. A

       akbiyik

       2023-12-07

       Selected A, C1 upvote

       A) Test( ) { } B) String str = "Java"; C) package p1; D) import java.io.*; E) for (int ival = 0 ; ival <= 5; ival++) { }

    4. S

       shivkumarx

       2023-09-13

       2 upvotes

       Correct answer is A and C You can't declare a for loop outside main method

21. Question 21 of 25

    Given the following main method:

    

    What is the result?

    • A5 4 3 2 1 0
    • B5 4 3 2 1
    • C4 2 1
    • D5
    • ENothing is printed

    Show answer

    Correct answer: D

    How others answered

    • D100%

    Explanation

    Antwort D ; public static void main(String[] args) { int num = 5; do { System.out.println(num-- + " "); }while (num == 0); } } 5

    Adapted from community discussion by dya45792 · 2020-12-12 · 7 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

    View community discussion4

    Candidate context

    Question 21 discussion

    These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

    1. D

       dya45792

       2020-12-12

       7 upvotes

       Antwort D ; public static void main(String[] args) { int num = 5; do { System.out.println(num-- + " "); }while (num == 0); } } 5

    2. D

       ddpk

       2023-01-26

       2 upvotes

       Answer D Do while loop works at least one time and (num==0) always return false.

    3. T

       TheeOne

       2025-01-21

       Selected D1 upvote

       Do while executes atleast once, --order--: print first then check the condition, in this case num will never be 0

    4. D

       duydn

       2024-08-25

       Selected D1 upvote

       do while always exec for the first loop, and the condition always false -> D is correct

22. Question 22 of 25

    Select 2

    Which two code fragments cause compilation errors? (Choose two.)

    • Adouble y1 = 203.22; float fit = y1;
    • Bfloat fit = (float) 1_11.00;
    • CFloat fit = 100.00;
    • Dint y2 = 100; float fit = (float) y2;
    • Efloat fit = 100.00F;

    Show answer

    Correct answer: A, C

    How others answered

    • AC100%

    Explanation

    Answer is AC. Options that do not compile are A and C. Here is similar variables to test: public static void main(String[] args) { double y1 = 203.22; float fit = y1; // A --- error float fit2 = (float) 1_11.00; // B Float fit3 = 100.00; // C --- error int y2 = 100; float fit4 = (float) y2; // D float fit5 = 100.00F; // E }

    Adapted from community discussion by carloswork · 2023-04-28 · 2 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

    View community discussion4

    Candidate context

    Question 22 discussion

    These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

    1. C

       carloswork

       2023-04-28

       Selected A, C2 upvotes

       Answer is AC. Options that do not compile are A and C. Here is similar variables to test: public static void main(String[] args) { double y1 = 203.22; float fit = y1; // A --- error float fit2 = (float) 1_11.00; // B Float fit3 = 100.00; // C --- error int y2 = 100; float fit4 = (float) y2; // D float fit5 = 100.00F; // E }

    2. I

       iSnover

       2023-04-10

       Selected A, C1 upvote

       The answer is AC, I knew but just in case I tested it and here is the explanation of each one: A -> Wrong, because a float cannot receive a double B -> Correct, how are you doing a cast to float compiles C -> Wrong, a float cannot receive a value of 100.00, it is out of its range D -> Compile, cast to float And -> Compiles too, when you put F at the end inform that it is a floating point number, then compile.

    3. S

       ShaileshRaj

       2024-04-05

       Selected A, C2 upvotes

       These answer will give compilation error as question is asking, other option will give no compilation error

    4. B

       BelloMio

       2024-12-06

       1 upvote

       Someone gave wrong explanations. The reason why A and C are wrong is because we cannot implicitly cast double values to float. They need to be explicitly casted, but it is possible.

23. Question 23 of 25

    Select 2

    Given the code fragment:

    

    Which two modifications, when made independently, enable the code to print joe:true: 100.0? (Choose two.)

    

    • AOption A
    • BOption B
    • COption C
    • DOption D
    • EOption E

    Show answer

    Correct answer: A, C

    How others answered

    • AC70%
    • AD20%

    Only 70% agreed on this answer — treat it as contested rather than settled.

    Explanation

    Here is the code for testing: public class Employee { String name; Boolean contract; double salary; Employee(){ // Line n1 - Option C /* this.name = new String("Joe"); this.contract = new Boolean(true); this.salary = new Double(100); */ } public String toString() { return name + " : " + contract + " : " + salary; } public static void main(String [] args) { Employee e = new Employee(); // Line n2 - Option A /* e.name = "Joe"; e.contract = true; e.salary = 100; */ // Line n2 - Option B /* this.name = "Joe"; this.contract = true; this.salary = 100; */ System.out.print(e); } }

    Adapted from community discussion by carloswork · 2022-10-19 · 3 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

    View community discussion4

    Candidate context

    Question 23 discussion

    These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

    1. C

       carloswork

       2022-10-19

       3 upvotes

       Here is the code for testing: public class Employee { String name; Boolean contract; double salary; Employee(){ // Line n1 - Option C /* this.name = new String("Joe"); this.contract = new Boolean(true); this.salary = new Double(100); */ } public String toString() { return name + " : " + contract + " : " + salary; } public static void main(String [] args) { Employee e = new Employee(); // Line n2 - Option A /* e.name = "Joe"; e.contract = true; e.salary = 100; */ // Line n2 - Option B /* this.name = "Joe"; this.contract = true; this.salary = 100; */ System.out.print(e); } }

    2. C

       carloswork

       2022-10-19

       Selected A, C2 upvotes

       Tested. Answer is A and C. Comments: * Option C works even though it has been deprecated since version 9 (1z0-808 is about java 8). ** Option B would only be correct if it is inside the constructor, on line n1. As option B is pointing to line n2, it will throws an error.

    3. I

       iSnover

       2022-10-03

       3 upvotes

       Wrong, you can't put "this" in static methods. The correct one is A and C.

    4. I

       iSnover

       2022-10-03

       Selected A, C1 upvote

       The correct options are A and C, as all others contain syntax errors. It cannot be the letter B because "this" cannot be instantiated in static methods.

24. Question 24 of 25

    Given the code fragment:

    

    Which code fragment, when inserted at line 9, enables the code to print true?

    • AString str2 = str1;
    • BString str2 = new String(str1);
    • CString str2 = sb1. toString();
    • DString str2 = "Duke";

    Show answer

    Correct answer: A

    How others answered

    • A100%

    Explanation

    The answer is letter A, even the string containing the content inside them as they are not considered equal, to "circumvent" this, we must instantiate the second String taking the first as a reference as "str2 = str1;" and with that we can print true with the boolean "str2 == str1".

    Adapted from community discussion by iSnover · 2022-10-04 · 4 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

    View community discussion4

    Candidate context

    Question 24 discussion

    These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

    1. I

       iSnover

       2022-10-04

       Selected A4 upvotes

       The answer is letter A, even the string containing the content inside them as they are not considered equal, to "circumvent" this, we must instantiate the second String taking the first as a reference as "str2 = str1;" and with that we can print true with the boolean "str2 == str1".

    2. D

       dya45792

       2019-12-14

       3 upvotes

       Antwort A, ist richtig: StringBuilder sb1 = new StringBuilder("Duke"); String str1 = sb1.toString(); String str2 = str1; System.out.println(str1 == str2); } } true

    3. S

       Srivani

       2021-02-18

       2 upvotes

       A is the correct answer because == checks the reference equalities where as equals checks the character equality.

    4. A

       Anton2020

       2021-11-25

       Selected A1 upvote

       if you want the object location in memory for two Strings to be the same, they need to be set equal with ==

25. Question 25 of 25

    Select 2

    Given the code fragment:

    

    And given the requirements: ✑ If the value of the qty variable is greater than or equal to 90, discount = 0.5 ✑ If the value of the qty variable is between 80 and 90, discount = 0.2 Which two code fragments can be independently placed at line n1 to meet the requirements? (Choose two.)

    

    • AOption A
    • BOption B
    • COption C
    • DOption D
    • EOption E

    Show answer

    Correct answer: A, C

    How others answered

    • AC100%

    Explanation

    A and C are the correct options because in Option B we are using the discount variable twice, and thus only the second discount's value will be the final updated value in Option D only the case where qty >= 90 will work in Option E everytime the output printed will be 0.2 no matter if the condition is 1st or 2nd.

    Adapted from community discussion by kkaayyyy · 2022-10-06 · 3 community upvotes. Third-party contribution; reviewed by our quality filter, not independently verified.

    View community discussion4

    Candidate context

    Question 25 discussion

    These are quality-filtered excerpts from third-party community discussion. They can be useful reasoning signals, but they are not official answers and have not been independently verified by 1Z0-808 or CertExamHQ.

    1. K

       kkaayyyy

       2022-10-06

       3 upvotes

       A and C are the correct options because in Option B we are using the discount variable twice, and thus only the second discount's value will be the final updated value in Option D only the case where qty >= 90 will work in Option E everytime the output printed will be 0.2 no matter if the condition is 1st or 2nd.

    2. C

       carloswork

       2022-11-01

       Selected A, C1 upvote

       Answer is AC. It boring to test... It is necessary to change the value of the variable 'qty' to perform the test. The code can be compiled on the command line and the value passed by "args" or simply change the value of this variable directly in the code. That's it. To test: public static void main(String[] args) { double discount = 0; int qty = Integer.parseInt(args[0]); qty=90; // change here // Answer A if (qty >= 90) { discount = 0.5; } System.out.println(discount); if (qty > 80 && qty = 90) ? 0.5 : (qty > 80)? 0.2: 0; System.out.println(discount); }

    3. B

       BelloMio

       2024-06-06

       2 upvotes

       Ok I got why D is wrong. it is wrong in the case if qty is 85 for example. it will go in the first if condition which will make discount = 0.2 all good. Then it will also go in the second if condition where it will go into the else statement and assign discount = 0, making the code not correct

    4. V

       vic88

       2024-11-30

       Selected A, C1 upvote

       D is not right because it's double checked again after if (qty >=90) {discount = 0.5), reset the discount to 0.

Keyboard:J Kmove between questionsA–Dselect an answerRshow or hide the answer← →change page

1Z0-808 practice, question 1 of 25