To answer this question, let's go through each option to understand why it is correct or incorrect:
Option A) As the code stands, it will not compile - This option is incorrect. The code provided does compile without any errors.
Option B) As the code stands, the output will be 2 - This option is incorrect. The output will not be 2 because the equals()
method in the ToDos
class is not properly overridden.
Option C) As the code stands, the output will be 3 - This option is correct. The output will be 3 because the equals()
method in the ToDos
class is comparing the day
field of the objects. Since t1
and t2
have the same value for day
("Monday"), they are considered equal, but they are distinct objects, so both of them are added to the map. Therefore, the map contains 3 entries: t1
-> "doLaundry", t2
-> "payBills", and t3
-> "cleanAttic".
Option D) If the hashCode()
method is uncommented, the output will be 2 - This option is correct. If the hashCode()
method is uncommented, the output will be 2 because the hashCode()
method is used to determine the bucket in which to store the key-value pair in the HashMap. Since the hashCode()
method is not overridden in the ToDos
class, it uses the default implementation from the Object
class, which generates a unique hash code for each object. As a result, t1
and t2
will generate different hash codes, and the map will consider them as distinct keys, resulting in only two entries: t2
-> "payBills" and t3
-> "cleanAttic".
Option E) If the hashCode()
method is uncommented, the output will be 3 - This option is incorrect. If the hashCode()
method is uncommented, the output will be 2 as explained in option D.
Option F) If the hashCode()
method is uncommented, the code will not compile - This option is incorrect. Uncommenting the hashCode()
method will not cause any compilation errors.
The correct answer is Option C and Option D. These options are correct because they accurately describe the behavior of the code provided.