class base { base() { System.out.println("base"); } base(int i1) { } } class Super extends base { Super() { System.out.println("super"); super(1); } public static void main(String [] a) { base b1=new Super(); } }

technology programming languages

  1. compile time error

  2. prints base and super

  3. prints super and base

  4. none of the above

Show answer
Correct Option: A

AI Explanation

To answer this question, we need to understand the concepts of constructors and inheritance in Java.

In the given code, there are two classes: base and Super. The Super class extends the base class.

The base class has two constructors:

  1. The default constructor base() which prints "base".
  2. The parameterized constructor base(int i1).

The Super class has one constructor Super(). Inside this constructor, the super() statement is used to invoke the default constructor of the base class.

In the main method, an object b1 is created using the new keyword. The object is of type base and is assigned as a reference to the Super class.

Now, let's go through each option to determine the correct answer:

Option A) Compile time error - This option is correct. When the Super class extends the base class, it inherits the default constructor of the base class. However, when a class has a parameterized constructor and the default constructor is not explicitly defined, the default constructor is no longer available. In this case, the base class does not have an explicit default constructor, so the super() statement in the Super class constructor will result in a compile-time error.

Option B) Prints base and super - This option is incorrect because the code will not compile, and hence, nothing will be printed.

Option C) Prints super and base - This option is incorrect because the code will not compile, and hence, nothing will be printed.

Option D) None of the above - This option is incorrect because the code will not compile, and hence, nothing will be printed.

Therefore, the correct answer is A) Compile time error.

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