Java Programming - Declarations and Access Control - Discussion

Why the option A is wrong ?
The compiler will implicitly put "public" in front of Test(). Then why this option is wrong.

And wht do you mean by prototype ?

Option A is wrong. The default constructor has the same access modifier as the class. Here the is defined public so default constructor must be public. And Test() is in default access level.

The compiler will not implicitly put "public" in front of Test().

If the class having the access modifier as private. Then is it the default constructor of class is private?

//I have written simple program
public class Outer
{
Outer(){

System.out.println(" This less access specifier");
}
public static void main(String[] argv)
{
new Outer();
}
}

/*OUTPUT is
This less access specifier
*/
/*This means option A is also correct */
If I am wrong please clarifiy me.

What is the prototype of constructor in before creating an object?

@Sushil.

In your program you are creating an object and overriding default constructor. Try this, create object without constructor and compile it.

Here compiler creates an constructor with the access modifier same as class access modifier. To check this, use javap command in your command prompt.

//Your program:

public class Outer
{
public static void main(String[] argv)
{ new Outer();
}
}

//compile this source code as
-> javac Outer.java
-> javap Outer

@Jothi Pandiyan.

What is the meaning of this?

javac Outer.java
java Outer

Another way of looking at this is that you can overwrite the default constructor to be default or protected, (and likely even private). I've done this below.

However, if you do *NOT* overwrite the default constructor its public and therefore accessible from anywhere.

public class NewTest
{
public int i = 0;

// protected NewTest() // This compiles even though its protected (not public as class is).
NewTest() // This alsp compiles even though its default (not public as class is).
{
System.out.println("NewTest() new called.");
i = 99;
}

public static void main(String args[])
{
NewTest sub = new NewTest();
System.out.println(sub.i);
}
}

Option A is correct.

And I agree @Jothi Pandiyan.

I think Option A is correct.