What will happen when you attempt to compile and run the following code

class Base{ 
 public void Base(){ 
 System.out.println("Base"); 
 } 
} 
public class In extends Base{ 
 public static void main(String argv[]){ 
 In i = new In(); 
 } 
} 

My thinking as follows: The subclass will have default no-arg contructor, which will try calling super().

2. The difference between a method having the same name as the class and the constructor of the class is the return type. Methods MUST have return type and constructors MUST not have return type in their corresponding declarations.

3. The compiler automatically provides a default (no-arg) constructor if and only if there is no other constructors defined by you. If you decide to provide a constructor taking some args then you HAVE TO define the no-arg constructor. Compiler WILL not give one in this later case.

4. Defining a method with the same name as the class is NO WAY affects the decision made by the compiler whether to provide/not to provide no-arg constructor.

So the correct answer for the above qstn is 2)

regds
maha anna

I agree with all your above points.
According to your point 3....my posting(question) should give compilation error, right ?
Because the Base class will not provide no-arg constructor since it has got method. The subclass In() will look for no-arg constructor in the super!!!!

CODE:
class Base{
public void Base(){
System.out.println("Base");
}
//public base(){} This constructor will be provided by compiler since Base doesn't contain any explicitly defined constructors.
}
public class In extends Base{
public static void main(String argv[]){
In i = new In();
//In() {super();} Same here.This constructor will be provided by compiler since In doesn't contain any explicitly defined constructors.
}
}

Finally since both the constructors Base(){} and In(){} doesn't contain anything it will compile just fine but without giving any output.