What's the access modifier of the default constructor in java?

Question

We all know that if we don't specifically define a constructor, the compiler inserts an invisible zero-parameter constructor. I thought its access modifier was public, but in dealing with an inner class issue, I found maybe I was wrong. Here is my code:

public class Outer {
  protected class ProtectedInner {
    // adding a public constructor will solve the error in SubOuterInAnotherPackage class
    //public ProtectedInner() {}
  }
}

And there is a subclass of Outer in another package:

public class SubOuterInAnotherPackage extends Outer {
  public static void main(String[] args) {
    SubOuterInAnotherPackage.ProtectedInner protectedInner 
      = new SubOuterInAnotherPackage().new ProtectedInner(); // Error!! Can't access the default constructor
  }
}

You will get an error in the main() method, but if you add a public constructor to the ProtectedInner class, that error is solved. That's why I'm thinking that the modifier of the default constructor is not public! So could anyone tell me what the access modifier of the default constructor is?

score 53 · Accepted Answer · answered Feb 25 '14 at 07:16

I thought its access modifier is public, but when I deal with a inner class issue, I found maybe I was wrong.

Yup. Indeed, I found myself in the same situation a couple of years ago. I was surprised by an error (through Guice injection, which made it slightly harder to find).

The key is to check the spec, in this case section 8.8.9:

In a class type, if the class is declared public, then the default constructor is implicitly given the access modifier public (§6.6); if the class is declared protected, then the default constructor is implicitly given the access modifier protected (§6.6); if the class is declared private, then the default constructor is implicitly given the access modifier private (§6.6); otherwise, the default constructor has the default access implied by no access modifier.

So in this case, your constructor is implicitly protected.

thanks, I forgot the spec, I only search in Tutorials, thanks a lot for reminding and sharing! — jason, Feb 25 '14 at 07:21
that drove me crazy :) because in I could access type of nested classes but not the constructor till I declare explicit public constructor for this protected nested class .. this spec explains what happened.. ,,thnx — Maher Abuthraa, May 03 '14 at 22:45

score 1 · Answer 2 · answered Jan 18 '16 at 13:33

In addition to what Jon pretty well stated, here is an image example, for the visual guys.

If there is no constructor in a class, compiler automatically creates a default constructor.

Here is an example that successfully depicts the above rule:

For further reference, please refer here.

score 1 · Answer 3 · answered Sep 03 '17 at 05:55

1

I would like to point out one more thing that I recently got. If you define a default constructor for your class then it's acess specifier will be what you assign. For example,

 public class A{
       A(){
       // do some stuff
       }
    }

Here the access specifier of the default constructor is package access and not public access (that of the class). However

 public class A{
       // no constructor is defined
    }

Here the compiler will sympathize with you and give you a default constructor whose access specifier will be same as the class , that is public.

answered Sep 03 '17 at 05:55

Surendra Sharma

33
1
6

1

Defining a no-arg constructor and calling it a default constructor isn't really right. A default constructor is generated by the compiler, when a programmer provides a no-arg constructor, it isn't the same as a default constructor(even if their functionalities can be same, it is still incorrect to call them by the same name. One is provided by the compiler - *default*, the other is a *customised* constructor provided by the programmer). – Ayush Jul 01 '20 at 14:20

What's the access modifier of the default constructor in java?

3 Answers3

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