Question

Why can’t this() and super() both be used together in a constructor?

What is the reason for incorporating such a thing?

Answer 1

this(...) will call another constructor in the same class whereas super() will call a super constructor. If there is no super() in a constructor the compiler will add one implicitly.

Thus if both were allowed you could end up calling the super constructor twice.

Example (don’t look for a sense in the parameters):

class A {
  public A() {
    this( false );
  }

  public A(boolean someFlag) {
  }
}

class B extends A {
  public B() {
    super();
  }

  public B( boolean someFlag ) {
    super( someFlag );
  }

  public B ( int someNumber ) {
    this(); //
  }
} 

Now, if you call new B(5) the following constructors are invoked:

     this( false);
A() ---------------> A(false)
^
|
| super(); 
|
|     this();
B() <--------------- B(5)  <--- you start here

Update:

If you were able to use this() and super() you could end up with something like this:

(Attention: this is meant to show what could go wrong, if you were allowed to do that – which you fortunately aren’t)

     this( false);
A() ---------------> A(false)
^                    ^
|                    |
| super();           | super( true ); <--- Problem: should the parameter be true or false? 
|                    |
|     this();        |
B() <--------------- B(5)  <--- you start here

As you can see, you’d run into a problem where the A(boolean) constructor could be invoked with different parameters and you’d now have to somehow decide which should be used. Additionally the other constructors (A() and B()) could contain code that now might not get called correctly (i.e. out of order etc.) since the call to super( true ) would circumvent them while this() wouldn’t.