When do you need to explicitly call a superclass constructor?

JavaConstructorSubclassSuperclass

Java Problem Overview


So say I have a subclass that extends a superclass. In what scenarios do I need to explicitly type super() to get the superclass constructor to run?

I'm looking at an example in a book about abstract classes and when they extend it with a non-abstract subclass, the subclass's default constructor is blank and there's a comment that says the superclass's default constructor will be called. At the same time I've also seen instances on here where someone's problem was not explicitly calling super().

Is the distinction from calling the superclass's default/non-default constructor from the subclass's default/non-default constructor?

Java Solutions


Solution 1 - Java

You never need just

super();

That's what will be there if you don't specify anything else. You only need to specify the constructor to call if:

  • You want to call a superclass constructor which has parameters
  • You want to chain to another constructor in the same class instead of the superclass constructor

You claim that:

> At the same time I've also seen instances on here where someone's problem was not explicitly calling super().

Could you give any examples? I can't imagine how that's possible...

Solution 2 - Java

If you don't explicitly call a super constructor the argument less constructor (super()) will be called. This means you have to call a specific constructor yourself if there's no reachable argument-less constructor of the super class.

But often enough you want a different constructor anyways even if you could use the default constructor - depends on your code.

Also note that if no constructor is declared the compiler generates a public default constructor automatically, but as soon as you write your own constructor this does not happen anymore.

Solution 3 - Java

The super() method is always called in constructors of sub-classes, even if it is not explicitly written in code.

The only time you need to write it, is if there are several super(...) methods in the super-class with different initialization parameters.

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Questionjhlu87View Question on Stackoverflow
Solution 1 - JavaJon SkeetView Answer on Stackoverflow
Solution 2 - JavaVooView Answer on Stackoverflow
Solution 3 - JavaOmnaestView Answer on Stackoverflow