What does super() mean in a constructor?

Question

What does the code

super();

do inside of a constructor?

For example, here is a constructor for my class

public abstract class Rectangle extends AbstractShape
{

private double height, width;

// Constructors...  
public Rectangle()
{
    super(); //this is how i inherit that point!
    height = -1;
    width = -1;
}

Does it have something to do with super-classes?

Question was well stated, however. So, I am surprised he now has a negative 7 vote on the question. — finneycanhelp, May 01 '13 at 16:27
@finneycanhelp: As at:Nirbhay already mentioned, it would have been easy to find an answer to this question by just doing a bit of research (googling)... IMO this is the reason for the downvotes. — home, May 01 '13 at 16:34
ok. So, in the future the google keywords to use are: super java constructor — finneycanhelp, May 01 '13 at 20:21

score 4 · Accepted Answer · answered May 01 '13 at 16:00

4

It calls the parent class's constructor

answered May 01 '13 at 16:00

KyleM

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Is using super class a self declaration even if it is not added in the initialization? – David Dimalanta Oct 09 '14 at 02:22
1

@DavidDimalanta "If a constructor does not explicitly invoke a superclass constructor, the Java compiler automatically inserts a call to the no-argument constructor of the superclass. If the super class does not have a no-argument constructor, you will get a compile-time error." http://docs.oracle.com/javase/tutorial/java/IandI/super.html – KyleM Oct 09 '14 at 14:45

score 3 · Answer 2 · answered May 01 '13 at 16:05

3

It goes to the parent class and calls the constructor.

This link over at oracle should help

answered May 01 '13 at 16:05

g-aldrighetti

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super() is a reference which invokes SuperClass Constructor.. – Shubhamhackz Aug 30 '17 at 11:50

score 1 · Answer 3 · answered May 01 '13 at 16:00

The call super(); explicitly invokes the superclass constructor, here the parameterless constructor for AbstractShape. If this call is not present in a constructor, then Java inserts an implicit super(); for you.

Quoting the JLS, section 8.8.7:

If a constructor body does not begin with an explicit constructor invocation and the constructor being declared is not part of the primordial class Object, then the constructor body implicitly begins with a superclass constructor invocation "super();", an invocation of the constructor of its direct superclass that takes no arguments.

score 1 · Answer 4 · edited May 23 '17 at 12:13

1

It calls the constructor of the superclass. See the following questions:

edited May 23 '17 at 12:13

Community

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answered May 01 '13 at 16:02

Matthias Herlitzius

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What does super() mean in a constructor?

4 Answers4

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