CODEWITHSUNDEEP
java
Jan Kl.
Jan Kl.

Reputation: 89

Do I have to call super every time I extend a class and need its constructor to trigger?

Is there any way I can trigger class constructor automatically while extending from the class?

I have a class TestSet.java

public class TestSet {
    public TestSet(String name) {
        Logger.msg("Test set: " + name);
    }
}

And I would like for the constructor to trigger every time that I extend from this class.

It seems that I need to call the constructor "manually" with:

public class TC1SendAnEmail extends TestSet{

    //I have to type this every single time again and again...
    //---------------------------------------------------------
    public TC1SendAnEmail(String name) {
        super(name);
    }
    //---------------------------------------------------------

    public void run() {
        new EmailLogin().run();
        ...
    }

}

Which I would like to avoid. (Because I will be creating possibly hundreds/thousands of those extended classes.)

From what I have managed to research, I guess that this function is not implemented in Java. But it just seems weird that I would have to "copy-paste" the constructor again, again and again...

Maybe there is another solution that I dont see? (Maybe without even using the constructor to "do something every time an instance of a class that extends my TestSet class is created".)

EDIT: Yes, I can see why you think that creating hundreds/thousands of subclasses is wrong. I am creating a big automation project. Every class of this type will be a "test". And there will be thousands of tests...

EDIT#2: The point of this question was that I needed to trigger the superclass constructor every time I extend from it. My mistake was adding parameter to the superclass constructor. If you don´t add a parameter to the constructor, then it is triggered automatically while extending.

Upvotes: 1

Views: 1143

Answers (3)

Lino
Lino

Reputation: 19910

From your comments I kinda understand what you're trying to achieve. (Automatically printing the name of the current running test)

So what about the following snippet:

public class TestSet {
     public TestSet(){
         Logger.msg("Test set: " + getClass().getSimpleName());
     }
}

This baseclass just prints the name of the implementing class when it is created. E.g. when using the following class:

public class TC1SendAnEmail extends TestSet {
     // your methods
}

it prints:

Test set: TC1SendAnEmail

This works, because in java the default constructor (constructor with no arguments) of the superclass doesn't have to be overridden, because the compiler will generate it automatically.

Upvotes: 4

Bohdan Levchenko
Bohdan Levchenko

Reputation: 3561

What you can do is replace this copy-paste code with some sort of Template method pattern:

public class SO50215241 {

    public abstract static class TestSet {
        public TestSet() {
            System.out.println("Test set: " + getName());
        }

        abstract String getName();
    }

    public static class TC1SendAnEmail extends TestSet{

        @Override
        String getName() {
            return "TC1Name";
        }
    }

    public static void main(String[] args) {
        new TC1SendAnEmail();
    }
}

Prints:

Test set: TC1Name

Then you can try to extend some specific version of TestSet for concrete implementations instead of extending TestSet itself. Or calculate name dynamically inside getName method body.

Upvotes: 1

Roee Gavirel
Roee Gavirel

Reputation: 19453

Yes.

Otherwise, how will it know that you want to pass name into it and not some other parameter?

Upvotes: 2

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