Why doesn't super. call the superclass method in this code?

Question

When I execute this code, it only returns "Eating" instead of "Studying" AND "Eating." Can anyone explain it?

public class Student {
    public String getFood() {
        return "Pizza";
    }
    public String getInfo(){
        return "Studying";
    }
}

public class GradStudent extends Student {
    public String getFood(){
        return "Taco";
    }
    public String getInfo(){
        super.getInfo();
        return "Eating";
    }

    public static void main(String[] args){
        Student s = new GradStudent();
        System.out.println(s.getInfo());
    }
}

Because you're doing nothing with the return value from `super.getInfo()` — Federico klez Culloca, Mar 23 '23 at 13:13
I see, I thought return works like print or println, my bad. — BenNye09, Mar 23 '23 at 13:41
It's a common mistake for beginners. See [here](https://stackoverflow.com/questions/7129285/what-is-the-purpose-of-the-return-statement-how-is-it-different-from-printing) for an explanation (it's for python, but the concept is the same). — Federico klez Culloca, Mar 23 '23 at 14:01

score 1 · Accepted Answer · answered Mar 23 '23 at 13:16

In the getInfo() method of the Student is getting invoked, but currently you are not using the result.

If, in the GradStudent class, you want to return a String that combines the result of the Student.getInfo method with the specific value from GradStudent you could do something like this:

public class GradStudent extends Student {
    public String getFood(){
        return "Taco";
    }
    public String getInfo(){
        String studentInfo = super.getInfo();
        return studentInfo + " Eating";
    }

    public static void main(String[] args){
        Student s = new GradStudent();
        System.out.println(s.getInfo());
    }
}

Why doesn't super. call the superclass method in this code?

1 Answers1