创建类时会调用Java构造函数吗？ 未创建对象时出现“无法将构造函数应用于给定类型”错误

Question

I'm reconstructing the following code from a tutorial. When I run complie it I get the following error:

constructor Vierbeiner in class Vierbeiner cannot be applied to given types; required: java.lang.String found: no arguments reason: actual and formal argument lists differ in length

My understanding is that this error occurs because the parent class Vierbeiner has a constructor that requires a String argument and no constructor requiring no argument. What I don't understand is why I get this error without creating an object of class Hund. Why is a constructor being called when I haven't created an object of class Hund? Every existing question I've seen about this error involves creating an object of the child class.

public class Vierbeiner {

    public static void main(String[] args){
         Vierbeiner hund = new Vierbeiner("Bello");
         hund.rennen();
    }

    String vierbeinerName;

    Vierbeiner(String pName) {
        vierbeinerName = pName;
    }

    public void rennen() {
        System.out.println(vierbeinerName+" rennt los!");
    }

}

class Hund extends Vierbeiner {
}

EDIT: In reponse to the answers I've gotten, what's still not clear to me is why the following (case 1) compiles with no problem:

public class Vierbeiner{
    public static void main(String[] args){

         Vierbeiner hund = new Vierbeiner();
         Hund hund2 = new Hund();  
         // Hund hund3 = new Hund("doggy");    
    }

   String vierbeinerName;

   Vierbeiner() {
       vierbeinerName = "test";
   };

   Vierbeiner(String pName){

   }

   }

class Hund extends Vierbeiner{

};

In which I create a Hund object seemingly using the no argument constructor which I defined in the Vierbeiner class.

But if I uncomment the following (case 2) I get a compilation error:

Hund hund3 = new Hund("doggy");

In case 1, the compiler looks to the parent class Vierbeiner to find a no argument constructor to create the hund2 object. In case 2 I would expect the compiler to do the same thing, that is go to the parent class Vierbeiner to find a string argument constructor to create the hund3 object, but this doesn't seem to be the case without using "super" and I don't understand why.

Answer 1

You are receiving an error during compilation, not an error or exception during runtime.

This means that the code is not being executed, so no class instances have been created - there is no invocation of any constructor.

During compilation, it is easy to detect that you are extending a class which has a single string argument constructor, which the extending class must call (in order to create the underlying base class).

Note that once you have implemented a constructor, you can no longer rely upon the autogenerated no-parameter default constructor of your base class - you must now invoke the parameterized constructor of your base class in your deriving class.

Since this sort of implementation is lacking in your code - the error you described occurs.

A simple possible solution would be to provide a default constructor implementation which will invoke the single parameter constructor of the base class:

class Hund extends Vierbeiner{
    Hund() {
        super("DefaultName");
    }    

};

This addition to your code solves the compilation time error.

A more practical approach would be to make sure that your deriving class has a constructor with a string parameter as well, which can then be passed to the base class constructor:

class Hund extends Vierbeiner{
    Hund(String name) {
        super(name);
    }    

};

References:

• Java Compiler definition
• Java Compilation time vs Runtime errors
• Java common compilation time errors
• Calling a base class constructor from derived class in Java
• Java Specifications - Default Constructor

Answer 2

You have a compile time error. You need to add the constructor in your derived class calling the base clase constructor using the super keyword.

class Hund extends Vierbeiner {

    public Hund(String pName) {
        super(pName);
    }

};