为带有父对象的子类编写Java构造函数

Question

Is it possible to use a parent object to instantiate a child object? The class Length2 extends Length1 with the addition of an instance variable. Common I tried to make a copy constructor that takes the Length1 as a param and sets the extra ivar to 0, but I'm still being told I cannot convert from Legnth1 to Length2. I suppose there may be some implicit casting taking place... have a look.

public class Length1 extends Length implements Comparable<Length1>{

    protected int miles;
    protected int yards;

    public Length1(){ //default constructor
        miles = 0;
        yards = 0;
    }

    public Length1(int m, int y){ //constructor
        miles = m;
        yards = y;
    }

    public void setYards(int x){
        yards = x;
    }
    public void setMiles(int x){
        miles = x;
    }
    public int getYards(){
        return yards;
    }
    public int getMiles(){
        return miles;
    }

...

public class Length2 extends Length1 {

    protected int feet;

    public Length2(){ //default constructor
        super(0,0);
        setFeet(0); 
    }
    public Length2(int m, int y, int f){ // constructor
        super(m,y); 
        setFeet(f); 
    }
    public Length2(Length1 x){ 
        super(x.miles,x.yards);
        setFeet(0); 
    }
    public void setFeet(int x){
        feet = x;
    }
    public int getFeet(){
        return feet;
    }

...

public class LengthTest{

    public static void main(String[] args) {
        Length1 a,b,c,d,e,f;
        Length2 g,h,i,j;

        a = new Length1(78,1610);
        b = new Length1(77,1694);
        c = new Length1();
        d = new Length1();

        g = new Length2(32,1022,1);
        h = new Length2(31,1700,2);
        i = new Length2();
        j = new Length2();


        j = c; //problem occurs here


    }

}

Answer 1

Like Leo said, if L1 is the super type of L2, that implies L2 has all of L1 and then some. So if you try to create an L2 from an L1, information will be missing.

Answer 2

My answer is no, you can't unfortunately. But you can try to push all useful information from L1 into L2 using some method like L2 = L2.valueOf(L1) that copies all the known fields from L1 into L2, leaving the missing ones empty. You have to write this method by hand. Maybe you can save some time reusing some L1 constructor in L2.

Answer 3

You can't do that. Length2 IS A Length1, so you can always assign a Length2 object to a Length1 reference (you never have to cast it). You can also assign a Length1 reference that points to a Length2 object to another reference declared as Length2 . In this case you need an explicit cast because the compiler doesn't know at compile time that the Length1 reference actually is a Length2 . But what you are trying to do is not possible, since Length1 IS NOT A Length2 .

It's like saying that Car extends Vehicle . You can always point to a Car and say Car .

Car c = new Car();

It's always OK. And you can always point to one and say Vehicle :

Vehicle v = new Car();

No problem. The only problem you'll have here is if you decide to call, say, the printLicensePlate() method of Car which doesn't exist in Vehicle, which has a melhod called printSpeed() . Since a Car IS A `Vehicle, you can always call:

c.printSpeed();
c.printLicensePlate();

and

v.printSpeed();

But you can't call:

v.printLicensePlate();

even though you know that v is a Car . This is the case where you use a cast, to convert the reference. This is illegal:

c = v; // WRONG

because the at compile time we don't know what's really in c , since new Car() only happens at runtime . So you tell the compiler "trust me, this is really a Car " by doing:

c = (Car) v;

But what you did was something like:

Car c = new Vehicle(); // WRONG

You pointed to a generic Vehicle and said Car . While a Car is always a Vehicle , a Vehicle may not be a car. (Using the c reference, the compiler believes you can legally call c.printLicensePlate() , but at runtime the object you find has no such method. So even if you could pass this by the compiler by casting the reference, it wouldn't work and would produce a ClassCastException at runtime.)