Java Puzzler - 任何人都可以解释这种行为吗？

Question

abstract class AbstractBase {
    abstract void print();

    AbstractBase() {
        // Note that this call will get mapped to the most derived class's method
        print();
    }
}

class DerivedClass extends AbstractBase {
    int value = 1;

    @Override
    void print() {
        System.out.println("Value in DerivedClass: " + value);
    }   
}

class Derived1 extends DerivedClass {
    int value = 10;

    @Override
    void print() {
        System.out.println("Value in Derived1: " + value);
    }
}

public class ConstructorCallingAbstract {

    public static void main(String[] args) {
        Derived1 derived1 = new Derived1();
        derived1.print();
    }
}

上面的程序产生以下输出:

Value in Derived1: 0
Value in Derived1: 10

我不明白为什么print()in AbstractBase构造函数总是被映射到派生程度最高的类(这里Derived1)print()

为什么不来DerivedClass的print()？有人能帮助我理解这个吗？

Answer 1

因为所有非显式super调用的Java方法调用都被调度到最派生类,即使在超类构造函数中也是如此.这意味着超类可以获得子类行为的好处,但这也意味着理论上可以在该类中的构造函数之前调用重写方法.