我在超类的构造函数中写了一次,因为我希望它强制它用于每个实现.

public abstract class SupahClass {
    public SupahClass() {
        doStuff(); // It IS executed when the subclass constructor is called
        init(); // NOT executed,even though it's implemented
    }

    private void doStuff() { ... }        

    protected abstract void init();
}

public class SomeSubClass extends SupahClass {

    // The problem lies HERE: this is executed AFTER init() ... so it gets NULL again
    private TextBox myTextBox = null;

    public SomeSubClass() {
        super(); // invokes the super constructor,so init() should be called
        // I could call init(); here EACH time i create a new subclass... but no :)
    }

    @Override
    public void init() {
        this.myTextBox = new TextBox(); // Executed BEFORE its declared as null above
    }
}

当然,超类不能真正调用它,因为它是一个抽象(如此未定义)的方法,但它是一个ABSTRACT类,所以它不能实例化,它必须将任务委托给它的子类,所以为什么它们不能调用抽象但现在实施方法？

编辑
请参阅子类属性myTextBox和init()实现

你认为我应该采取哪种方法？删除属性声明中的= null(duhhh)

或删除超类中的init()并在子类构造函数中显式调用它(这是我想要避免的,因为我必须在100％的时间内编写它.)