Parcelable继承问题与继承人类getCreator [英] Parcelable Inheritance issue with getCreator of heir class

问题描述

我有一个parcelable类 A 和 B 扩展 A

I have a parcelable class A and B that extends A

例如：

A类

public abstract class A implements Parcelable {
    private int a;

    protected A(int a) {
        this.a = a;
    }

    public int describeContents() {
        return 0;
    }

    public static Creator<A> getCreator() {
        return CREATOR;
    }

    public static final Parcelable.Creator<A> CREATOR = new Parcelable.Creator<A>() {
        public A createFromParcel(Parcel in) {
            return new A(in);
        }

        public A[] newArray(int size) {
            return new A[size];
        }
    };

    public void writeToParcel(Parcel out, int flags) {
        out.writeInt(a);
    }

    protected A(Parcel in) {
        a = in.readInt();
    }
}

继承人B类

public class B extends A {
    private int b;

    public B(int a, int b) {
        super(a);
        this.b = b;
    }

    public static Creator<B> getCreator() {
        return CREATOR;
    }

    public static final Parcelable.Creator<B> CREATOR = new Parcelable.Creator<B>() {
        public B createFromParcel(Parcel in) {
            return new B(in);
        }

        public B[] newArray(int size) {
            return new B[size];
        }
    };

    public int describeContents() {
        return 0;
    }

    public void writeToParcel(Parcel out, int flags) {
        super.writeToParcel(out, flags);
        out.writeInt(b);
    }

    private B(Parcel in) {
        super(in);
        b = in.readInt();
    }
}

在B I得到错误的的返回类型是A.getCreator（）不兼容的的在

In B I get the error "the return type is incompatible with A.getCreator()" in

public static Creator<B> getCreator() {
            return CREATOR;
}

显然，如果我尝试 getCreator 的类 B 的类型更改为造物主＆LT; A方式＆gt; ，不起作用，因为Parcelable创作者的类型是 B

clearly if I try to change the type of getCreator of the class B to Creator<A>, doesn't work because the type of Parcelable creator is B.

我怎么能解决这个问题呢？

how could I solve this issue?

推荐答案

下面是我如何实现这一点。我创建了一个抽象父类，允许使用的父类 A ，您应该增加两个抽象方法：
     保护抽象无效writeChildParcel（包裹PC，诠释标志）和保护抽象无效readFromParcel（包裹PC）

Here is how I implemented this. I created an abstract parent class, lets use your parent class A, where you should add two abstract methods: protected abstract void writeChildParcel(Parcel pc, int flags) and protected abstract void readFromParcel(Parcel pc)

然后你需要一个静态方法来创建 A 的正确实例。在我的情况下，它是有意义的有一个类型的属性（可以使用枚举），在那里我可以识别它们。这样我们就可以有一个静态的newInstance（整型）方法，像这样：

Then you need a static method to create the right instance of A. In my case it made sense to have a type attribute (you can use an enum) where I could identify each of them. This way we can have a static newInstance(int type) method like so:

public static A newInstance(int type) {
    A a = null;
    switch (type) {
    case TYPE_B:
            a = new B();
            break;
        ...
    }
    return a;
 }

public static A newInstance(Parcel pc) {
    A a = A.newInstance(pc.readInt()); //
    //call other read methods for your abstract class here
    a.readFromParcel(pc);
    return a;
 }

public static final Parcelable.Creator<A> CREATOR = new Parcelable.Creator<A>() {
    public A createFromParcel(Parcel pc) {
        return A.newInstance(pc);
    }
    public A[] newArray(int size) {
        return new A[size];
    }
};

然后，写你的 writeToParcel 如下：

public void writeToParcel(Parcel out, int flags) {
    out.writeInt(type);
    //call other write methods for your abstract class here
    writeChildParcel(pc, flags);
}

现在摆脱你的 CREATOR 和其他 Parcelable 方法乙和刚实施 writeChildParcel 和 readFromParcel 。你应该是好去！

Now get rid of your CREATOR and all other Parcelable methods in B and just implement writeChildParcel and readFromParcel in it. You should be good to go!

希望它帮助。

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