Question

假设我有

public interface MyInterface {
    public void aMethod();
}

public class MyClass implements MyInterface {
    @Override
    public void aMethod() {
        Log.d("aMethod");
    }
}

我想声明一个实现MyInterface的对象，所以我可以在构造函数中实例化它，如下所示

public class AnotherClass { 
    Class <? extends MyInterface> mObjectThatImplements;

    public AnotherClass() {
        // Says it cannot convert MyClass to Class <? extends MyInterface>
        mObjectThatImplements = new MyClass();
    }
}

Answer 1

您只需更改声明类型即可。将其作为接口，您可以为其分配任何实现。

public class AnotherClass { 
    MyInterface mObjectThatImplements;

    public AnotherClass() {           
        mObjectThatImplements = new MyClass();
    }
}

声明一个实现接口的对象，而不显式使用特定的类

1 个答案: