[英]RxJava Single/Observable return implementation of interface
I have faced with interesting problem 我遇到了有趣的问题
I have the following method 我有以下方法
@Override
public Single<IResults> getResults() {
return Single.just(new ConcreteResults());
}
IResults is an interface and ConcreteResults is the implementation. IResults是接口, ConcreteResults是实现。
However I am getting an error 但是我遇到一个错误
Error:(149, 27) error: incompatible types: Single<ConcreteResults> cannot be converted to Single<IResults>
How to solve this issue ? 如何解决这个问题? I need to have an interface to abstract return type. 我需要一个接口来抽象返回类型。
However, this dirty trick seems to work :) 但是,这个肮脏的把戏似乎有效:)
Single.just(new ConcreteResults())
.map(new Function<ConcreteResults, IResults>() {
@Override
public IResults apply(@NonNull ConcreteResults results) throws Exception {
return results;
}
});
Use: 采用:
@Override public Single<? extends IResults> getResults() {
return Single.just(new ConcreteResults());
}
Please remember that hieritance with genetics is a bit different. 请记住,遗传学上的嘲笑有些不同。
Although IResults is a sub type of ConcreteResults, Single of ConcreteResults is NOT a sub type of Single of IResults . 虽然IResults是ConcreteResults的子类型,但是Single ConcreteResults不是IResults的Single的子类型。 You need to use a wildcard to create a relationship between then 您需要使用通配符在
You can search for more information about it here: https://docs.oracle.com/javase/tutorial/java/generics/subtyping.html 您可以在此处搜索有关它的更多信息: https : //docs.oracle.com/javase/tutorial/java/generics/subtyping.html
You need to create a variable before returning the value. 您需要先创建变量,然后再返回值。 The compiler will figure out, that IResult is the over-type of ConcretResult. 编译器会发现IResult是ConcretResult的过度类型。 If you return it immediately the compiler does not get it. 如果立即返回,编译器将无法获取。
@Test
void name() {
Single<IResult> result = getResult();
}
private Single<IResult> getResult() {
Single<IResult> just = Single.just(new Result());
return just;
}
interface IResult {
}
class Result implements IResult {
}
For 1_7 Settings you must use: 对于1_7设置,您必须使用:
compileOptions {
sourceCompatibility JavaVersion.VERSION_1_7
targetCompatibility JavaVersion.VERSION_1_7
}
You have to give the compiler pointers, because in 1_7 the compiler does not inference the type from definition. 您必须提供编译器指针,因为在1_7中,编译器不会从定义中推断类型。
private Single<IResult> getResult() {
Single<IResult> just = Single.<IResult>just(new Result());
return just;
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.