Java 如何从Kotlin使用Flowable.generate

下面是Flowable.generate的一次失败存根尝试（类型注释比我通常使用的多）：

val xs:Flowable=Flowable.generate(
java.util.concurrent.Callable{->0}，
io.reactivex.functions.BiConsumer{t1，t2->}
)

我想使用的Java签名是：

public static <T, S> Flowable<T> generate(Callable<S> initialState, final BiConsumer<S, Emitter<T>> generator)

publicstaticflowablegenerate（可调用的initialState，最终双消费者生成器）

我得到的错误是：

Error:(145, 12) None of the following functions can be called with the arguments supplied:
@CheckReturnValue @BackpressureSupport @SchedulerSupport public final fun <T : Any!, S : Any!> generate(p0: (() -> (???..???))!, p1: (((???..???), Emitter<(???..???)>!) -> Unit)!): Flowable<(???..???)>! defined in io.reactivex.Flowable
@CheckReturnValue @BackpressureSupport @SchedulerSupport public final fun <T : Any!, S : Any!> generate(p0: (() -> (???..???))!, p1: ((???, Emitter<(???..???)>) -> ???)!): Flowable<(???..???)>! defined in io.reactivex.Flowable
@CheckReturnValue @BackpressureSupport @SchedulerSupport public open fun <T : Any!, S : Any!> generate(p0: Callable<(???..???)>!, p1: BiConsumer<(???..???), Emitter<String!>!>!): Flowable<String!>! defined in io.reactivex.Flowable
@CheckReturnValue @BackpressureSupport @SchedulerSupport public open fun <T : Any!, S : Any!> generate(p0: Callable<Int!>!, p1: BiFunction<Int!, Emitter<(???..???)>!, Int!>!): Flowable<(???..???)>! defined in io.reactivex.Flowable

Error:（145，12）使用提供的参数无法调用以下函数：
@CheckReturnValue@Backpressures Support@Scheduler Support public final fun generate（p0:（（）->（？？…？）！，p1:（（（？？…？），发射器！）->单位）！：可流动！在io.reactivex.Flowable中定义
@检查返回值@BackpressureSupport@scheduler支持公共最终乐趣生成（p0:（（）->（？？？…？）！，p1:（（？？，发射器）->？）！：可流动！在io.reactivex.Flowable中定义
@CheckReturnValue@BackpressuresSupport@SchedulerSupport public open fun generate（p0:Callable！、p1:BiConsumer！）：可流动！在io.reactivex.Flowable中定义
@CheckReturnValue@BackpressuresSupport@SchedulerSupport public open fun generate（p0:Callable！、p1:BiFunction！）：可流动！在io.reactivex.Flowable中定义

---

我应该向编译器提供什么？

生成（）的类型应该是

，

双消费者的类型应该是

val xs:Flowable=Flowable.generate(
java.util.concurrent.Callable{->0}，
io.reactivex.functions.BiConsumer{t1，t2->}
)
我发誓我已经三次检查了类型参数的顺序，但显然没有。
Error:(145, 12) None of the following functions can be called with the arguments supplied:
@CheckReturnValue @BackpressureSupport @SchedulerSupport public final fun <T : Any!, S : Any!> generate(p0: (() -> (???..???))!, p1: (((???..???), Emitter<(???..???)>!) -> Unit)!): Flowable<(???..???)>! defined in io.reactivex.Flowable
@CheckReturnValue @BackpressureSupport @SchedulerSupport public final fun <T : Any!, S : Any!> generate(p0: (() -> (???..???))!, p1: ((???, Emitter<(???..???)>) -> ???)!): Flowable<(???..???)>! defined in io.reactivex.Flowable
@CheckReturnValue @BackpressureSupport @SchedulerSupport public open fun <T : Any!, S : Any!> generate(p0: Callable<(???..???)>!, p1: BiConsumer<(???..???), Emitter<String!>!>!): Flowable<String!>! defined in io.reactivex.Flowable
@CheckReturnValue @BackpressureSupport @SchedulerSupport public open fun <T : Any!, S : Any!> generate(p0: Callable<Int!>!, p1: BiFunction<Int!, Emitter<(???..???)>!, Int!>!): Flowable<(???..???)>! defined in io.reactivex.Flowable

val xs: Flowable<String> = Flowable.generate<String, Int>(
        java.util.concurrent.Callable<Int> { -> 0 },
        io.reactivex.functions.BiConsumer<Int, Emitter<String>> { t1, t2 ->  }
)