如何使用Flowable.generate从科特林

Question

这里是在Flowable.generate一个失败的短线尝试（与更多类型的注释比我正常使用）：

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

Java的签名我想使用是：

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

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 

什么应该我正在编译器？

Answer 1

generate()的类型应该是<String, Int>，而BiConsumer的类型应该是<Int, Emitter<String>>。

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