从 Future 顺序执行 IO

Question

我有两个 IO 的简单组合，可以向 Telegram 机器人发送消息

def send:Future[Message] = request(SendMessage(chatID, msg))

如何在greeting中组合IO每次在"first"之后得到输出"second"。我尝试使用 *>, flatMap, IO.fromFuture 但结果不同。

first
second

second 
first

  def greeting(chatId: Long): IO[Unit] =
    IO(request(SendMessage(chatId, "first"))) *>
    IO(request(SendMessage(chatId, "second")))

  override def onMessage(message: Message) = message.text match {
    case Some(_) => greeting(message.chat.id)
      .unsafeRunAsyncAndForget()
  }

Answer 1

以下代码片段有效，它不会开始第二个 IO / Future 直到第一个完成。

import scala.concurrent.Future
import scala.concurrent.ExecutionContext.Implicits.global
import cats.effect.IO

def send(message: => String): Future[Unit] =
  Future {
    Thread.sleep(2000);
    println(message)
  }

def greeting: IO[Unit] =
  for {
   _ <- IO.fromFuture(IO(send("First")))
   _ <- IO.fromFuture(IO(send("Second")))
  } yield ()

greeting.unsafeRunAsyncAndForget()

从 Future 顺序执行 IO

Sequential execution of IO from Future

scala

cats-effect