scala - 从 future 链初始化的值

Question

考虑以下代码片段:

class MyClass {
  var a = _
  var b = _
  ...
  var y = _

  val z = (for {
    a0 <- someMethodReturningFuture
    b0 <- someMethodReturningFuture(a0)
    ...
    z0 <- someMethodReturningFuture(y0)
  } yield {
    a = a0
    b = b0
    ...
    y = y0
    someCalculation(y)
  }).result
}

是否可以执行这样的初始化但使用 val s 而不是 var s(不知何故)？关键是，我不想阻塞并等待每个中间结果，只是为了最后一个。

Answer 1

除非你有超过 22 个变量:

val (a, b, c, d, ..., v) = (for {
  a0 <- someMethodReturningFuture
  b0 <- someMethodReturningFuture(a0)
  ...
  u0 <- someMethodReturningFuture(t0)
} yield {
  (a0, b0, ..., u0, someCalculation(u0))
}).result