Question

当使用Specs2（v 2.3.13）等待时，我注意到在某些情况下它不会失败。具体来说，我观察到如果“await”在Scope内部超时，那么它就不会失败。

"Awaiting a failed future should fail" in {
  Future.failed(throw new Exception()).map(_ => success).await // Fails Correctly
}
"Awaiting a timed out future should fail" in {
  Future(Thread.sleep(5000)).map(_ => success).await // Fails Correctly
}
"Awaiting a failed scope should fail" in new Scope {
  failure // Fails Correctly
}
"Awaiting a failed future in a scope should fail" in new Scope {
  Future.failed(throw new Exception()).map(_ => success).await // Fails Correctly
}
"Awaiting a timed out future in a scope should fail" in new Scope {
  Future(Thread.sleep(5000)).map(_ => success).await  // DOES NOT FAIL
}

我误解了用法，还是这个错误？

Answer 1

.await隐含的Future[T]只会创建Result，如果超时，则会失败。但是不会抛出任何异常，因此结果不会逃脱Scope特征。

要解决此问题，您需要使用Matchers并在T中指定您对Future[T]类型值的期望：

Future(Thread.sleep(5000)).map(_ => 1) must be_==(1).await

这将正确显示

[info] x Awaiting a timed out future in a scope should fail
[error]  Timeout after 1 second (TestSpec.scala:35)

规范2等待在范围内不会失败

1 个答案: