是否可以为Future类型的方法参数提供默认值？
我们可以做到：
def myMethod(test: String = "") {...}
但是如果我们有未来的话
def myMethod(test: Future[List[Color]]) {...} //How can I set default for this?
答案 0 :(得分：4)
<?php
date_default_timezone_set('Europe/Berlin');
echo "Europe/Berlin:\n";
echo "gmdate: ".gmdate('d.m.y H:i', 0) . "\n";
echo "date: ".date('d.m.y H:i', 0) . "\n";
date_default_timezone_set('America/Los_Angeles');
echo "\nAmerica/Los_Angeles:\n";
echo "gmdate: ".gmdate('d.m.y H:i', 0) . "\n";
echo "date: ".date('d.m.y H:i', 0) . "\n";
/* OUTPUT:
Europe/Berlin:
gmdate: 01.01.70 00:00
date: 01.01.70 01:00
America/Los_Angeles:
gmdate: 01.01.70 00:00
date: 31.12.69 16:00
*/
将包含
Future[A]类型的成功结果示例：
A这意味着您可以在示例中将默认值分配为scala> import scala.concurrent.Future
import scala.concurrent.Future
scala> val success = Future.successful("some data")
success: scala.concurrent.Future[String] = Future(Success(some data))
scala> val failed = Future.failed(throw new Exception("failed"))
java.lang.Exception: failed
... 30 elided
，
Future.successful示例：
def myMethod(test: Future[List[Color]] = Future.successful(List.empty[Color])) {...}
在代码库中，始终将已计算的结果标记为scala> implicit val ec = scala.concurrent.ExecutionContext.global
ec: scala.concurrent.ExecutionContextExecutor = scala.concurrent.impl.ExecutionContextImpl$$anon$3@3c8802cf[Running, parallelism = 8, size = 0, active = 0, running = 0, steals = 0, tasks = 0, submissions = 0]
scala> def addLoyaltyPoints(previousLoyalty: Future[Int] = Future.successful(0)): Future[Int] = previousLoyalty.map(_ + 2)
addLoyaltyPoints: (previousLoyalty: scala.concurrent.Future[Int])scala.concurrent.Future[Int]
scala> addLoyaltyPoints()
res16: scala.concurrent.Future[Int] = Future(<not completed>)
scala> res16
res17: scala.concurrent.Future[Int] = Future(Success(2))
而不是Future.successful(a)总是一个好习惯，因为Future(a)将在线程上运行。为预先计算的结果分配线程是没有意义的。