斯卡拉 – varargs拼图?
发布时间:2020-12-16 18:40:08 所属栏目:安全 来源:网络整理
导读:我确定答案很简单,但我陷入了困境: Welcome to Scala version 2.7.1.final (Java HotSpot(TM) 64-Bit Server VM,Java 1.6.0_14).Type in expressions to have them evaluated.Type :help for more information.scala def f(x:Int*)=0f: (Int*)Intscala val
我确定答案很简单,但我陷入了困境:
Welcome to Scala version 2.7.1.final (Java HotSpot(TM) 64-Bit Server VM,Java 1.6.0_14). Type in expressions to have them evaluated. Type :help for more information. scala> def f(x:Int*)=0 f: (Int*)Int scala> val xs:Seq[Int]=1::2::3::4::Nil xs: Seq[Int] = List(1,2,3,4) scala> f (xs) <console>:7: error: type mismatch; found : Seq[Int] required: Int f (xs) ^ 我如何构建’Int *’? 解决方法
要将序列解压缩到参数列表中,请使用_ *
scala> f(xs: _*) res1: Int = 0 (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |