将Seq [String]传递给超类

Question

在Scala中,我有一个必须以下面的形式初始化的类:

val box = new Box("foo", "bar", "tut")

这是定义它的类:

class Box(boxMembers: String*) {
    val members = boxMembers
}

如果我想扩展基类以具有命名参数,如下所示:

class FlexibleBox(big: String, small: String, otherMembers: String*) 
    extends Box(big + small + otherMembers) {} //pseudo code

如何将3个参数(大,小和其他familyMembers)传递给它的超级构造函数？

Answer 1

类似于mz的答案,但没有转换为List:

class FlexibleBox(big: String, small: String, otherMembers: String*)
    extends Box(big +: small +: otherMembers : _*) {}