scala - 如何调用带有短参数列表的 Scala 函数？

Question

假设我有一个带有更新方法的 Scala 类:

case class User (id: Integer, name: String, surname: String, address: String) {
    def update (name: String, surname: String, address: String): Unit = {
        copy (name, surname, address)
    }        
}

Scala 是否可以使用一个或两个参数调用更新函数并委托(delegate)调用复制函数？像这样:

val user: User = User (1, "name", "surname", "address")
user.update(name: "name") // Here it delegate to call copy (name)

或者我是否必须为每种情况复制并粘贴每个更新函数？

Answer 1

考虑

case class User (id: Integer, name: String, surname: String, address: String)

其中成员类型跟在成员名称后面。

然后像这样直接调用copy，

val userA = User(1,"a","b","c")
userA: User(1,a,b,c)

val userA2 = a.copy(2)
userA2: User(2,a,b,c)

val userBB = a.copy(name = "bb")
userBB: User(1,bb,b,c)