Scala 3.3.1
我有这个代码：

class Person(
    val firstName: String,
    val lastName: String,
    val birthYear: Int = -1,
    val address: String = ""
) {}

@main
def main() = {
  val p1 = Person("John", "Doe")
  val p2 = Person("Jane", "Doe", 1900)
  val p3 = Person("Josh", "Doe", 1912, "Main Street")
}

字符串
这是正确的，我可以创建没有new的对象，使用默认值。
现在，如果我引入一个二级构造函数，它会中断：

class Person(
    val firstName: String,
    val lastName: String,
    val birthYear: Int = -1,
    val address: String = ""
):
  def this() = this("John", "Doe")

型
它检测到main的前两行错误：None of the overloaded alternatives of method apply in object Person with types (): Person (firstName: String, lastName: String, birthYear: Int, address: String): Person match arguments (("John" : String), ("Doe" : String))和None of the overloaded alternatives of method apply in object Person with types (): Person (firstName: String, lastName: String, birthYear: Int, address: String): Person match arguments (("Jane" : String), ("Doe" : String), (1900 : Int))。第三行正常工作。如果我用new创建对象，一切正常。
奇怪的是，如果我定义了一个二级构造函数，它的参数数量和我试图使用的构造函数的参数数量相同（但类型完全不同），它就可以工作。例如：

class Person(
    val firstName: String,
    val lastName: String,
    val birthYear: Int = -1,
    val address: String = ""
):
  def this(x: Int, y: Int) = this("John1", "Doe1")
  def this(x: Array[List[HashMap[Int, String]]], y: Boolean, z: Int) =
    this("John2", "Doe2")

@main
def main() = {
  val p1 = Person("John", "Doe")
  val p2 = Person("Jane", "Doe", 1900)
  val p3 = Person("Josh", "Doe", 1912, "Main Street")
} // no errors

型
有人能解释这种行为吗？