Kotlin - 有一个参数不同的辅助构造函数

Question

我有一段 Kotlin 代码片段，其中第一个和第二个构造函数略有不同，请参见下文

class InstructionPrototype constructor(
      val iname: String,
      val opcode: Int,
      val mnemonicExample: String,
      val numericExample: Int,
      val description: String,
      val format: Format,
      val pattern: Pattern,
      var type: Type? = null,
      var rt: Int? = null,
      var funct: Int? = null,
      var conditions: Array<(n: Int) -> String?>? = null) {
  constructor(
        iname: String,
        opcode: Int,
        mnemonicExample: String,
        numericExample: Int,
        description: String,
        format: Format,
        pattern: Pattern,
        type: Type?,
        rt: Int?,
        funct: Int?,
        condition: (n: Int) -> String?
  ): this(iname, opcode, mnemonicExample, numericExample, description, 
        format, pattern, type, rt, funct, arrayOf(condition)) {

  }

是否可以通过某种语言结构来减少冗长的内容？我正在考虑代数数据类型，但感觉不太合适 - 它给人的印象是“hacky”。​​

Answer 1

Variable number of arguments (vararg)似乎非常适合您的用例，但前提是您可以放弃 null 作为 conditions 的默认值，因为 vararg 不能为空(例如使用emptyArray()):

class InstructionPrototype constructor(
      val iname: String,
      val opcode: Int,
      val mnemonicExample: String,
      val numericExample: Int,
      val description: String,
      val format: Format,
      val pattern: Pattern,
      var type: Type? = null,
      var rt: Int? = null,
      var funct: Int? = null,
      vararg var conditions: (n: Int) -> String? = emptyArray())

在使用现场，你可以传递单个(n: Int) -> String?，它会被打包到一个数组中，并且，除了传递多个用逗号分隔的函数之外，你还可以可以使用扩展运算符来传递数组:

f(vararg a: String) { }

f("a")
f("a", "b", "c")

val array = arrayOf("a", "b", "c")
f(*array) // any array of the correct type can be passed as vararg

此外，conditions 之前的几个参数也有默认值，除了使用 named arguments 之外，没有其他方法可以跳过它们并传递 conditions和扩展运算符:

fun f(x: Int = 5, vararg s: String) { }

f(5, "a", "b", "c") // correct
f(s = "a") // correct
f(s = "a", "b", "c") // error
f(s = *arrayOf("a", "b", "c") // correct