我有一段 Kotlin 代码,其中第一个和第二个构造函数有细微的不同,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)) {

  }

是否有可能通过某种语言结构来减少它的冗长?我在考虑代数数据类型,但感觉不太合适 - 它被认为是“hacky”。

最佳答案

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,您可以传递单个 (n: Int) -> String? ,它将被打包成一个数组,并且,除了传递多个以逗号分隔的函数之外,您还可以使用 扩展运算符 传递一个数组:
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 和扩展运算符之外,没有其他方法可以跳过它们并传递 conditions:
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

关于Kotlin - 一个参数不同的辅助构造函数,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/40320920/

10-11 22:31
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