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/