如何在Kotlin中实现Builder模式？

Question

嗨,我是Kotlin世界的新手.我喜欢到目前为止所看到的并开始考虑将我们在Java应用程序中使用的一些库转换为Kotlin.

这些库充满了带有setter,getter和Builder类的Pojos.现在我用谷歌搜索了什么是在Kotlin中实施Builders的最佳方法,但没有成功.

第二次更新:问题是如何在Kotlin中使用一些参数为简单的pojo编写Builder设计模式？下面的代码是我尝试编写java代码然后使用eclipse-kotlin-plugin转换为Kotlin.

class Car private constructor(builder:Car.Builder) {
    var model:String? = null
    var year:Int = 0
    init {
        this.model = builder.model
        this.year = builder.year
    }
    companion object Builder {
        var model:String? = null
        private set

        var year:Int = 0
        private set

        fun model(model:String):Builder {
            this.model = model
            return this
        }
        fun year(year:Int):Builder {
            this.year = year
            return this
        }
        fun build():Car {
            val car = Car(this)
            return car
        }
    }
}

Answer 1

首先,在大多数情况下,您不需要在Kotlin中使用构建器,因为我们有默认和命名参数.这使您可以写

class Car(val model: String? = null, val year: Int = 0)

并像这样使用它:

val car = Car(model = "X")

如果你绝对想要使用构建器,那么你可以这样做:

使构建器成为一个companion object没有意义,因为objects是单例.而是将其声明为嵌套类(默认情况下在Kotlin中是静态的).

将属性移动到构造函数,以便也可以以常规方式实例化对象(如果不应该将构造函数设为私有),并使用将构建器和委托作为主构造函数的辅助构造函数.代码如下:

class Car( //add private constructor if necessary
        val model: String?,
        val year: Int
) {

    private constructor(builder: Builder) : this(builder.model, builder.year)

    class Builder {
        var model: String? = null
            private set

        var year: Int = 0
            private set

        fun model(model: String) = apply { this.model = model }

        fun year(year: Int) = apply { this.year = year }

        fun build() = Car(this)
    }
}

用法: val car = Car.Builder().model("X").build()

通过使用构建器DSL,可以另外缩短此代码:

class Car (
        val model: String?,
        val year: Int
) {

    private constructor(builder: Builder) : this(builder.model, builder.year)

    companion object {
        inline fun build(block: Builder.() -> Unit) = Builder().apply(block).build()
    }

    class Builder {
        var model: String? = null
        var year: Int = 0

        fun build() = Car(this)
    }
}

用法: val car = Car.build { model = "X" }

如果某些值是必需的并且没有默认值,则需要将它们放在构建器的构造函数中以及build我们刚刚定义的方法中:

class Car (
        val model: String?,
        val year: Int,
        val required: String
) {

    private constructor(builder: Builder) : this(builder.model, builder.year, builder.required)

    companion object {
        inline fun build(required: String, block: Builder.() -> Unit) = Builder(required).apply(block).build()
    }

    class Builder(
            val required: String
    ) {
        var model: String? = null
        var year: Int = 0

        fun build() = Car(this)
    }
}

用法: val car = Car.build(required = "requiredValue") { model = "X" }

Answer 2

因为我使用Jackson库来解析JSON中的对象,所以我需要一个空构造函数,而且我不能有可选字段.所有字段都必须是可变的.然后我可以使用这个与Builder模式相同的漂亮语法:

val car = Car().apply{ model = "Ford"; year = 2000 }

Answer 3

一种方法是执行以下操作：

class Car(
  val model: String?,
  val color: String?,
  val type: String?) {

    data class Builder(
      var model: String? = null,
      var color: String? = null,
      var type: String? = null) {

        fun model(model: String) = apply { this.model = model }
        fun color(color: String) = apply { this.color = color }
        fun type(type: String) = apply { this.type = type }
        fun build() = Car(model, color, type)
    }
}

用法样本：

val car = Car.Builder()
  .model("Ford Focus")
  .color("Black")
  .type("Type")
  .build()

Answer 4

我个人从未见过Kotlin的建筑师,但也许只是我.

所有验证都需要在init块中发生:

class Car(val model: String,
          val year: Int = 2000) {

    init {
        if(year < 1900) throw Exception("...")
    }
}

在这里,我冒昧地猜测你并不是真的想要model和year变化.这些默认值似乎没有任何意义,(特别是null对于name)但我留下了一个用于演示目的.

意见: Java中使用的构建器模式作为没有命名参数的生存方式.在具有命名参数的语言(如Kotlin或Python)中,使用具有长列表(可能是可选的)的构造函数是一个好习惯.

Answer 5

我见过许多将额外乐趣声明为构建器的示例。我个人喜欢这种方法。节省编写构建器的工作量。

package android.zeroarst.lab.koltinlab

import kotlin.properties.Delegates

class Lab {
    companion object {
        @JvmStatic fun main(args: Array<String>) {

            val roy = Person {
                name = "Roy"
                age = 33
                height = 173
                single = true
                car {
                    brand = "Tesla"
                    model = "Model X"
                    year = 2017
                }
                car {
                    brand = "Tesla"
                    model = "Model S"
                    year = 2018
                }
            }

            println(roy)
        }

        class Person() {
            constructor(init: Person.() -> Unit) : this() {
                this.init()
            }

            var name: String by Delegates.notNull()
            var age: Int by Delegates.notNull()
            var height: Int by Delegates.notNull()
            var single: Boolean by Delegates.notNull()
            val cars: MutableList<Car> by lazy { arrayListOf<Car>() }

            override fun toString(): String {
                return "name=$name, age=$age, " +
                        "height=$height, " +
                        "single=${when (single) {
                            true -> "looking for a girl friend T___T"
                            false -> "Happy!!"
                        }}\nCars: $cars"
            }
        }

        class Car() {

            var brand: String by Delegates.notNull()
            var model: String by Delegates.notNull()
            var year: Int by Delegates.notNull()

            override fun toString(): String {
                return "(brand=$brand, model=$model, year=$year)"
            }
        }

        fun Person.car(init: Car.() -> Unit): Unit {
            cars.add(Car().apply(init))
        }

    }
}

我还没有找到一种方法可以强制在 DSL 中初始化某些字段，例如显示错误而不是抛出异常。如果有人知道，请告诉我。