Question

只需学习Kotlin，在下面的第一个代码中，其他代码中没有val关键字，如果省略了val和var，有什么区别？

class Person(val firstName: String, val lastName: String) {
}

class Person(firstName: String, lastName: String) {
}

Answer 1

如果在构造函数中省略val或var，则唯一可访问这些参数的位置是在构造时评估的初始化语句。参见https://kotlinlang.org/docs/reference/classes.html

当您想在存储前用值做某事时，这很有用。在Java中，您可以将该代码放入构造函数正文中

class Person(firstName: String, lastName: String) {
    // directly in val / var declarations
    val firstName = firstName.capitalize()
    val lastName = lastName

    // in init blocks
    val fullName: String
    init {
        fullName = "$firstName $lastName"
    }

    // secondary constructors can only see their own parameters
    // and nothing else can access those
    constructor(fullName: String) : this("", fullName)
}

但它也适用于delegation using by

interface Named {
    fun getName(): String
}
class Human(private val fname: String, private val lname: String) : Named {
    override fun getName() = "$fname + $lname" // functions need val since
                                               // value is resolved after construction
}
class Person2(firstName: String, lastName: String) : Named by Human(firstName, lastName)

class Person3(human: Human) : Named by human {
    constructor(firstName: String, lastName: String): this(Human(firstName, lastName))
}

或者在财产委派中

class Person4(firstName: String, lastName: String) {
    val fullName: String by lazy { "$firstName $lastName" }
}

注意：闭包是在初始化时捕获的，因此，当lazy最终求值时，这些值仍然可用。

Answer 2

如果省略val或var，则它们将不是属性，而是将参数传递给构造函数。在构造函数之外，您将无法使用它们。

为什么要将val或var放在kotlin类构造函数中

2 个答案: