空重载构造函数

时间:2016-01-08 02:28:40

标签: android kotlin

我有以下课程。

class Student(id: String, name: String) {

    var id: String? = null
    var name: String? = null
    var grade: String? = null

    constructor(id: String, name: String, grade: String) : this(id,name) {
        this.grade = grade
    }
}

使用:

  var student = Student("AB001","Smith","N/A")

  prinln(student.id + student.name + student.grade)

输出:

nullnullN / A

任何人都可以解释为什么我从默认构造函数中获取输出null吗?

3 个答案:

答案 0 :(得分:5)

class Student {
    var id: String? = null
    var name: String? = null
    var grade: String? = null

    constructor(id: String, name: String) {
        this.id=id
        this.name=name
    }

    constructor(id: String, name: String, grade: String) : this(id, name) {
        this.grade = grade
    }
}

class Student(var id: String?, var name: String?) {
    var grade: String? = null

    constructor(id: String, name: String, grade: String) : this(id, name) {
        this.grade = grade
    }
}

答案 1 :(得分:4)

除了其他答案,还有另一种方法。使用主构造函数参数直接初始化属性:

class Student(id: String, name: String) {

    var id: String? = id
    var name: String? = name
    var grade: String? = null

    constructor(id: String, name: String, grade: String) : this(id,name) {
        this.grade = grade
    }
}

注意,由于idname始终使用不可为空的值进行初始化,因此您可以省略?。除此之外,您可以使用主构造函数中的默认值省略辅助构造函数:

class Student(id: String, name: String, grade: String? = null) {

    var id: String = id
    var name: String = name
    var grade: String? = grade

}

但是现在我们只剩下一个构造函数,所以我们可以将属性直接拉到​​构造函数中:

class Student(
        var id: String,
        var name: String, 
        var grade: String? = null
)

因为类的主体现在是空的,所以我也省略了花括号。

答案 2 :(得分:1)

或者

class Student(id: String, name: String) {
    var id: String? = null
    var name: String? = null
    var grade: String? = null

    init {
        this.id = id
        this.name = name
    }

    constructor(id: String, name: String, grade: String) : this(id,name) {
        this.grade = grade
    }
}

init块是你在主构造函数中使用参数的方式,当它们没有被标记为val或vars时必须完成。

换句话说,由于您没有将主构造函数中的参数(类名旁边的括号)标记为val或vars,因此它们不会自动指定为属性。为了在主构造函数中使用参数,特别是那些没有被标记为vars和val的参数,你需要一个init块。

我不禁想到你真正想要的是这个但是:

class Student (var id: String, var name: String, var grade: String? = null)

甚至可能将它们改为val而不是vars。

相关问题