scala类构造函数参数

Question

有什么区别:

class Person(name: String, age: Int) {
  def say = "My name is " + name + ", age " + age
}

和

class Person(val name: String, val age: Int) { 
  def say = "My name is " + name + ", age " + age
}

我可以将参数声明为vars,并在以后更改它们的值吗？例如,

class Person(var name: String, var age: Int) {

  age = happyBirthday(5)

  def happyBirthday(n: Int) {
    println("happy " + n + " birthday")
    n
  }
}

Answer 1

对于第一部分,答案是范围:

scala> class Person(name: String, age: Int) {
     |   def say = "My name is " + name + ", age " + age
     | }

scala> val x = new Person("Hitman", 40)

scala> x.name
<console>:10: error: value name is not a member of Person
              x.name

如果您使用参数前缀val,则var它们将在课外显示,否则,它们将是私有的,您可以在上面的代码中看到.

是的,你可以改变var的值,就像通常一样.

Answer 2

这个

class Person(val name: String, val age: Int)

使字段的用户可以在外部使用,例如,您可以稍后进行

val p = new Person("Bob", 23)
val n = p.name

如果将args指定为var,则范围与for相同val,但字段是可变的.

Answer 3

如果您熟悉Java,可以从这个示例中获得想法:

class Person(name: String, age: Int)

类似于

class Person {
  public Person(String name, int age) {
  }
}

而

class Person(var name: String, var age: Int) // also we can use 'val'

类似于

class Person {
  String name;
  int age;

  public Person(String name, int age) {
     this.name = name;
     this.age = age;
  }
}

直觉是没有var/val,变量只能在构造函数中访问.如果添加了var/val,则该类将具有相同名称的成员变量.