scala 类构造函数参数 [英] scala class constructor parameters

问题描述

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，然后再更改它们的值吗?例如，

Can I declare parameters as vars, and change their values later? For instance,

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

  age = happyBirthday(5)

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

推荐答案

第一部分的答案是范围:

For the first part the answer is scope:

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 val、var 作为参数前缀，它们将从类外部可见，否则，它们将是私有的，如您在上面的代码中所见.

If you prefix parameters with val, var they will be visible from outside of class, otherwise, they will be private, as you can see in code above.

是的，您可以像往常一样更改 var 的值.

And yes, you can change value of the var, just like usually.

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