Question

任何人都可以解释以下输出。我有一个像这样的简单Scala代码..

object compOrNotTest {
  def main(args: Array[String]) {
    var emp = new Employee("tom", 20)
    println(emp)
    println(Employee.adult(emp))
    Employee.printName("Roland", 38)

    var emp2 = new Employee("Harry", 37)
    Employee.printName(emp2)
  }
}

class Employee(name: String, age: Int) {
  val ageOfEmplyedd: Int = age
  val nameEmp: String = name

  override def toString() = this.name + " age : " + this.age

  def printName() {
    println("name is in Class " + nameEmp)
  }
}

object Employee {
  def adult(emp: Employee) = {
    if (emp.ageOfEmplyedd > 18)
      true
    else
      false
  }

  def printName(name: String, age: Int) = {
    val emp1 = new Employee(name, age)
    println("Name is : " + emp1.printName())
  }

  def printName(emp1: Employee) = {
    //val emp1 = new Employee(name, age)
    println("Name is :  "+ emp1.printName())
  }
}

我得到的输出是

tom age : 20
true
name is in Class Roland
Name is : ()
name is in Class Harry
Name is : ()

我的问题是，为什么，当我从Companion对象调用时，我只得到Name is : ()。我期待Name is : name is in Class Roland之类的东西。请帮助我了解scala在这种情况下的工作原理。非常感谢

Answer 1

Employee.printName（在课程Employee中）的返回类型为Unit。这是因为此函数是使用过程语法（一个没有=符号的函数声明which has been deprecated声明的，并且将来不再支持< em> Scala ），其关联的返回类型为Unit。返回的Unit值在Scala中表示为()。

以下函数声明是等效的：

// Using (soon-to-be-deprecated) procedure syntax.
def printName() {
  println("name is in Class " + nameEmp)
}

// Using an explicit return type.
def printName(): Unit = {
  println("name is in Class " + nameExp)
}

// Using an inferred return type (note "=" in declaration). The last statement is the call
// to println, which returns Unit, so a return type of Unit is inferred.
def printName() = {
  println("name is in Class " + nameExp)
}

如果你想返回打印的字符串，你需要这样的东西：

def printName() = {
  val s = "name is in Class " + nameEmp
  println(s)
  s
}

或者，使用显式返回类型，而不是从最后一个语句推断它：

def printName(): String = {
  val s = "name is in Class " + nameEmp
  println(s)
  s
}

scala中的伴随对象

1 个答案: