Scala（7）深入理解函数

/**

• 深入的理解什么是函数
• 函数的本质就是一个引用类型，有多重表现形式，即创建函数时可以用下面的多中方式。
• 本质上就是new FunctionN的实现类（实例），就是在堆内存中开辟的一块空间。函数名（变量名）指向堆内存中的空间。

*/

 object FunctionInDeep {


  val f1 = (x: Int, y: Double) => (y, x)

  val f2: (Int, Double) => (Double, Int) = (x: Int, y: Double) => (y, x)

  val i = 10
  val j: Int = 10

  val f3: (Int, Double) => (Double, Int) = (x, y) => (y, x)


  //函数本质是引用类型
  val f4 = new Function2[Int, Double, (Double, Int)] {
    override def apply(v1: Int, v2: Double): (Double, Int) = {
      (v2, v1)
    }
  }

  val f5 = new ((Int, Double) => (Double, Int)) {
    override def apply(v1: Int, v2: Double): (Double, Int) = {
      (v2, v1)
    }
  }

  val f6: Function2[Int, Double, (Double, Int)] = new Function2[Int, Double, (Double, Int)] {
    override def apply(v1: Int, v2: Double): (Double, Int) = {
      (v2, v1)
    }
  }

  val f7: (Int, Double) => (Double, Int) = new ((Int, Double) => (Double, Int)) {
    override def apply(v1: Int, v2: Double): (Double, Int) = {
      (v2, v1)
    }
  }

  val f8: Function2[Int, Double, (Double, Int)]  = new ((Int, Double) => (Double, Int)) {
    override def apply(v1: Int, v2: Double): (Double, Int) = {
      (v2, v1)
    }
  }

  val f9: Function2[Int, Double, Tuple2[Double, Int]]  = new ((Int, Double) => (Double, Int)) {
    override def apply(v1: Int, v2: Double): (Double, Int) = {
      (v2, v1)
    }
  }


  def main(args: Array[String]): Unit = {

    val r = f4(5, 6.6)

    println(r)
  }



}