Which characters can I omit in Scala?

后端 未结 2 812
一生所求
一生所求 2020-12-05 07:33

In Scala, why can I omit the dot and brakets in T m 0 (instead of T.m(0)) in the following?

scala> object T { def m(i:Int) = 0 =         


        
相关标签:
2条回答
  • 2020-12-05 07:58

    I believe the operator syntax style works only when you've got an explicit object on the left-hand side. The syntax is intended to let you express "operand operator operand" style operations in a natural way.

    0 讨论(0)
  • 2020-12-05 08:00

    The former example, T m 0, is an example of "operator notation". Scala has three types of operator notations, prefix (called unary), infix and postfix. Let's see examples of all three in action here:

    class MyByte(val n : Int) {
      require(n >= 0 && n <= 255)
      def unary_! = new MyByte(n ^ 0xff)
      def +(m : MyByte) = new MyByte(n + m.n)
      def bits = (math.log(n) / math.log(2) + 1).toInt
      override def toString = "0" * (8 - bits) + n.toBinaryString
    }
    

    An here it is in use:

    scala> val a = new MyByte(5)
    a: MyByte = 00000101
    
    scala> val b = new MyByte(10)
    b: MyByte = 00001010
    
    scala> ! a  // Prefix/Unary
    res58: MyByte = 11111010
    
    scala> a + b  // Infix
    res59: MyByte = 00001111
    
    scala> b bits  // Postfix
    res60: Int = 4
    

    While infix and postfix notations accept any valid Scala identifier, though there is talk of restricting postfix notation, only four identifiers can be used as prefix: ~, !, - and +.

    Now, when you try "m 0", Scala discards it being a unary operator, on the grounds of not being a valid one (~, !, - and +). It finds that "m" is a valid object -- it is a function, not a method, and all functions are objects.

    As "0" is not a valid Scala identifier, it cannot be neither an infix nor a postfix operator. Therefore, Scala complains that it expected ";" -- which would separate two (almost) valid expressions: "m" and "0". If you inserted it, then it would complain that m requires either an argument, or, failing that, a "_" to turn it into a partially applied function.

    0 讨论(0)
提交回复
热议问题