What is the Scala type-programming analogy for the `this` keyword? -


I'm trying to work in my own way to understand type-programming in Scala, and I've found that Most of us need to know about typing programming. There is an equivalent equivalent in price programming as reflected in it, however, I do not have to get it according to the this key word or self-type. It is expected of such things Rsha does not mean to, but I thought I'd ask.

For example, I can write the following to represent Boolean as values ​​in the run time: 

  Sealed attribute BoolVal {def not: BoolVal def or (Key: BoolVal): BoolVal def and (Key: BoolVal) = (this.not or that.not) def Imp .not (that: BoolVal) = this.not Or that} case object TrueVal extends BoolVal {do not dominate domes = FalseVal def override or (Key: BoolVal) = TrueVal} case object FalseVal extends BoolVal {Val = No TrueVal override def or o Ride (Key: BoolVal) that =}   
 Here my  and  and  imp  are able to take advantage of the fact that this is someone It does not matter if I have the right thing to define the wrong thing or the right object, my code can get the same code as  TrueVal  and  FalseVal  objects.  
 I can create similar type-level programming, but I do not understand how  and  and  IM  are my base attributes To define in   "seal" feature is not a type of type type: type type:  BoolType] = True Type} Seal Attribute FalseType Extends BoolType {Not Dominate Type = TrueType Override Type or [that is & lt; : BoolType] = that}   
 I can see that it does not come from the type of my type,  and  and  Is there a way to define> my  BoolType  in Implements , or should I Each is to define  and  wrong type  Symptoms?   
 
 
 You can always define an abstract type of your Boolean base type as follows:  
  attribute MyBool extends BoolType {Type is   
 It should be good to go along, after which you can use the rules of Demorgon to do the following tasks  
 ! (X & Y) == (! X ||! Y)   
 Then you can run the  and  status by a double negative:  
 ! (! X ||! Y) == !! (X & amp; y) == (X & amp; nbsp;)

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