scala - What's the rule to implement an method in trait? -


I have defined an attribute: 

  attribute A {def hello (name: someone Also): any one}   
 Then apply it to a class X:  
  class X one {def hello (name: any ) Expands: Any = {}}   
 It has been compiled after which I change the type of return in the subclass:  
  in class X The extension is {def hello (name: any): string = "hello"}   
 also compiled Change the parameter type:  
  square X extends a {def hello (name: string): any = {}}   
 it does not compile This time, the error is:  
  Error: Class X must be intact, because the type of method is in Hello (name: any) None is defined (note that None matches string: The class string in the package string is a subclass of class, in the package scal E, but the method parameter types match exactly.)   
 It seems that the parameters must match exactly but the return type can be a subtype of the sub-class?  
 
 Update: @ Mik378, thanks for your reply, but why the following examples can not work? I think it does not break Liskov:  
  attribute A {def hello (name: string): any square X one {def hello (name: any) extends: no Also = {}}    
 
 
 This is exactly like Java, to keep you  Can not override  a method with a more accurate parameter  
 In fact, if your code is related to the  A  type, under the hood  X  refers to type  a  Accordingly, you  can give anyone you want , but  B  only  String  allow. Therefore = & gt; Boom  
 Logically, with a single argument, a more accurate return type is allowed as it is in case of any code related to the  A  class , will cover.  
 You can see those parts:  
 and  
  
 UPDATE ------------ Specialty A {def hello (name: string): Any square X expands a {def hello (name: any): any = {}}  / code>  
 This will act as a full overloading, not overriding.

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