scala - Using fold on Option without having x => x -


given:

val personsopt:option[list[person]] = ???

i prefer: 

persons = personsopt.fold(list[person]()){person => person}

to this:

persons = personsopt.getorelse(list[person]())

for type safety reasons. example not compile:

persons = personsopt.fold(nil){person => person}

is there simple way type safety not have {person => person}?

edit: 2 things concretely understood:

  1. there nothing un-type-safe getorelse. instance not compile: personsopt.getorelse("")
  2. nil list() , if type can't inferred compiler ask explicit. there can no type issues using nil

i couldn't find link now, did (incorrectly) read getorelse somehow less type safe using fold option.

there function identity defined in predef:

persons = personsopt.fold(list[person]())(identity)

i find lot less readable using getorelse, , using not make code more type-safe using getorelse. note passing nil getorelse make return correct type:

scala> case class person(name: string)  scala> val personsopt:option[list[person]] = none personsopt: option[list[person]] = none  scala> val persons = personsopt.getorelse(nil) persons: list[person] = list()

note persons list[person].


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