Functional way to mutate state inside higher order functions scala












0














Consider the following code,



 1 var ip = ArrayBuffer[String]()

 2 ip += "1"

 3 println(ip)

 4 ip.clear()

 5 (1 to 10).foreach(ip += ("1"))

 6 println(ip)


Here in line no: 5 using variable ip inside higher order function results in exception. I do know using var is not advisable but I want to know how to use vars inside higher order functions. Or is there an alternative to mange state.






scala scala-collections







share|improve this question






















  • If you want to go the mutable route, just use a for loop: for (_ <- 1 to 10) ip += "1"
    – Lasf
    Nov 12 '18 at 16:15
















0














Consider the following code,



 1 var ip = ArrayBuffer[String]()

 2 ip += "1"

 3 println(ip)

 4 ip.clear()

 5 (1 to 10).foreach(ip += ("1"))

 6 println(ip)


Here in line no: 5 using variable ip inside higher order function results in exception. I do know using var is not advisable but I want to know how to use vars inside higher order functions. Or is there an alternative to mange state.






scala scala-collections







share|improve this question






















  • If you want to go the mutable route, just use a for loop: for (_ <- 1 to 10) ip += "1"
    – Lasf
    Nov 12 '18 at 16:15














0












0








0







Consider the following code,



 1 var ip = ArrayBuffer[String]()

 2 ip += "1"

 3 println(ip)

 4 ip.clear()

 5 (1 to 10).foreach(ip += ("1"))

 6 println(ip)


Here in line no: 5 using variable ip inside higher order function results in exception. I do know using var is not advisable but I want to know how to use vars inside higher order functions. Or is there an alternative to mange state.






scala scala-collections







share|improve this question













Consider the following code,



 1 var ip = ArrayBuffer[String]()

 2 ip += "1"

 3 println(ip)

 4 ip.clear()

 5 (1 to 10).foreach(ip += ("1"))

 6 println(ip)


Here in line no: 5 using variable ip inside higher order function results in exception. I do know using var is not advisable but I want to know how to use vars inside higher order functions. Or is there an alternative to mange state.






scala scala-collections




scala scala-collections






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 12 '18 at 16:01









Hariharan

4271821




4271821












  • If you want to go the mutable route, just use a for loop: for (_ <- 1 to 10) ip += "1"
    – Lasf
    Nov 12 '18 at 16:15


















  • If you want to go the mutable route, just use a for loop: for (_ <- 1 to 10) ip += "1"
    – Lasf
    Nov 12 '18 at 16:15
















If you want to go the mutable route, just use a for loop: for (_ <- 1 to 10) ip += "1"
– Lasf
Nov 12 '18 at 16:15




If you want to go the mutable route, just use a for loop: for (_ <- 1 to 10) ip += "1"
– Lasf
Nov 12 '18 at 16:15












1 Answer
1






active

oldest

votes


















3














The following works:



(1 to 10).foreach(_ => ip += "1")


foldLeft is more functional and you can dispense with mutable state:



(1 to 10).foldLeft(List.empty[String]){

  case (acc, _) => "1" :: acc

}


Output:



List[String] = List(1, 1, 1, 1, 1, 1, 1, 1, 1, 1)





share|improve this answer























  • How foldLeft works in parallel?
    – Hariharan
    Nov 12 '18 at 17:01






  • 1




    stackoverflow.com/questions/38734007/…
    – Terry Dactyl
    Nov 12 '18 at 17:05











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














The following works:



(1 to 10).foreach(_ => ip += "1")


foldLeft is more functional and you can dispense with mutable state:



(1 to 10).foldLeft(List.empty[String]){

  case (acc, _) => "1" :: acc

}


Output:



List[String] = List(1, 1, 1, 1, 1, 1, 1, 1, 1, 1)





share|improve this answer























  • How foldLeft works in parallel?
    – Hariharan
    Nov 12 '18 at 17:01






  • 1




    stackoverflow.com/questions/38734007/…
    – Terry Dactyl
    Nov 12 '18 at 17:05
















3














The following works:



(1 to 10).foreach(_ => ip += "1")


foldLeft is more functional and you can dispense with mutable state:



(1 to 10).foldLeft(List.empty[String]){

  case (acc, _) => "1" :: acc

}


Output:



List[String] = List(1, 1, 1, 1, 1, 1, 1, 1, 1, 1)





share|improve this answer























  • How foldLeft works in parallel?
    – Hariharan
    Nov 12 '18 at 17:01






  • 1




    stackoverflow.com/questions/38734007/…
    – Terry Dactyl
    Nov 12 '18 at 17:05














3












3








3






The following works:



(1 to 10).foreach(_ => ip += "1")


foldLeft is more functional and you can dispense with mutable state:



(1 to 10).foldLeft(List.empty[String]){

  case (acc, _) => "1" :: acc

}


Output:



List[String] = List(1, 1, 1, 1, 1, 1, 1, 1, 1, 1)





share|improve this answer














The following works:



(1 to 10).foreach(_ => ip += "1")


foldLeft is more functional and you can dispense with mutable state:



(1 to 10).foldLeft(List.empty[String]){

  case (acc, _) => "1" :: acc

}


Output:



List[String] = List(1, 1, 1, 1, 1, 1, 1, 1, 1, 1)






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 12 '18 at 16:34

























answered Nov 12 '18 at 16:28









Terry Dactyl

1,104412




1,104412












  • How foldLeft works in parallel?
    – Hariharan
    Nov 12 '18 at 17:01






  • 1




    stackoverflow.com/questions/38734007/…
    – Terry Dactyl
    Nov 12 '18 at 17:05


















  • How foldLeft works in parallel?
    – Hariharan
    Nov 12 '18 at 17:01






  • 1




    stackoverflow.com/questions/38734007/…
    – Terry Dactyl
    Nov 12 '18 at 17:05
















How foldLeft works in parallel?
– Hariharan
Nov 12 '18 at 17:01




How foldLeft works in parallel?
– Hariharan
Nov 12 '18 at 17:01




1




1




stackoverflow.com/questions/38734007/…
– Terry Dactyl
Nov 12 '18 at 17:05




stackoverflow.com/questions/38734007/…
– Terry Dactyl
Nov 12 '18 at 17:05


















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