How Can I Make a Simple Exit in Haskell? -

this code:

main :: io() main =   {          putstrln ("meniu: ");          putstrln ("1. menu 1");          putstrln ("2. menu 2");          putstrln ("3. menu 3");          putstrln ("4. menu 4");          putstrln ("5. exit - iesire)");          putstrln ("-------------------------");          putstr ("enter option: ");          opt <- getline;          if(opt == "1")          {             code 1 etc             main          }             else if(opt == "2")          {             code 2 etc             main          }             else if(opt == "3")          {             code 3 etc             main          }          else if(opt == "4")          {             code 4 etc             main           }          else if(opt == "5")          {             ??????????? ()          }          else putstrln "option not exist"; } 

the problem: in option 5 (opt == 5) need make code stop menu, dont know how can this. tried find more examples on google , stackoverflow, can't find solution.

return () work here. in case, return behaves in procedural language (but watch out, not true).

note on style: chains of if...else equality comparisons un-idiomatic in haskell. right way case:

main =        putstrln "meniu: "      sequence_ [ putstrln $ [n]++". menu "++[n] | n<-['1'..'5'] ]      putstrln "-------------------------"      putstr "enter option: "       opt <- getline      case opt of       "1" ->         code 1 etc         main       "2" ->         code 2 etc         main       "3" ->         code 3 etc         main       "4" ->         code 4 etc         main       "5" -> return ()       _ -> putstrln "option not exist" 

braces , semicolons aren't needed if correctly indent code.

what return () here simply... nothing @ all, it's no-op. because main ends after case switch, program end if don't recurse main in other options.