Scala: map with two or more Options -

basically i'm looking scala-like way following:

def sum(value1: option[int], value2: option[int]): option[int] =    if(value1.isdefined && value2.isdefined) some(value1.get + value2.get)   else if(value1.isdefined && value2.isempty) value1   else if(value1.isempty && value2.isdefined) value2   else none 

this gives correct output:

sum(some(5), some(3))  // result = some(8) sum(some(5), none)     // result = some(5) sum(none, some(3))     // result = some(3) sum(none, none)        // result = none 

yet sum more 2 options i'd have use way many ifs or use sort of loop.

edit-1:

while writing question came sort of answer:

def sum2(value1: option[int], value2: option[int]): option[int] =    value1.tolist ::: value2.tolist reduceleftoption { _ + _ } 

this 1 looks idiomatic inexperienced eye. work more 2 values. yet possible same without converting lists?

edit-2:

i ended solution (thanks ziggystar):

def sum(values: option[int]*): option[int] =    values.flatten reduceleftoption { _ + _ } 

edit-3:

another alternative landei:

def sum(values: option[int]*): option[int] =    values collect { case some(n) => n } reduceleftoption { _ + _ } 

how about:

scala> def sum(values: option[int]*): option[int] = values.flatten match {      | case nil => none                                                         | case l => some(l.sum)                                                    | } sum: (values: option[int]*)option[int]  scala> sum(some(1), none) res0: option[int] = some(1)  scala> sum(some(1), some(4)) res1: option[int] = some(5)  scala> sum(some(1), some(4), some(-5)) res3: option[int] = some(0)  scala> sum(none, none)                 res4: option[int] = none 

edit

maybe sane return 0 if arguments none. in case function reduce values.flatten.sum.