matrix - Adding 2 Int Lists Together F# -

i working on homework , problem 2 int lists of same size, , add numbers together. example follows.

    vecadd [1;2;3] [4;5;6];; return [5;7;9] 

i new , need keep code pretty simple can learn it. have far. (not working)

    let rec vecadd l k =          if l <> [] vecadd ((l.head+k.head)::l) k else [];; 

i want replace first list (l) added numbers. have tried code different way using match cases.

    let rec vecadd l k =        match l          |[]->[]          |h::[]-> l          |h::t -> vecadd ((h+k.head)::[]) k 

neither of them working , appreciate can get.

first, idea modifying first list instead of returning new 1 misguided. mutation (i.e. modifying data in place) number 1 reason bugs today (used goto, that's been banned long time now). making every operation produce new datum rather modify existing ones much, safer. , in cases may more performant, quite counterintuitively (see below).

second, way you're trying it, you're not doing think you're doing. double-colon doesn't mean "modify first item". means "attach item in front". example:

let = [1; 2; 3] let b = 4 ::    // b = [4; 1; 2; 3] let c = 5 :: b    // c = [5; 4; 1; 2; 3] 

that's how lists built: start empty list , prepend items it. [1; 2; 3] syntax you're using syntactic sugar that. is, [1; 2; 3] === 1::2::3::[].

so how modify list, ask? answer is, don't! f# lists immutable data structures. once you've created list, can't modify longer.

this immutability allows interesting optimization. take @ example posted above, 1 3 lists a, b, , c. how many cells of memory think these 3 lists occupy? first list has 3 items, second - 4, , third - 5, total amount of memory taken must 12, right? wrong! total amount of memory taken these 3 lists actual 5 cells. because list b not block of memory of length 4, rather number 4 paired pointer list a. number 4 called "head" of list, , pointer called "tail". similarly, list c consists of 1 number 5 (its "head") , pointer list b, "tail".

if lists not immutable, 1 couldn't organize them this: if modifies tail? lists have copied every time (google "defensive copy").

so way lists return new one. you're trying can described this: if input lists empty, result empty list; otherwise, result sum of tails prepended sum of heads. can write down in f# verbatim:

let rec add b =     match a, b     | [], [] -> []   // sum of 2 empty list empty list     | a::atail, b::btail -> (a + b) :: (add atail btail)  // sum of non-empty lists sum of tails prepended sum of heads 

note program incomplete: doesn't specify result should when 1 input empty , other not. compiler generate warning this. i'll leave solution exercise reader.