recursion - Standard ML - Return the index of the occurrences of a given value in a list -

i'm trying figure out how return list of indexes of occurrences of specific value in list. i.e. indexes(1, [1,2,1,1,2,2,1]); val = [1,3,4,7] int list

i'm trying figure out how lists work , trying better @ recursion don't want use list.nth (or library functions) , don't want move pattern matching quiet yet.

this have far

fun index(x, l) = if null l 0 else if x=hd(l)      1 else      1 + index(x,tl l);  fun inde(x, l) = if null l [] else if x=hd(l)      index(x, tl l) :: inde(x, tl l) else     inde(x, tl l);  index(4, [4,2,1,3,1,1]);  inde(1,[1,2,1,1,2,2,1]); 

this gives me [2, 1, 3, 0]. guess i'm having hard time incrementing things index. index function works correctly though.

instead make 2 passes on list: first add index each element in list, , second grap index of right elements:

fun addindex (xs, i) =     if null xs []     else (hd xs, i) :: addindex(tl xs, i+1)  fun fst (x,y) = x fun snd (x,y) = y fun indexi(n, xs) =     if fst(hd xs) = n ... :: indexi(n, tl xs)     else indexi(n, tl xs) 

(i left out part of indexi exercise.) addindex([10,20,30],0) gives [(10,0),(20,1),(30,2)]. can use addindex , indexi implement original index function:

fun index(n, xs) = indexi(n, addindex(xs, 0)) 

when work, can try merge addindex , indexi 1 function both.

however, want write pattern matching, see instance addindex written using patterns:

fun addindex ([], _) = []   | addindex (x::xs, i) = (x,i) :: addindex(xs, i+1)