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Home/ Questions/Q 7183183
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T17:55:41+00:00 2026-05-28T17:55:41+00:00

I’m new at OCaml (and still a novice in learning programming in general) and

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I’m new at OCaml (and still a novice in learning programming in general) and I have a quick question about checking what kind of string the next element in the string list is.

I want it to put a separator between each element of the string (except for the last one), but I can’t figure out how to make the program ‘know’ that the last element is the last element.

Here is my code as it is now:

let rec join (separator: string) (l : string list) : string = 
 begin match l with
    | []->""
    | head::head2::list-> if head2=[] then head^(join separator list) else head^separator^(join separator list)
 end


let test () : bool =
 (join "," ["a";"b";"c"]) = "a,b,c"
;; run_test "test_join1" test 

Thanks in advance!

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  1. Editorial Team
    Editorial Team
    2026-05-28T17:55:41+00:00Added an answer on May 28, 2026 at 5:55 pm

    You’re almost there. The idea is breaking down the list in three cases where it has 0, 1 or at least 2 elements. When the list has more than one element, you’re safe to insert separator into the output string:

    let rec join (separator: string) (l : string list) : string =   
       begin match l with
        | [] -> ""
        | head::[] -> head
        | head::list-> head^separator^(join separator list)  
       end
    

    I have several comments about your function:

    • Type annotation is redundant. Because (^) is string concatenation operator, the type checker can infer types of separator, l and the output of the function easily.
    • No need to use begin/and pair. Since you have only one level of pattern matching, there is no confusion to the compiler.
    • You could use function to eliminate match l with part.

    Therefore, your code could be shortened as:

    let rec join sep l = 
        match l with
        | [] -> ""
        | x::[] -> x
        | x::xs -> x ^ sep ^ join sep xs
    

    or even more concise:

     let rec join sep = function
        | [] -> ""
        | x::[] -> x
        | x::xs -> x ^ sep ^ join sep xs 
    
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