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[Caml-list] How to find out if a function is tail recursive?
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Date: -- (:)
From: Neel Krishnaswami <neelk@a...>
Subject: Re: [Caml-list] How to find out if a function is tail recursive? 
Richard Jones writes:
> I was writing the section on tail recursion in the OCaml tutorial, and
> was surprised to find out that the range function (below) isn't tail
> recursive. Or at least it causes a stack overflow on a
> large-but-not-unreasonable input value.
> 
> let rec range a b =
>   if a > b then []
>   else a :: range (a+1) b
>   ;;
> 
> let list = range 1 1000000;;
> 
> Printf.printf "length = %d\n" (List.length list);;
> 
> Can you tell me why this function isn't tail recursive, and share any
> useful tips on how to tell whether a function is or is not tail
> recursive?

A function call is a tail call if it is the last thing that the
function does before returning. In this example:

 let rec range a b =
   if a > b then
      []  
   else
      a :: range (a+1) b

The two expressions '[]' and 'a :: range (a+1) b' are in tail
position. The recursive call to range is *not* in tail position,
because you need to do the 'a :: <value>' before returning.

You can identify 'tail position' as a purely syntactic criterion, and
then a 'tail call' is any function call in tail position.

With the function definition 

  let f x = <expr>

<expr> is an expression in tail position.

If you have an expression <expr> in tail position, then

If <expr> = <f> <x>, then:
 
  o neither subexpression <f> nor <x> is in tail position,
  o the call '<f> <x>' is in tail position

If <expr> = if <test>
            then <e_1>
            else <e_2>, then:

  o <test> is not in tail position
  o <e_1> and <e_2> are in tail position

If <expr> = match <m> with
            | pat -> <e_1>
            | ...
            | pat -> <e_n>, then:

  o <m> is not in tail position
  o <e_1> ... <e_n> are in tail position.

If <expr> = begin
              <e_1>;
              ...
              <e_n-1>;
              <e_n>
            end, then:

  o <e1> ... <e_n-1> are not in tail position
  o <e_n> is in tail position

If <expr> = try <body> with exn -> <handler>, then:

  o <body> is not in tail position
  o <handler> is in tail position

-- 
Neel Krishnaswami
neelk@alum.mit.edu

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