Verify that an OCaml function is tail-recursive

Question

How can I tell if OCaml recognizes a particular function as tail-recursive? In particular, I want to find out if the OCaml compiler recognizes Short-circuited operators and tail recursion

---

Thanks to Jeffrey's answer below, I tried this with the simple function

let rec check_all l =
    match l with
    | [] -> true
    | hd :: tl ->
        hd && check_all tl

and indeed, it does optimize to:

camlTest__check_all_1008:
        .cfi_startproc
.L102:
        cmpl    $1, %eax
        je      .L100
        movl    (%eax), %ebx
        cmpl    $1, %ebx
        je      .L101
        movl    4(%eax), %eax
        jmp     .L102
        .align  16
.L101:
        movl    $1, %eax
        ret

Answer 1

Starting from OCaml 4.03, and despite the typo in the Changes file, you can use @tailcall in a function application and the compiler will warn if it is not the case.

(f [@tailcall]) x y warns if f x y is not a tail-call

Example:

$ cat tailrec.ml
let rec f a x = if x <= 1 then a else (f [@tailcall]) (a * x) (x - 1)

let rec g x = if x <= 1 then 1 else x * (g [@tailcall]) (x - 1)

$ ocaml tailrec.ml

File "tailrec.ml", line 3, characters 40-63:
Warning 51: expected tailcall

Answer 2

Many others are wiser than I am about OCaml internals, but for simple functions it's pretty easy to see tail recursion in the generated assembly code of ocamlopt:

$ cat tailrec.ml
let rec f a x = if x <= 1 then a else f (a * x) (x - 1)

let rec g x = if x <= 1 then 1 else x * g (x - 1)
$ ocamlopt -c -S tailrec.ml

If you ignore a lot of extra output you see this for f:

_camlTailrec__f_1008:
        .cfi_startproc
.L101:
        cmpq    $3, %rbx
        jg      .L100
        ret
        .align  2
.L100:
        movq    %rbx, %rdi
        addq    $-2, %rdi
        sarq    $1, %rbx
        decq    %rax
        imulq   %rbx, %rax
        incq    %rax
        movq    %rdi, %rbx
        jmp     .L101
        .cfi_endproc

The compiler has changed the recursive call into a loop (i.e., the function is tail recursive).

Here's what you get for g:

        .cfi_startproc
        subq    $8, %rsp
        .cfi_adjust_cfa_offset  8
.L103:
        cmpq    $3, %rax
        jg      .L102
        movq    $3, %rax
        addq    $8, %rsp
        .cfi_adjust_cfa_offset  -8
        ret
        .cfi_adjust_cfa_offset  8
        .align  2
.L102:
        movq    %rax, 0(%rsp)
        addq    $-2, %rax
        call    _camlTailrec__g_1011
.L104:
        movq    %rax, %rbx
        sarq    $1, %rbx
        movq    0(%rsp), %rax
        decq    %rax
        imulq   %rbx, %rax
        incq    %rax
        addq    $8, %rsp
        .cfi_adjust_cfa_offset  -8
        ret
        .cfi_adjust_cfa_offset  8
        .cfi_endproc

The recursion is handled by an actual recursive call (not tail recursive).

As I say, there may be better ways to figure this out if you understand the OCaml intermediate forms better than I do.