Will OCaml compiler deal with boolean operators to make recursion tail-recursive? -

let's have @ following function is_prime:

let is_prime n =   let rec div_check = * > n || (n mod <> 0 && div_check (i+1))   in   n >= 2 && div_check 2 

so if n mod <> 0 false, stop.

but if n mod <> 0 true, recursively continue.

my question if continues, has ocaml optimised compiler return div_check(i+1) directly, i.e., tail-recursive? or still hold true && part , wait div_check return?

ps function taken http://www.cs.cornell.edu/courses/cs3110/2014sp/lectures/2/lec02.html

the answer yes. i've simplified example, make compiler output more understandable:

let rec is_prime n =   let rec div_check =     * > n || (i + < n && div_check (i+1)) in   div_check n 

given input, compiler emit following assembly (i've added annotations):

camlis_prime__div_check_1010: .l102:     movq    16(%rbx), %rdi     movq    %rax, %rsi     sarq    $1, %rsi     movq    %rax, %rdx     decq    %rdx     imulq   %rsi, %rdx           ; *     incq    %rdx     cmpq    %rdi, %rdx           ; if * <= n return     jle .l101     movq    $3, %rax     ret .l101:     movq    16(%rbx), %rdi     leaq    -1(%rax, %rax), %rsi ; +     cmpq    %rdi, %rsi           ; + ? n     jge .l100                    ; if + >= n return     addq    $2, %rax             ; else := + 1     jmp .l102                    ;      div_check .l100:     movq    $1, %rax     ret 

but cautious, work vanilla ocaml's short-circuit operators. such harmless change as, let (&&) = (&&) break it, , jmp .l102, substituted call camlis_prime__div_check_....

also, work if call right of short-circuit operator. left expression marked non tail-recursive. i've tested and, indeed, putting div_check left of && operator emit non tail-recursive code.

if you're in doubt whether call tail or not, can add -annot flag compiler, , @ corresponding *.annot file, call (tail) annotations. there tooling support in merlin that, i've not figured out on how use properly. , keep in mind, final judge still assembly output.