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submitted 6 months ago by [email protected] to c/[email protected]

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[–] [email protected] 76 points 6 months ago (15 children)

GNU is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX is not UNIX[Maximum call stack size exceeded]

[–] [email protected] 8 points 6 months ago (5 children)

A tail-recursive version written in OCaml that should not reach stack limits easily. (Not an expert in OCaml, so this might be stupid. But I tried it with 10000 iterations, and it worked without any issues.)

let gnu =
    let rec aux s = function
    | 0 -> s
    | n -> aux (s^" is Not Unix") (n-1)
in aux "GNU";;

[–] [email protected] 8 points 6 months ago* (last edited 6 months ago) (4 children)

Not an OCaml expert either but that looks tail recursive, you're never going to blow the stack.

You can tell by how after the recursive call within aux, its result does not get used within the function. That means that the compiler doesn't need to push a return address to the stack as the only code that would be at that address is instructions to pop another address and return there, we can short-circuit all that and jump from the base case (0) directly to where aux(10000) is supposed to return to instead of taking 10000 dumb steps (like practically all procedural languages do because they don't have tail call optimisation).

This would've been different if you had concatenated the string not as an argument to aux.

[–] [email protected] 3 points 6 months ago* (last edited 6 months ago) (1 children)

I thought Tail recursion just gets turned into an iterative loop by the compiler? Hence why you won’t get a stack overflow. And since in procedural languages you can just use a loop in place of a tail recursive function you would never run into this problem, right? At least this is how it was taught to me when I was learning about it in lisp.

[–] [email protected] 3 points 6 months ago

Yes you still need the loop part I skipped over that one, only focussing on the "why no return address on the stack" part. It's what you need to focus on to see whether a recursive call is in a tail position and if it is the compiler does the rest no need to worry about that part.

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