Lecture 9: Proper Tail Calls:   Using the stack better
1 A motivating example
2 A simpler example
3 Defining tail position
4 Examining the stack
5 Strategy
6 Implementation pitfalls
6.1 Reusing arguments
6.2 Changing arities
6.3 Different stack-frame sizes
7 Testing
8 Applicability

Lecture 9: Proper Tail Calls: Using the stack better🔗

1 A motivating example🔗

A compiler’s job is to faithfully translate the semantics of one language into another; this much is obvious. However, not all translations are equal: some can be drastically more efficient than others, to the point where they change which programs we can actually, effectively run. This is not a claim about optimization, though. Optimizations typically improve the performance by some constant factor, meaning the amount of a particular resource (time, memory, IO, etc.) is lowered by some fraction. Our concern here is about compiling a ubiquitous feature of our language—function calls—in such a way that it provides an asymptotic improvement in our program.

Suppose we had a list of some arbitrary length, and we wanted to detect the presence of a value within it. We would naturally write,

let rec member (haystack : int list) (needle : int) : bool =
  match haystack with
  | [] -> false
  | first::rest ->
    if (needle == first) then true else (member rest needle)

What are the practical limits of this function? Conceptually, it should work for any list we’ve constructed. But in practice, this code might crash on lengthy lists, simply because the recursion progressed too deeply: we encounter a stack overflow, because we can’t allocate a stack frame for the next recursive call. This is dissatisfying: our language semantics don’t include any arbitrary limits on the depth of recursion. And clearly, whatever machine we ran this program on was capable of building a long list; it seems capricious to then be unable to process it!

Looking more carefully at the program, though, this failure is even more disappointing. When the if condition is false, we fall through to the else-branch and start evaluating the recursive call to member. But notice that when that call returns...the function simply passes the return value back as its answer, with no further processing needed! It makes sense that we’d need a stack frame to handle the recursive call, but the current stack frame is basically no longer needed. Perhaps we could consolidate the two somehow, and not need any extra space?

2 A simpler example🔗

Let’s consider a program in Diamondback that has a similar recursive structure. Instead of working through a list data structure (which we do not yet have), let’s compute the factorial function:

def fact_v1(n):
  if n <= 1: 1
  else: n * fact_v1(n - 1)

At first glance, this doesn’t match the structure of member since here we have additional work to do after the recursive call to fact(n - 1). But we’ve learned ways to transform this program into a similar one, such that the answer to the recursive call simply is the overall answer: we use an accumulator parameter, and rewrite the code as follows:

def fact-tail(n, acc):
  if n <= 1: acc
  else: fact-tail(n - 1, n * acc)

def fact_v2(n):
  fact-tail(n, 1)

Compare and contrast the evaluation order of these two functions, using the substitution model of evaluation that we began the course with:

fact_v1(4) ==> if 4 <= 1: 1 else 4 * fact_v1(3)
           ==> 4 * fact_v1(3)
           ==> 4 * (if 3 <= 1: 1 else 3 * fact_v1(2))
           ==> 4 * (3 * fact_v2(2))
           ==> 4 * (3 * (if 2 <= 1: 1 else 2 * fact_v1(1)))
           ==> 4 * (3 * (2 * fact_v1(1)))
           ==> 4 * (3 * (2 * (if 1 <= 1: 1 else 1 * fact_v1(0))))
           ==> 4 * (3 * (2 * (1)))
           ==> 4 * (3 * 2)
           ==> 4 * 6
           ==> 24

fact_v2(4) ==> fact_tail(4, 1)
           ==> if 4 <= 1: 1 else fact-tail(4 - 1, 4 * 1)
           ==> fact_tail(3, 4)
           ==> if 3 <= 1: 1 else fact-tail(3 - 1, 3 * 4)
           ==> fact_tail(2, 12)
           ==> if 2 <= 1: 1 else fact-tail(2 - 1, 2 * 12)
           ==> fact_tail(1, 24)
           ==> if 1 <= 1: 1 else fact-tail(1 - 1, 1 * 24)
           ==> 24

The initial version keeps a bunch of multiplications pending, until the innermost function call returns. It is reasonable to think that in our compiled code, each of those will correspond to one stack frames, and we clearly still need to keep track of the intermediate values of n in order to compute the final answer.

The second version of the function, though, never has more than one call to fact-tail pending. Nothing in this evaluation sequence “looks like” it needs a deep call stack. Can we achieve this?

3 Defining tail position🔗

What distinguishes the recursive calls to fact_v1 from the calls to fact_tail (or, for that matter, member)? Intuitively, we described them as “the last thing to be done”, before returning from the function. We say that such expressions are in tail position, and we can define such positions explicitly, looking at each expression form in our language:

  1. The expression of our program is in tail position.

  2. The body of a function is in tail position.

  3. If a let-binding is in tail position, then (a) its body is in tail position, but (b) the bindings themselves are not.

  4. If a conditional is in tail position, then (a) its branches are in tail position, but (b) the condition itself is not.

  5. The operands to an operator are never in tail position.

Visually, green expressions are always in tail position, yellow expressions are potentially in tail position, and red expressions are never in tail position:

type 'a aprogram = (* whole programs *)
  | AProgram of ~hl:2:s~'a adecl list~hl:2:e~ * ~hl:3:s~'a aexpr~hl:3:e~ * 'a      (* Rule 1 *)
and 'a adecl = (* function declarations *)
  | ADFun of string * string list * ~hl:3:s~'a aexpr~hl:3:e~ * 'a  (* Rule 2 *)
and 'a aexpr = (* anf expressions *)
  | ALet of string * ~hl:2:s~'a cexpr~hl:2:e~ * ~hl:1:s~'a aexpr~hl:1:e~ * 'a      (* Rule 3 *)
  | ACExpr of ~hl:1:s~'a cexpr~hl:1:e~
and 'a cexpr = (* compound expressions *)
  | CIf of ~hl:2:s~'a immexpr~hl:2:e~ * ~hl:1:s~'a aexpr~hl:1:e~ * ~hl:1:s~'a aexpr~hl:1:e~ * 'a   (* Rule 4 *)
  | CPrim1 of prim1 * ~hl:2:s~'a immexpr~hl:2:e~ * 'a              (* Rule 5 *)
  | CPrim2 of prim2 * ~hl:2:s~'a immexpr~hl:2:e~ * ~hl:2:s~'a immexpr~hl:2:e~ * 'a (* Rule 5 *)
  | CImmExpr of ~hl:1:s~'a immexpr~hl:1:e~
and 'a immexpr = (* immediate expressions *)
  | ImmNum of ~hl:1:s~int~hl:1:e~ * 'a
  | ImmBool of ~hl:1:s~bool~hl:1:e~ * 'a
  | ImmId of ~hl:1:s~string~hl:1:e~ * 'a

We can codify this, if we so choose, as a kind of tagging operation,

mark_tails : ('a aprogram) -> bool aprogram

In practice we probably don’t need to, and instead can just carry along a boolean flag through our compile function that keeps track of our tail-position status:

let rec compile_prog (prog : 'a aprog) =
  match prog with
  | AProg(decls, body, _) ->
    ... List.map compile_decl decls ...
    ... compile_aexp body ~hl:3:s~true~hl:3:e~ ...  (* Rule 1 *)
and compile_decl (decl : 'a adecl) =
  match decl with
  | ADFun(name, args, body, _) ->
    ... compile_aexp body ~hl:3:s~true~hl:3:e~ ...  (* Rule 2 *)
and compile_aexp (aexp : 'a aexpr) (tail_pos : boolean) =
  match aexp with
  | ALet(name, bind, body, _) ->
    ... compile_cexp bind ~hl:2:s~false~hl:2:e~ ... (* Rule 3b *)
    ... compile_aexp body ~hl:1:s~tail_pos~hl:1:e~  (* Rule 3a *)
  | ACExpr(e, _) -> compile_cexp e tail_pos
and compile_cexp (cexp : 'a cexpr) (tail_pos : boolean) =
  match cexp with
  | CIf(c, t, f, _) ->
    ... compile_imm c ~hl:2:s~false~hl:2:e~ ...     (* Rule 4a *)
    ... compile_aexp t ~hl:1:s~tail_pos~hl:1:e~ ... (* Rule 4b *)
    ... compile_aexp f ~hl:1:s~tail_pos~hl:1:e~ ... (* Rule 4b *)
  | CPrim1(op, arg, _) ->
    ... compile_imm arg ~hl:2:s~false~hl:2:e~ ...   (* Rule 5 *)
  | CPrim2(op, l, r, _) ->
    ... compile_imm l ~hl:2:s~false~hl:2:e~ ...     (* Rule 5 *)
    ... compile_imm r ~hl:2:s~false~hl:2:e~ ...     (* Rule 5 *)
  | CImmExpr(i, _) ->
    ... compile_imm i ~hl:1:s~tail_pos~hl:1:e~ ...

Do Now!

Extend this definition to include CApp expressions.

4 Examining the stack🔗

Let’s consider what the stack looks like while evaluating fact_v1. In this diagram, colors indicate which stack frame uses a particular value on the stack, while the brackets indicate which stack frame created a particular value on the stack. In this example we are using the stack-only calling convention, where all function arguments are pushed onto the stack. See Lecture 10 for similar examples using the x64 calling convention.

In top-level


In fact_v1(4)


In fact_v1(3)


In fact_v1(2)


In fact_v1(1)

At fact_v1(4)


At fact_v1(3)


At fact_v1(2)


At fact_v1(1)


About to return










Now let’s examine the stacks for fact_v2, assuming we compile our code exactly as we’ve always been. We’ll include the local variables, this time:

In top-level


In fact_tail(4, 1)


In fact_tail(3, 4)


In fact_tail(2, 12)


In fact_tail(1, 24)

At fact_tail(4, 1)


At fact_tail(3, 4)


At fact_tail(2, 12)


At fact_tail(1, 24)


About to return










Because the recursive calls here are all in tail-position, the next four instructions are all going to be ret instructions, which means the entirety of this stack can effectively be eliminated in one step. In other words, once the olive stack frame makes the call to the dark green frame, we never need to access an olive stack slot again. Looking carefully at the stack, we see that the next values for n and acc are precisely the local values computed in the previous stack frame, and moreover, each stack frame has exactly the same shape. If instead of creating a new stack frame, we simply reused the existing one, then we wouldn’t need more than constant stack depth to provide arbitrary call depth!

5 Strategy🔗

Rather than pushing the next set of arguments onto the stack, simply move them into the existing stack slots at RBP + 16, RBP + 24, etc. Once we’ve done that, we need to re-enter our existing function, but we can’t use call to do it.

Do Now!

Why not?

The meaning of call is to push a return address onto the stack and jump to the destination address. But we already have the necessary return address sitting on the stack! We also have a saved RBP on the stack too, which means that the function prologue we normally execute isn’t really needed here. So instead, we’ll simply jump directly to the next instruction in our code. The compiled assembly for fact_tail would then look roughly like this (ignoring all tag checks, and simplifying the condition slightly):

  push RBP
  mov RBP, RSP
  sub RSP, 16           ; reserve stack slots
  mov RAX, [RBP + 16]   ; load n
  cmp RAX, 2            ; compare to representation of 1
  jg keep_going
  mov RAX, [RBP + 24]   ; load acc into answer
  mov RSP, RBP          ; and return directly
  pop RBP               ; to the original
  ret                   ; caller
  mov RAX, [RBP + 16]   ; \
  sub RAX, 2            ; | compute n - 1
  mov [RBP - 8], RAX    ; /
  mov RAX, [RBP + 16]   ; \
  sar RAX, 1            ; |
  imul RAX, [RBP + 24]  ; | compute n * acc
  mov [RBP - 16], RAX   ; /
  mov RAX, [RBP - 8]    ; \
  mov [RBP + 16], RAX   ; / OVERWRITE argument n
  mov RAX, [RBP - 16]   ; \
  mov [RBP + 24], RAX   ; / OVERWRITE argument acc
  jmp fact_tail_body    ; AND RESTART fact_tail

This code is almost legible enough that we could turn it into C code pretty easily:

int fact_tail(int n, int acc) {
  while (true) {
    if (n <= 1) { return acc; }
    else {
      int temp1 = n - 1;
      int temp2 = n * acc;
      n = temp1;
      acc = temp2;

We’ve turned our (tail-)recursive function into a while-loop, and eliminated all the function calls!

6 Implementation pitfalls🔗

6.1 Reusing arguments🔗

Consider the following code:

def max(x, y):
  if y >= x: y
  else: max(y, x)

This is clearly tail-recursive, so we can apply the same technique above. Since we have no intermediate expressions (again, simplifying the conditional), we don’t even need to move RSP at all; all our values are already on the stack:

  push RBP
  mov RBP, RSP
  mov RAX, [RBP + 24]   ; load y
  cmp RAX, [RBP + 16]   ; compare to x
  jl keep_going
  mov RAX, [RBP + 24]   ; load y into answer
  mov RSP, RBP          ; and return directly
  pop RBP               ; to the original
  ret                   ; caller
  mov RAX, [RBP + 24]   ; \
  mov [RBP + 16], RAX   ; / OVERWRITE argument x
  mov RAX, [RBP + 16]   ; \
  mov [RBP + 24], RAX   ; / OVERWRITE argument y
  jmp max_body          ; AND RESTART max

Do Now!

What went horribly wrong?


Try to fix it.

Try tracing through two simple calls to max, to test both branches of the if expression, and carefully step through the generated assembly. If we call max(10, 20), then we fall through the jl instruction, and end up returning [RBP + 12], which is y as expected. But suppose we try max(20, 10). then we fall through to keep_going, where we load the current value of [RBP + 12] and overwrite [RBP + 8] with it, which effectively copies y into x. Then we load the current value of [RBP + 8] and copy it into [RBP + 12], in an attempt to copy the current value of x into y but at this point, the value of x is gone! So the effect of our tail-call of max(y, x) is to call (10, 10), which then executes the first branch of the conditional and returns 10.

(Note that if we updated our arguments in the other order, such that we overwrote y before we overwrote x, we would have an even more insidious problem: this particular function call would compute the correct answer! Our call to max(10, 20) would effectively call max(20, 20) and return 20 purely coincidentally the correct answer. If we changed our program to compute the minimum instead, then this reversed argument-replacement order would once again cause problems.)

The problem is that our new arguments to the call reside in addresses that we are about to overwrite, and we’ve managed to create a cycle from the address of y, to the value of the new argument of x, to the address of x to the value of the new argument of y. Our naive strategy of simply moving arguments was too simple. Instead, we can try any of the following strategies in increasing sophistication (or others, in a similar spirit):

The last strategy above is optimal: it never uses more that one extra stack slot at a time, and it uses the minimum number of movs and stack operations. But it’s also clearly the most complicated, and therefore the hardest to test and guarantee correct. The next-to-last strategy strikes a good balance between efficiency and simplicity: the safety condition is easy to check, and both the push/pop-based code and the mov-based code handle all arguments in a uniform manner, making it much easier to test. In a production-quality compiler, we’d obviously like to implement the most efficient version, and take on the maintenance and correctness burden that goes with it.

6.2 Changing arities🔗

The technique above is not limited to self-recursion; it works for tail-calls between functions as well, meaning that mutually recursive functions can also be compiled to essentially a while-loop with a few conditions inside it.

However, the technique above works smoothly only for tail calls to callees whose arities are no greater than their callers’ arities. Suppose function F calls function G, whose arity is \(A_G\). Suppose G then tail-calls another function H with arity \(A_H > A_G\). We have two problems:

Obviously, these difficulties are not insurmountable, but they do require some clever thought...

6.3 Different stack-frame sizes🔗

If we tail-call from one function into another, such that the second function uses more local variables than the first, then RSP is too low to reserve sufficient space. We could accommodate this by restoring RSP to RBP, and moving the instruction that reserves stack space (i.e. sub RSP, 8 * #vars) after the body label. This adds two instructions to every tail-call that might not be necessary, but the space savings from the stack should more than make up for it.

7 Testing🔗

Testing tail calls is not much more difficult than testing regular calls, and requires just as much dilligence about covering all cases. It is trivial to convert any tail-call into a non-tail-call, e.g. by adding 0 or by or’ing with false. Construct tail-recursive test programs whose recursion depth should otherwise overflow the stack, then use one of these gimmicks to convert the tail calls into non-tail calls, and confirm that only the tail-call program runs to completion.

Alternatively, we might implement a new primitive printStack that outputs us a “stack trace” of the current program, and confirm that the tail-recursive stack trace is appropriately short, while the non-tail-recursive one is inordinately long.

As was noted in the aliasing section, we must carefully test that our argument-replacement code never introduces unintentional cycles that produce the wrong results. The difficulty of testing this depends on the complexity of your heuristic for dealing with these cases. Unless the utmost efficiency is paramount, it may make sense to choose a slightly suboptimal compilation strategy and trade off a slight bit of performance for a greater confidence in correctness.

8 Applicability🔗

Do Now!

Does this matter in practice?

Yes. We’ve changed the performance of our compiled code from \(O(n)\) to \(O(1)\), which means we no longer have an artifical limit on the size of problems we can solve with a recursive function.

Do Now!

Ok, fine, but surely this is just an academic problem, a design flaw in functional languages! Surely real languages like OO don’t need to do all this work, right?

Yes, they do.