Implement recursion in ASM without procedures - recursion

Question

I'm trying to implement functions and recursion in an ASM-like simplified language that has no procedures. Only simple jumpz, jump, push, pop, add, mul type commands.
Here are the commands:
(all variables and literals are integers)
set (sets the value of an already existing variable or declares and initializes a new variable) e.g. (set x 3)
push (pushes a value onto the stack. can be a variable or an integer) e.g. (push 3) or (push x)
pop (pops the stack into a variable) e.g. (pop x)
add (adds the second argument to the first argument) e.g. (add x 1) or (add x y)
mul (same as add but for multiplication)
jump (jumps to a specific line of code) e.g. (jump 3) would jump to line 3 or (jump x) would jump to the line # equal to the value of x
jumpz (jumps to a line number if the second argument is equal to zero) e.g. (jumpz 3 x) or (jumpz z x)
The variable 'IP' is the program counter and is equal to the line number of the current line of code being executed.
In this language, functions are blocks of code at the bottom of the program that are terminated by popping a value off the stack and jumping to that value. (using the stack as a call stack) Then the functions can be called anywhere else in the program by simply pushing the instruction pointer onto the stack and then jumping to the start of the function.
This works fine for non-recursive functions.
How could this be modified to handle recursion?
I've read that implementing recursion with a stack is a matter of pushing parameters and local variables onto the stack (and in this lower level case, also the instruction pointer I think)
I wouldn't be able to do something like x = f(n). To do this I'd have some variable y (that is also used in the body of f), set y equal to n, call f which assigns its "return value" to y and then jumps control back to where f was called from, where we then set x equal to y.
(a function that squares a number whose definition starts at line 36)
1 - set y 3
2 - set returnLine IP
3 - add returnLine 2
4 - push returnLine
5 - jump 36
6 - set x y
...
36 - mul y 2
37 - pop returnLine
38 - jump returnLine
This doesn't seem to lend itself to recursion. Arguments and intermediate values would need to go on the stack and I think multiple instances on the stack of the same address would result from recursive calls which is fine.

Answer 1

Next code raises the number "base" to the power "exponent" recursively in "John Smith Assembly":
1 - set base 2 ;RAISE 2 TO ...
2 - set exponent 4 ;... EXPONENT 4 (2^4=16).
3 - set result 1 ;MUST BE 1 IN ORDER TO MULTIPLY.
4 - set returnLine IP ;IP = 4.
5 - add returnLine 4 ;RETURNLINE = 4+4.
6 - push returnLine ;PUSH 8.
7 - jump 36 ;CALL FUNCTION.
.
.
.
;POWER FUNCTION.
36 - jumpz 43 exponent ;FINISH IF EXPONENT IS ZERO.
37 - mul result base ;RESULT = ( RESULT * BASE ).
38 - add exponent -1 ;RECURSIVE CONTROL VARIABLE.
39 - set returnLine IP ;IP = 39.
40 - add returnLine 4 ;RETURN LINE = 39+4.
41 - push returnLine ;PUSH 43.
42 - jump 36 ;RECURSIVE CALL.
43 - pop returnLine
44 - jump returnLine
;POWER END.
In order to test it, let's run it manually :
LINE | BASE EXPONENT RESULT RETURNLINE STACK
------|---------------------------------------
1 | 2
2 | 4
3 | 1
4 | 4
5 | 8
6 | 8
7 |
36 |
37 | 2
38 | 3
39 | 39
40 | 43
41 | 43(1)
42 |
36 |
37 | 4
38 | 2
39 | 39
40 | 43
41 | 43(2)
42 |
36 |
37 | 8
38 | 1
39 | 39
40 | 43
41 | 43(3)
42 |
36 |
37 | 16
38 | 0
39 | 39
40 | 43
41 | 43(4)
42 |
36 |
43 | 43(4)
44 |
43 | 43(3)
44 |
43 | 43(2)
44 |
43 | 43(1)
44 |
43 | 8
44 |
8 |
Edit : parameter for function now on stack (didn't run it manually) :
1 - set base 2 ;RAISE 2 TO ...
2 - set exponent 4 ;... EXPONENT 4 (2^4=16).
3 - set result 1 ;MUST BE 1 IN ORDER TO MULTIPLY.
4 - set returnLine IP ;IP = 4.
5 - add returnLine 7 ;RETURNLINE = 4+7.
6 - push returnLine ;PUSH 11.
7 - push base ;FIRST PARAMETER.
8 - push result ;SECOND PARAMETER.
9 - push exponent ;THIRD PARAMETER.
10 - jump 36 ;FUNCTION CALL.
...
;POWER FUNCTION.
36 - pop exponent ;THIRD PARAMETER.
37 - pop result ;SECOND PARAMETER.
38 - pop base ;FIRST PARAMETER.
39 - jumpz 49 exponent ;FINISH IF EXPONENT IS ZERO.
40 - mul result base ;RESULT = ( RESULT * BASE ).
41 - add exponent -1 ;RECURSIVE CONTROL VARIABLE.
42 - set returnLine IP ;IP = 42.
43 - add returnLine 7 ;RETURN LINE = 42+7.
44 - push returnLine ;PUSH 49.
45 - push base
46 - push result
47 - push exponent
48 - jump 36 ;RECURSIVE CALL.
49 - pop returnLine
50 - jump returnLine
;POWER END.

Answer 2

Your asm does provide enough facilities to implement the usual procedure call / return sequence. You can push a return address and jump as a call, and pop a return address (into a scratch location) and do an indirect jump to it as a ret. We can just make call and ret macros. (Except that generating the correct return address is tricky in a macro; we might need a label (push ret_addr), or something like set tmp, IP / add tmp, 4 / push tmp / jump target_function). In short, it's possible and we should wrap it up in some syntactic sugar so we don't get bogged down with that while looking at recursion.
With the right syntactic sugar, you can implement Fibonacci(n) in assembly that will actually assemble for both x86 and your toy machine.
You're thinking in terms of functions that modify static (global) variables. Recursion requires local variables so each nested call to the function has its own copy of local variables. Instead of having registers, your machine has (apparently unlimited) named static variables (like x and y). If you want to program it like MIPS or x86, and copy an existing calling convention, just use some named variables like eax, ebx, ..., or r0 .. r31 the way a register architecture uses registers.
Then you implement recursion the same way you do in a normal calling convention, where either the caller or callee use push / pop to save/restore a register on the stack so it can be reused. Function return values go in a register. Function args should go in registers. An ugly alternative would be to push them after the return address (creating a caller-cleans-the-args-from-the-stack calling convention), because you don't have a stack-relative addressing mode to access them the way x86 does (above the return address on the stack). Or you could pass return addresses in a link register, like most RISC call instructions (usually called bl or similar, for branch-and-link), instead of pushing it like x86's call. (So non-leaf callees have to push the incoming lr onto the stack themselves before making another call)
A (silly and slow) naively-implemented recursive Fibonacci might do something like:
int Fib(int n) {
if(n<=1) return n; // Fib(0) = 0; Fib(1) = 1
return Fib(n-1) + Fib(n-2);
}
## valid implementation in your toy language *and* x86 (AMD64 System V calling convention)
### Convenience macros for the toy asm implementation
# pretend that the call implementation has some way to make each return_address label unique so you can use it multiple times.
# i.e. just pretend that pushing a return address and jumping is a solved problem, however you want to solve it.
%define call(target) push return_address; jump target; return_address:
%define ret pop rettmp; jump rettmp # dedicate a whole variable just for ret, because we can
# As the first thing in your program, set eax, 0 / set ebx, 0 / ...
global Fib
Fib:
# input: n in edi.
# output: return value in eax
# if (n<=1) return n; // the asm implementation of this part isn't interesting or relevant. We know it's possible with some adds and jumps, so just pseudocode / handwave it:
... set eax, edi and ret if edi <= 1 ... # (not shown because not interesting)
add edi, -1
push edi # save n-1 for use after the recursive call
call Fib # eax = Fib(n-1)
pop edi # restore edi to *our* n-1
push eax # save the Fib(n-1) result across the call
add edi, -1
call Fib # eax = Fib(n-2)
pop edi # use edi as a scratch register to hold Fib(n-1) that we saved earlier
add eax, edi # eax = return value = Fib(n-1) + Fib(n-2)
ret
During a recursive call to Fib(n-1) (with n-1 in edi as the first argument), the n-1 arg is also saved on the stack, to be restored later. So each function's stack frame contains the state that needs to survive the recursive call, and a return address. This is exactly what recursion is all about on a machine with a stack.
Jose's example doesn't demonstrate this as well, IMO, because no state needs to survive the call for pow. So it just ends up pushing a return address and args, then popping the args, building up just some return addresses. Then at the end, follows the chain of return addresses. It could be extended to save local state across each nested call, doesn't actually illustrate it.
My implementation is a bit different from how gcc compiles the same C function for x86-64 (using the same calling convention of first arg in edi, ret value in eax). gcc6.1 with -O1 keeps it simple and actually does two recursive calls, as you can see on the Godbolt compiler explorer. (-O2 and especially -O3 do some aggressive transformations). gcc saves/restores rbx across the whole function, and keeps n in ebx so it's available after the Fib(n-1) call. (and keeps Fib(n-1) in ebx to survive the second call). The System V calling convention specifies rbx as a call-preserved register, but rbi as call-clobbered (and used for arg-passing).
Obviously you can implement Fib(n) much faster non-recursively, with O(n) time complexity and O(1) space complexity, instead of O(Fib(n)) time and space (stack usage) complexity. It makes a terrible example, but it is trivial.

Answer 3

Margaret's pastebin modified slightly to run in my interpreter for this language: (infinite loop problem, probably due to a transcription error on my part)
set n 3
push n
set initialCallAddress IP
add initialCallAddress 4
push initialCallAddress
jump fact
set finalValue 0
pop finalValue
print finalValue
jump 100
:fact
set rip 0
pop rip
pop n
push rip
set temp n
add n -1
jumpz end n
push n
set link IP
add link 4
push link
jump fact
pop n
mul temp n
:end
pop rip
push temp
jump rip
Successful transcription of Peter's Fibonacci calculator:
String[] x = new String[] {
//n is our input, which term of the sequence we want to calculate
"set n 5",
//temp variable for use throughout the program
"set temp 0",
//call fib
"set temp IP",
"add temp 4",
"push temp",
"jump fib",
//program is finished, prints return value and jumps to end
"print returnValue",
"jump end",
//the fib function, which gets called recursively
":fib",
//if this is the base case, then we assert that f(0) = f(1) = 1 and return from the call
"jumple base n 1",
"jump notBase",
":base",
"set returnValue n",
"pop temp",
"jump temp",
":notBase",
//we want to calculate f(n-1) and f(n-2)
//this is where we calculate f(n-1)
"add n -1",
"push n",
"set temp IP",
"add temp 4",
"push temp",
"jump fib",
//return from the call that calculated f(n-1)
"pop n",
"push returnValue",
//now we calculate f(n-2)
"add n -1",
"set temp IP",
"add temp 4",
"push temp",
"jump fib",
//return from call that calculated f(n-2)
"pop n",
"add returnValue n",
//this is where the fib function ultimately ends and returns to caller
"pop temp",
"jump temp",
//end label
":end"
};