Find limit of recursion

Find limit of recursion
You are encouraged to solve this task according to the task description, using any language you may know.

Find limit of recursion is part of Short Circuit's Console Program Basics selection.

Find the limit of recursion.

ACL2

<lang Lisp>(defun recursion-limit (x)

  (if (zp x)
      0
      (prog2$ (cw "~x0~%" x)
              (1+ (recursion-limit (1+ x))))))</lang>

Output:

(trimmed)

87195
87196
87197
87198
87199
87200
87201

***********************************************
************ ABORTING from raw Lisp ***********
Error:  Stack overflow on value stack.
***********************************************

Ada

<lang Ada>with Ada.Text_IO; use Ada.Text_IO;

procedure Test_Recursion_Depth is

  function Recursion (Depth : Positive) return Positive is
  begin
     return Recursion (Depth + 1);
  exception
     when Storage_Error =>
        return Depth;
  end Recursion;

begin

  Put_Line ("Recursion depth on this system is" & Integer'Image (Recursion (1)));

end Test_Recursion_Depth;</lang> Note that unlike some solutions in other languages this one does not crash (though usefulness of this task is doubtful).

In Ada Storage_Error exception is propagated when there is no free memory to accomplish the requested action. In particular it is propagated upon stack overflow within the task where this occurs. Storage_Error can be handled without termination of the task. In the solution the function Recursion calls itself or else catches Storage_Error indicating stack overflow.

Note that this technique requires some care, because there must be enough stack space for the handler to work. In this case it works because the handler just return the current call depth. In real-life Storage_Error is usually fatal.

Output:

Recursion depth on this system is 524091

ALGOL 68

The depth of recursion in Algol 68 proper is unlimited. Particular implementations will reach a limit, if only through exhaustion of storage and/or address space and/or time before power failure. If not time limited, the depth reached depends very much on what the recursive routine needs to store on the stack, including local variables if any. The simplest recursive Algol68 program is: <lang algol68>PROC recurse = VOID : recurse; recurse</lang> This one-liner running under Algol68 Genie and 64-bit Linux reaches a depth of 3535 with the shell's default stack size of 8Mbytes and 28672 when set to 64Mbytes, as shown by the following output. From this we can deduce that Genie does not implement tail recursion. The --trace option to a68g prints a stack trace when the program crashes; the first two commands indicate the format of the trace, the third counts the depth of recursion with the default stack size and the fourth shows the result of octupling the size of the stack.

Output:

pcl@anubis ~/a68/Rosetta $ a68g --trace Recurse.a68 | head
genie: frame stack 6144k, expression stack 2048k, heap 49152k, handles 8192k
      BEGIN MODE DOUBLE = LONG REAL, QUAD = LONG LONG REAL;
      -                                                    
1     PROC recurse = VOID : recurse; recurse
      -                                     
genie_unit
1     PROC recurse = VOID : recurse; recurse
                                     -      
genie_unit
1     PROC recurse = VOID : recurse; recurse
pcl@anubis ~/a68/Rosetta $ a68g --trace Recurse.a68 | tail
1     PROC recurse = VOID : recurse; recurse
                            -               
genie_unit
1     PROC recurse = VOID : recurse; recurse
                            -               
genie_unit
1     PROC recurse = VOID : recurse; recurse
                     1                      
a68g: runtime error: 1: stack overflow (detected in particular-program).
Genie finished in 0.19 seconds
pcl@anubis ~/a68/Rosetta $ a68g --trace Recurse.a68 |  grep recurse | wc
   3535   28280  159075
pcl@anubis ~/a68/Rosetta $ prlimit --stack=67108864 a68g --trace Recurse.a68 | grep recurse | wc
  28672  229376 1290240
pcl@anubis ~/a68/Rosetta $ 

AutoHotkey

<lang AutoHotkey>Recurse(0)

Recurse(x) {

 TrayTip, Number, %x%
 Recurse(x+1)

}</lang>

Last visible number is 827.

AutoIt

<lang AutoIt>;AutoIt Version: 3.2.10.0 $depth=0 recurse($depth) Func recurse($depth)

  ConsoleWrite($depth&@CRLF)
  Return recurse($depth+1)

EndFunc</lang> Last value of $depth is 5099 before error. Error: Recursion level has been exceeded - AutoIt will quit to prevent stack overflow.

AWK

<lang AWK># syntax: GAWK -f FIND_LIMIT_OF_RECURSION.AWK

2. version depth messages
3. ------------------ ----- --------
4. GAWK 3.1.4 2892 none
5. XML GAWK 3.1.4 3026 none
6. GAWK 4.0 >999999
7. MAWK 1.3.3 4976 A stack overflow was encountered at
8. address 0x7c91224e.
9. TAWK-DOS AWK 5.0c 357 stack overflow
10. TAWK-WIN AWKW 5.0c 2477 awk stack overflow
11. NAWK 20100523 4351 Segmentation fault (core dumped)

BEGIN {

   x()
   print("done")

} function x() {

   print(++n)
   if (n > 999999) { return }
   x()

}</lang>

Axe

Warning: running this program will cause you to have to clear your RAM. You will lose any data stored in RAM.

In Axe 1.2.2 on a TI-84 Plus Silver Edition, the last line this prints before hanging is 12520. This should be independent of any arguments passed since they are not stored on the stack.

<lang axe>RECURSE(1) Lbl RECURSE .Optionally, limit the number of times the argument is printed Disp r₁▶Dec,i RECURSE(r₁+1)</lang>

BASIC

ZX Spectrum Basic

On the ZX Spectrum recursion is limited only by stack space. The program eventually fails, because the stack is so full that there is no stack space left to make the addition at line 110: <lang zxbasic> 10 LET d=0: REM depth 100 PRINT AT 1,1; "Recursion depth: ";d 110 LET d=d+1 120 GO SUB 100: REM recursion 130 RETURN: REM this is never reached 200 STOP </lang>

Output:

(from a 48k Spectrum)

{{{

Recursion depth: 13792
4 Out of memory, 110:1

}}}

Batch File

MUNG.CMD is a commandline tool written in DOS Batch language. It finds the limit of recursion possible using CMD /C.

<lang dos>@echo off set /a c=c+1 echo [Depth %c%] Mung until no good cmd /c mung.cmd echo [Depth %c%] No good set /a c=c-1</lang>

Result (abbreviated):

...
[Depth 259] Mung until no good
[Depth 260] Mung until no good
[Depth 261] Mung until no good
[Depth 261] No good
[Depth 260] No good
[Depth 259] No good
...

If one uses call rather than CMD/C, the call depth is much deeper but ends abruptly and can't be trapped.

<lang dos>@echo off set /a c=c+1 echo [Depth %c%] Mung until no good call mung.cmd echo [Depth %c%] No good set /a c=c-1</lang>

Result (abbreviated):

1240: Mung until no good
1241: Mung until no good
******  B A T C H   R E C U R S I O N  exceeds STACK limits ******
Recursion Count=1240, Stack Usage=90 percent
******       B A T C H   PROCESSING IS   A B O R T E D      ******

You also get the exact same results when calling mung internally, as below

<lang dos>@echo off set c=0

mung

set /a c=c+1 echo [Level %c%] Mung until no good call :mung set /a c=c-1 echo [Level %c%] No good</lang>

Setting a limit on the recursion depth can be done like this:

<lang dos>@echo off set c=0

mung

set /a c=%1+1 if %c%==10 goto :eof echo [Level %c%] Mung until no good call :mung %c% set /a c=%1-1 echo [Level %c%] No good</lang>

BBC BASIC

<lang bbcbasic> PROCrecurse(1)

     END
     
     DEF PROCrecurse(depth%)
     IF depth% MOD 100 = 0 PRINT TAB(0,0) depth%;
     PROCrecurse(depth% + 1)
     ENDPROC</lang>

Output:

from BBC BASIC for Windows with default value of HIMEM

     37400
No room

Bracmat

<lang bracmat>rec=.out$!arg&rec$(!arg+1)</lang>

Observed recursion depths:

 Windows XP command prompt: 6588
 Linux: 18276

Bracmat crashes when it tries to exceed the maximum recursion depth.

C

<lang c>#include <stdio.h>

void recurse(unsigned int i) {

 printf("%d\n", i);
 recurse(i+1); // 523756

}

int main() {

 recurse(0);
 return 0;

}</lang>

Segmentation fault occurs when i is 523756. (This was checked debugging with gdb rather than waiting the output: the printf line for the test was commented). It must be noted that the recursion limit depends on how many parameters are passed onto the stack. E.g. adding a fake double argument to recurse, the limit is reached at i == 261803. The limit depends on the stack size and usage in the function. Even if there are no arguments, the return address for a call to a subroutine is stored on the stack (at least on x86 and many more processors), so this is consumed even if we put arguments into registers.

The following code may have some effect unexpected by the unwary: <lang C>#include <stdio.h>

char * base; void get_diff() { char x; if (base - &x < 200) printf("%p %d\n", &x, base - &x); }

void recur() { get_diff(); recur(); }

int main() { char v = 32; printf("pos of v: %p\n", base = &v); recur(); return 0; }</lang> With GCC 4.5, if compiled without -O2, it segfaults quickly; if gcc -O2, crash never happens, because the optimizer noticed the tail recursion in recur() and turned it into a loop!

C#

<lang csharp>using System; class RecursionLimit {

 static void Main(string[] args)
 {
   Recur(0);
 }

 private static void Recur(int i) 
 {
   Console.WriteLine(i);
   Recur(i + 1);
 }

}</lang>

Through debugging, the highest I achieve is 14250.

Through execution (with Mono), another user has reached 697186.

Clojure

<lang clojure> => (def *stack* 0) => ((fn overflow [] ((def *stack* (inc *stack*))(overflow)))) java.lang.StackOverflowError (NO_SOURCE_FILE:0) => *stack* 10498 </lang>

COBOL

Works with: OpenCOBOL

<lang cobol>identification division. program-id. recurse. data division. working-storage section. 01 depth-counter pic 9(3). 01 install-address usage is procedure-pointer. 01 install-flag pic x comp-x value 0. 01 status-code pic x(2) comp-5. 01 ind pic s9(9) comp-5.

linkage section. 01 err-msg pic x(325).

procedure division. 100-main.

set install-address to entry "300-err".

call "CBL_ERROR_PROC" using install-flag install-address returning status-code.

if status-code not = 0 display "ERROR INSTALLING ERROR PROC" stop run

       end-if

	move 0 to depth-counter.

display 'Mung until no good.'. perform 200-mung. display 'No good.'. stop run.

200-mung. add 1 to depth-counter. display depth-counter. perform 200-mung. 300-err. entry "300-err" using err-msg. perform varying ind from 1 by 1 until (err-msg(ind:1) = x"00") or (ind = length of err-msg) continue end-perform

display err-msg(1:ind).

• > room for a better-than-abrupt death here.

exit program.</lang>

Compiled with

cobc -free -x -g recurse.cbl

gives, after a while,

...
249
250
251
252
253
Trapped: recurse.cob:38: Stack overflow, possible PERFORM depth exceeded
recurse.cob:50: libcob: Stack overflow, possible PERFORM depth exceeded

Without stack-checking turned on (achieved with -g in this case), it gives

...
249
250
251
252
253
254
255
256
257
Attempt to reference unallocated memory (Signal SIGSEGV)
Abnormal termination - File contents may be incorrect

which suggests that -g influences the functionality of CBL_ERROR_PROC

Thanks to Brian Tiffin for his demo code on opencobol.org's forum

A more 'canonical' way of doing it

from Richard Plinston on comp.lang.cobol

Works with: OpenCOBOL

<lang cobol> IDENTIFICATION DIVISION.

      PROGRAM-ID.          recurse RECURSIVE.
      DATA DIVISION.
      WORKING-STORAGE SECTION.
      01  Starter          PIC S9(8) VALUE 1.
      PROCEDURE DIVISION.
      Program-Recurse.
          CALL "recurse-sub" USING Starter
          STOP RUN.

      IDENTIFICATION DIVISION.
      PROGRAM-ID.          recurse-sub.
      DATA DIVISION.
      WORKING-STORAGE SECTION.
      LINKAGE SECTION.
      01  Countr                      PIC S9(8).
      PROCEDURE DIVISION USING Countr.
      Program-Recursive.
          DISPLAY Countr
          ADD 1   TO Countr
          CALL "recurse-sub" USING Countr

          EXIT PROGRAM.
      END PROGRAM recurse-sub.
      END PROGRAM recurse. </lang>

Compiled with

cobc -x -g recurse.cbl

gives

...
+00000959
+00000960
+00000961
+00000962
+00000963
+00000964
recurse.cbl:19: Attempt to reference unallocated memory (Signal SIGSEGV)
Abnormal termination - File contents may be incorrect

CoffeeScript

<lang coffeescript> recurse = ( depth = 0 ) ->

   try
       recurse depth + 1
   catch exception
       depth

console.log "Recursion depth on this system is #{ do recurse }" </lang>

Output:

Example on Node.js

    Recursion depth on this system is 9668

Common Lisp

<lang lisp> (defun recurse () (recurse)) (trace recurse) (recurse) </lang>

Output:

This test was done with clisp under cygwin

 3056. Trace: (RECURSE)
 3057. Trace: (RECURSE)
 3058. Trace: (RECURSE)
 3059. Trace: (RECURSE)
 
 *** - Lisp stack overflow. RESET

However, for an implementation of Lisp that supports proper tail recursion, this function will not cause a stack overflow, so this method will not work.

D

<lang d>import std.c.stdio;

void recurse(in uint i=0) {

   printf("%u ", i);
   recurse(i + 1);

}

void main() {

   recurse();

}</lang> With the DMD compiler, using default compilation arguments, the stack overflows at 51_002.

With DMD increasing the stack size using for example -L/STACK:1500000000 the stack overflows at 75_002_026.

Using -O compilation argument DMD performs tail call optimization, and the stack doesn't overflow.

Déjà Vu

This example is untested. Please check that it's correct, debug it as necessary, and remove this message.

<lang dejavu>rec-fun n: !. n rec-fun ++ n

rec-fun 0</lang> This continues until the memory is full, so I didn't wait for it to finish. Currently, it should to to almost 3 million levels of recursion on a machine with 1 GB free. Eliminating the n should give over 10 million levels on the same machine.

Delphi

Works with: Delphi version 2010 (and probably all other versions)

<lang delphi>program Project2; {$APPTYPE CONSOLE} uses

 SysUtils;

function Recursive(Level : Integer) : Integer; begin

 try
   Level := Level + 1;
   Result := Recursive(Level);
 except
   on E: EStackOverflow do
     Result := Level;
 end;

end;

begin

 Writeln('Recursion Level is ', Recursive(0));
 Writeln('Press any key to Exit');
 Readln;

end.</lang>

Output:

Recursion Level is 28781

DWScript

Recursion limit is a parameter of script execution, which can be specified independently from the stack size to limit execution complexity.

<lang delphi>var level : Integer;

procedure Recursive; begin

  Inc(level);
  try
     Recursive;
  except
  end;

end;

Recursive;

Println('Recursion Level is ' + IntToStr(level));</lang>

E

Outside of debugging access to other vats, E programs are (ideally) not allowed to observe recursion limits, because stack unwinding at an arbitrary point can break invariants of the code that was executing at the time. In particular, consider an attacker who estimates the stack size, nearly fills up the stack to that point, then invokes the victim — If the attacker is allowed to catch our hypothetical StackOverflowException from inside the victim, then there is a good chance of the victim then being in an inconsistent state, which the attacker can then make use of.

Elixir

same as "Erlang"

Emacs Lisp

<lang lisp>(defun my-recurse (n)

 (my-recurse (1+ n)))

(my-recurse 1) => enters debugger at (my-recurse 595), per the default max-lisp-eval-depth 600 in Emacs 24.1</lang>

Variable max-lisp-eval-depth[1] is the maximum depth of function calls and variable max-specpdl-size[2] is the maximum depth of nested let bindings. A function call is a let of the parameters, even if there's no parameters, and so counts towards max-specpdl-size as well as max-lisp-eval-depth.

The limits can be increased with setq etc globally, or let etc temporarily. Lisp code which knows it needs deep recursion might temporarily increase the limits. Eg. regexp-opt.el. The ultimate limit is memory or C stack.

Erlang

Erlang has no recursion limit. It is tail call optimised. If the recursive call is not a tail call it is limited by available RAM. Please add what to save on the stack and how much RAM to give to Erlang and I will test that limit.

F#

A tail-recursive function will run indefinitely without problems (the integer will overflow, though).

<lang fsharp>let rec recurse n =

 recurse (n+1)

recurse 0</lang>

The non-tail recursive function of the following example crashed with a StackOverflowException after 39958 recursive calls:

<lang fsharp>let rec recurse n =

  printfn "%d" n
  1 + recurse (n+1)

recurse 0 |> ignore</lang>

Forth

<lang forth>: munge ( n -- n' ) 1+ recurse ;

test 0 ['] munge catch if ." Recursion limit at depth " . then ;

test \ Default gforth: Recursion limit at depth 3817</lang>

Or you can just ask the system:

<lang forth>s" return-stack-cells" environment? ( 0 | potential-depth-of-return-stack -1 )</lang>

Full TCO is problematic, but a properly tail-recursive call is easy to add to any Forth. For example, in SwiftForth:

<lang forth>: recur; [ last 2 cells + literal ] @ +bal postpone again ; immediate

test dup if 1+ recur; then drop ." I gave up finding a limit!" ;

1 test</lang>

Fortran

<lang fortran>program recursion_depth

 implicit none

 call recurse (1)

contains

 recursive subroutine recurse (i)

   implicit none
   integer, intent (in) :: i

   write (*, '(i0)') i
   call recurse (i + 1)

 end subroutine recurse

end program recursion_depth</lang>

Output:

(snipped)

208914
208915
208916
208917
208918
208919
208920
208921
208922
208923
Segmentation fault (core dumped)

GAP

The limit is around 5000 : <lang gap>f := function(n)

 return f(n+1);

end;

1. Now loop until an error occurs

f(0);

1. Error message :
2. Entering break read-eval-print loop ...
3. you can 'quit;' to quit to outer loop, or
4. you may 'return;' to continue

n;

1. 4998

1. quit "brk mode" and return to GAP

quit;</lang> This is the default GAP recursion trap, see reference manual, section 7.10. It enters "brk mode" after multiples of 5000 recursions levels. On can change this interval : <lang gap>SetRecursionTrapInterval(100000);

1. No limit (may crash GAP if recursion is not controlled) :

SetRecursionTrapInterval(0);</lang>

gnuplot

<lang gnuplot># Put this in a file foo.gnuplot and run as

1. gnuplot foo.gnuplot

1. probe by 1 up to 1000, then by 1% increases

if (! exists("try")) { try=0 } try=(try<1000 ? try+1 : try*1.01)

recurse(n) = (n > 0 ? recurse(n-1) : 'ok') print "try recurse ", try print recurse(try) reread</lang>

Gnuplot 4.6 has a builtin STACK_DEPTH limit of 250, giving

try recurse 251
"/tmp/foo.gnuplot", line 2760: recursion depth limit exceeded

Gnuplot 4.4 and earlier has no limit except the C stack, giving a segv or whatever eventually.

Go

Go features stacks that grow as needed making the effective recursion limits relatively large.

Pre-Go 1.2 this could be all of memory and the program would grow without bounds until the system swap space was exhausted and the program was killed (either by the a run-time panic after an allocation failure or by the operating system killing the process).

Go 1.2 set a limit to the maximum amount of memory that can be used by a single goroutine stack. The initial setting is 1 GB on 64-bit systems, 250 MB on 32-bit systems. The default can be changed by SetMaxStack in the runtime/debug package. It is documented as "useful mainly for limiting the damage done by goroutines that enter an infinite recursion."

<lang go>package main

import ( "flag" "fmt" "runtime/debug" )

func main() { stack := flag.Int("stack", 0, "maximum per goroutine stack size or 0 for the default") flag.Parse() if *stack > 0 { debug.SetMaxStack(*stack) } r(1) }

func r(l int) { if l%1000 == 0 { fmt.Println(l) } r(l + 1) }</lang>

Run without arguments on a 64-bit system:

Output:

[…]
4471000
4472000
4473000
runtime: goroutine stack exceeds 1000000000-byte limit
fatal error: stack overflow

runtime stack:
runtime.throw(0x5413ae)
	/usr/local/go/src/pkg/runtime/panic.c:520 +0x69
runtime.newstack()
	/usr/local/go/src/pkg/runtime/stack.c:770 +0x486
runtime.morestack()
	/usr/local/go/src/pkg/runtime/asm_amd64.s:228 +0x61

goroutine 16 [stack growth]:
main.r(0x444442)
	[…]/rosetta/stack_size/stack.go:9 fp=0xc2680380c8 sp=0xc2680380c0
main.r(0x444441)
	[…]/rosetta/stack_size/stack.go:13 +0xc5 fp=0xc268038140 sp=0xc2680380c8
main.r(0x444440)
[…]
...additional frames elided...
created by _rt0_go
	/usr/local/go/src/pkg/runtime/asm_amd64.s:97 +0x120

goroutine 19 [finalizer wait]:
runtime.park(0x412a20, 0x542ce8, 0x5420a9)
	/usr/local/go/src/pkg/runtime/proc.c:1369 +0x89
runtime.parkunlock(0x542ce8, 0x5420a9)
	/usr/local/go/src/pkg/runtime/proc.c:1385 +0x3b
runfinq()
	/usr/local/go/src/pkg/runtime/mgc0.c:2644 +0xcf
runtime.goexit()
	/usr/local/go/src/pkg/runtime/proc.c:1445
exit status 2

Run with "-stack 262144000" (to simulate the documented 250 MB limit on a 32-bit system):

Output:

[…]
1117000
1118000
runtime: goroutine stack exceeds 262144000-byte limit
fatal error: stack overflow
[…]

On a 32-bit system an int is 32 bits so the maximum value to debug.SetMaxStack is 2147483647 (nearly 2 GB). On a 64-bit system an int is usually 64 bits so the maximum value will much larger, more than the available memory. Thus setting the maximum value will either exhaust all the system memory and swap (as with pre-Go 1.2) or will result in a allocation failure and run-time panic (e.g. 32-bit systems often have more memory and swap in total than the memory accessible to a single user program due to the limits of a 32 bit address space shared with the kernel).

Note, unlike with some other systems, increasing or changing this value only changes the allocation limit. The stack still starts out very small and only grows as needed, there is no large stack pre-allocation even when using a very high limit. Also note that this is per-goroutine, each goroutine can recurse independently and approach the limit.

The above code built pre-Go 1.2 (without the then non-existent debug.SetMaxStack call) and run on a 1 GB RAM machine with 2.5 GB swap filled available RAM quickly, at a recursion depth of about 10M. It took a several minutes to exhaust swap before exiting with this trace: (as you see, at a depth of over 25M.)

[…]
25611000
25612000
25613000
25614000
throw: out of memory (FixAlloc)

runtime.throw+0x43 /home/sonia/go/src/pkg/runtime/runtime.c:102
	runtime.throw(0x80e80c8, 0x1)
runtime.FixAlloc_Alloc+0x76 /home/sonia/go/src/pkg/runtime/mfixalloc.c:43
	runtime.FixAlloc_Alloc(0x80eb558, 0x2f)
runtime.stackalloc+0xfb /home/sonia/go/src/pkg/runtime/malloc.c:326
	runtime.stackalloc(0x1000, 0x8048c44)
runtime.newstack+0x140 /home/sonia/go/src/pkg/runtime/proc.c:768
	runtime.newstack()
runtime.morestack+0x4f /home/sonia/go/src/pkg/runtime/386/asm.s:220
	runtime.morestack()
----- morestack called from goroutine 1 -----
main.r+0x1a /home/sonia/t.go:9
	main.r(0x186d801, 0x0)
main.r+0x95 /home/sonia/t.go:13
	main.r(0x186d800, 0x0)
main.r+0x95 /home/sonia/t.go:13
	main.r(0x186d7ff, 0x0)
main.r+0x95 /home/sonia/t.go:13
	main.r(0x186d7fe, 0x0)
main.r+0x95 /home/sonia/t.go:13
	main.r(0x186d7fd, 0x0)

... (more of the same stack trace omitted)


----- goroutine created by -----
_rt0_386+0xc1 /home/sonia/go/src/pkg/runtime/386/asm.s:80

goroutine 1 [2]:
runtime.entersyscall+0x6f /home/sonia/go/src/pkg/runtime/proc.c:639
	runtime.entersyscall()
syscall.Syscall+0x53 /home/sonia/go/src/pkg/syscall/asm_linux_386.s:33
	syscall.Syscall()
syscall.Write+0x5c /home/sonia/go/src/pkg/syscall/zsyscall_linux_386.go:734
	syscall.Write(0x1, 0x977e4f18, 0x9, 0x40, 0x9, ...)
os.*File·write+0x39 /home/sonia/go/src/pkg/os/file_unix.go:115
	os.*File·write(0x0, 0x0, 0x9, 0x40, 0x9, ...)
os.*File·Write+0x98 /home/sonia/go/src/pkg/os/file.go:141
	os.*File·Write(0xbffe1980, 0x8, 0x9, 0x8048cbf, 0x186d6b4, ...)
----- goroutine created by -----
_rt0_386+0xc1 /home/sonia/go/src/pkg/runtime/386/asm.s:80

Gri

In Gri 2.12.23 the total depth of command calls is limited to an internal array size cmd_being_done_LEN which is 100. There's no protection or error check against exceeding this, so the following code segfaults shortly after 100,

<lang Gri>`Recurse' {

   show .depth.
   .depth. = {rpn .depth. 1 +}
   Recurse

} .depth. = 1 Recurse</lang>

Groovy

Translation of: Java

Solution: <lang groovy>def recurse; recurse = {

   try {
       recurse (it + 1)
   } catch (StackOverflowError e) {
       return it
   }

}

recurse(0)</lang>

Output:

387

Icon and Unicon

<lang Icon>procedure main() envar := "MSTKSIZE" write(&errout,"Program to test recursion depth - dependant on the environment variable ",envar," = ",\getenv(envar)|&null) deepdive() end

procedure deepdive() static d initial d := 0 write( d +:= 1) deepdive() end</lang> Note: The stack size environment variable defaults to about 50000 words. This terminates after approximately 3500 recursions (Windows). The interpreter should terminate with a 301 error, but currently this does not work.

Inform 7

<lang inform7>Home is a room.

When play begins: recurse 0.

To recurse (N - number): say "[N]."; recurse N + 1.</lang>

Using the interpreters built into Windows build 6F95, a stack overflow occurs after 6529 recursions on the Z-machine or 2030 recursions on Glulx.

J

Testing stack depth can be risky because the OS may shut down J in the limiting case. To portably test stack depth it's best to run jconsole and display a count as each stack frame is entered.

Note also that task assumes that all stack frames must be the same size, which is probably not the case.

 <lang J>(recur=: verb def 'recur smoutput N=:N+1')N=:0</lang>

This above gives a stack depth of 9998 on one machine.

Note also, that ^: can be used for induction, and does not have stack size limits, though it does require that the function involved is a mathematical function of known variables -- and this is not always the case (for example, Markov processes typically use non-functions or "functions of unknown variables").

Java

<lang Java> public class RecursionTest {

   private static void recurse(int i) {
       try {

recurse(i+1); } catch (StackOverflowError e) { System.out.print("Recursion depth on this system is " + i + "."); }

   }

   public static void main(String[] args) {
       recurse(0);
   }

} </lang>

Output:

Recursion depth on this system is 10473.

Settings:

Default size of stack is 320 kB.. To extend the memory allocated for stack can be used switch -Xss with the memmory limits.
For example: java -cp . -Xss1m RecursionTest (set the stack size to 1 MB).

JavaScript

<lang javascript> function recurse(depth) {

try
{
 return recurse(depth + 1);
}
catch(ex)
{
 return depth;
}

}

var maxRecursion = recurse(1); document.write("Recursion depth on this system is " + maxRecursion);</lang>

Output:

(Chrome)

Recursion depth on this system is 10473.

Output:

(Firefox 1.6.13)

Recursion depth on this system is 3000.

Output:

(IE6)

Recursion depth on this system is 2552.

jq

Recent versions of jq (after July 1, 2014, i.e. after version 1.4) include some "Tail Call Optimizations" (TCO). As a result, tail-recursive functions of arity 0 will run indefinitely in these later versions. The TCO optimizations also speed up other recursive functions.

Accordingly we present two test functions and show the results using jq 1.4 and using a version of jq with TCO optimizations.

Arity-0 Function <lang jq>def zero_arity:

 if (. % 1000000 == 0) then . else empty end, ((.+1)| zero_arity);

1|zero_arity</lang> Arity-1 Function <lang jq>def with_arity(n):

 if (n % 1000 == 0) then n else empty end, with_arity(n+1);

with_arity(1)</lang>

Results using jq 1.4 <lang sh>

1. Arity 0 - without TCO:

... 23000000 # 1.62 GB 25000000

• • • error: can't allocate region

user 0m54.558s sys 0m2.773s

1. Arity 1 - without TCO:

... 77000 # 23.4 MB ... 85000 # 23.7 MB 90000 # 25.4 MB 237000 # 47.4 MB (5h:08) 242000 # 50.0 MB (5h:14m)

1. [job cancelled manually after over 5 hours]

</lang> Results using jq with TCO

The arity-0 test was stopped after the recursive function had been called 100,000,000 (10^8) times. The memory required did not grow beyond 360 KB (sic). <lang sh> $ time jq -n -f Find_limit_of_recursions.jq ... 10000000 # 360 KB ... 100000000 # 360 KB

1. [job cancelled to get a timing]

user 2m0.534s sys 0m0.329s </lang>

The arity-1 test process was terminated simply because it had become too slow; at that point it had only consumed about 74.6K MB. <lang sh> ... 56000 # 9.9MB ... 95000 # 14.8 MB 98000 # 15.2 MB 99000 # 15.4 MB 100000 # 15.5 MB 127000 # 37.4 MB 142000 # 37.4 MB 254000 # 74.6 MB 287000 # 74.6 MB 406000 # 74.6 MB (8h:50m) 412000 # 74.6 MB (9h:05m)

1. [job cancelled manually after over 9 hours]</lang>

Discussion

Even without the TCO optimizations, the effective limits for recursive jq functions are relatively high:

1. the arity-0 function presented here proceeded normally beyond 25,000,000 (25 million) iterations;
2. the arity-1 function is more effectively constrained by performance than by memory: the test process was manually terminated after 242,000 iterations.

With the TCO optimizations:

1. the arity-0 function is not only unconstrained by memory but is fast and remains fast; it requires only 360 KB (that is KB).
2. the arity-1 function is, once again, more effectively constrained by performance than by memory: the test process was terminated after 412,000 iterations simply because it had become too slow; at that point it had only consumed about 74.6 MB.

Julia

This solution includes two versions of the function for probing recursion depth. The Clean version is perhaps more idiomatic. However the Dirty version, by using a global variable for the depth counter and minimizing the complexity of the called code reaches a significantly greater depth of recursion.

Clean <lang Julia> function divedivedive(d::Int)

   try
       divedivedive(d+1)
   catch
       return d
   end

end </lang> Dirty <lang Julia> function divedivedive()

   global depth
   depth += 1
   divedivedive()

end </lang> Main <lang Julia> depth = divedivedive(0) println("A clean dive reaches a depth of ", depth, ".")

depth = 0 try

   divedivedive()

end println("A dirty dive reaches a depth of ", depth, ".") </lang>

Output:

A clean dive reaches a depth of 21807.
A dirty dive reaches a depth of 174454.

Liberty BASIC

Checks for the case of gosub & for proper subroutine. <lang lb> 'subroutine recursion limit- end up on 475000

call test 1

sub test n

   if n mod 1000 = 0 then locate 1,1: print n
   call test n+1

end sub </lang>

<lang lb> 'gosub recursion limit- end up on 5767000 [test]

   n = n+1
   if n mod 1000 = 0 then locate 1,1: print n

gosub [test] </lang>

Logo

Like Scheme, Logo guarantees tail call elimination, so recursion is effectively unbounded. You can catch a user interrupt though to see how deep you could go.

<lang logo>make "depth 0

to recurse

 make "depth :depth + 1
 recurse

end

catch "ERROR [recurse]

 ; hit control-C after waiting a while

print error  ; 16 Stopping... recurse [make "depth :depth + 1] (print [Depth reached:] :depth)  ; some arbitrarily large number</lang>

LSL

I ran this twice and got 1891 and 1890; probably varies with the number Avatars on a Sim and other variables I can't control.

Originally I had it without the OwnerSay in the recursive function. Generally, if LSL has a Runtime Error it just shouts on the DEBUG_CHANNEL and skips to the next statement (which would have returned to the next statement in state_entry() said the highest number it had achieved) but, it just shouted "Script run-time error. Object: Stack-Heap Collision" on debug and quit running.

To test it yourself; rez a box on the ground, and add the following as a New Script. <lang LSL>integer iLimit_of_Recursion = 0; Find_Limit_of_Recursion(integer x) { llOwnerSay("x="+(string)x); iLimit_of_Recursion = x; Find_Limit_of_Recursion(x+1); } default { state_entry() { Find_Limit_of_Recursion(0); llOwnerSay("iLimit_of_Recursion="+(string)iLimit_of_Recursion); } } </lang>

Output:

[2012/07/07 18:40]  Object: x=0
[2012/07/07 18:40]  Object: x=1
[2012/07/07 18:40]  Object: x=2
   ...   ...   ...   ...   ...
[2012/07/07 18:41]  Object: x=1888
[2012/07/07 18:41]  Object: x=1889
[2012/07/07 18:41]  Object: x=1890
[2012/07/07 18:41]  Object: Object [script:New Script] Script run-time error
[2012/07/07 18:41]  Object: Stack-Heap Collision

Lua

<lang Lua> counter = 0

function test()

  print("Depth:", counter)
  counter = counter + 1
  test()

end

test() </lang>

Mathematica

The variable $RecursionLimit can be read for its current value or set to different values. eg <lang>$RecursionLimit=10^6</lang> Would set the recursion limit to one million.

MATLAB / Octave

The recursion limit can be 'get' and 'set' using the "get" and "set" keywords.

Sample Usage: <lang MATLAB>>> get(0,'RecursionLimit')

ans =

  500

>> set(0,'RecursionLimit',2500) >> get(0,'RecursionLimit')

ans =

       2500</lang>

Maxima

<lang maxima>f(p) := f(n: p + 1)$ f(0); Maxima encountered a Lisp error:

Error in PROGN [or a callee]: Bind stack overflow.

Automatically continuing. To enable the Lisp debugger set *debugger-hook* to nil.

n; 406</lang>

МК-61/52

<lang>П2 ПП 05 ИП1 С/П ИП0 ИП2 - x<0 20 ИП0 1 + П0 ПП 05 ИП1 1 + П1 В/О</lang>

Modula-2

<lang modula2>MODULE recur;

IMPORT InOut;

PROCEDURE recursion (a : CARDINAL);

BEGIN

 InOut.Write ('.');    (*  just count the dots....     *)
 recursion (a + 1)

END recursion;

BEGIN

 recursion (0)

END recur.</lang> Producing this: <lang Modula-2> jan@Beryllium:~/modula/rosetta$ recur >testfile Segmentation fault jan@Beryllium:~/modula/rosetta$ ls -l -rwxr-xr-x 1 jan users 20032 2011-05-20 00:26 recur* -rw-r--r-- 1 jan users 194 2011-05-20 00:26 recur.mod -rw-r--r-- 1 jan users 523264 2011-05-20 00:26 testfile jan@Beryllium:~/modula/rosetta$ wc testfile

    0      1 523264 testfile</lang>

So the recursion depth is just over half a million.

MUMPS

<lang MUMPS>RECURSE

IF $DATA(DEPTH)=1 SET DEPTH=1+DEPTH
IF $DATA(DEPTH)=0 SET DEPTH=1
WRITE !,DEPTH_" levels down"
DO RECURSE
QUIT</lang>

End of the run ...

1918 levels down
1919 levels down
1920 levels down
 DO RECURSE
 ^
<FRAMESTACK>RECURSE+4^ROSETTA
USER 72d0>

NetRexx

Like Java, NetRexx memory allocation is managed by the JVM under which it is run. The following sample presents runtime memory allocations then begins the recursion run. <lang NetRexx>/* NetRexx */ options replace format comments java crossref symbols binary

import java.lang.management.

memoryInfo() digDeeper(0)

/**

* Just keep digging
* @param level depth gauge
*/

method digDeeper(level = int) private static binary

 do
   digDeeper(level + 1)
 catch ex = Error
   System.out.println('Recursion got' level 'levels deep on this system.')
   System.out.println('Recursion stopped by' ex.getClass.getName())
 end
 return

/**

* Display some memory usage from the JVM
* @see ManagementFactory
* @see MemoryMXBean
* @see MemoryUsage
*/

method memoryInfo() private static

 mxBean = ManagementFactory.getMemoryMXBean()   -- get the MemoryMXBean
 hmMemoryUsage = mxBean.getHeapMemoryUsage()    -- get the heap MemoryUsage object
 nmMemoryUsage = mxBean.getNonHeapMemoryUsage() -- get the non-heap MemoryUsage object
 say 'JVM Memory Information:'
 say '      Heap:' hmMemoryUsage.toString()
 say '  Non-Heap:' nmMemoryUsage.toString()
 say '-'.left(120, '-')
 say
 return

</lang>

Output:

JVM Memory Information: 
      Heap: init = 0(0K) used = 2096040(2046K) committed = 85000192(83008K) max = 129957888(126912K) 
  Non-Heap: init = 24317952(23748K) used = 5375328(5249K) committed = 24317952(23748K) max = 136314880(133120K) 
------------------------------------------------------------------------------------------------------------------------ 
 
Recursion got 9673 levels deep on this system. 
Recursion stopped by java.lang.StackOverflowError 

Nim

<lang nim>proc recurse(i): int =

 echo i
 recurse(i+1)

echo recurse(0)</lang> Compiled without optimizations it would stop after 87317 recursions. With optimizations on recurse is translated into a tail-recursive function, without any recursion limit. Instead of waiting for the 87317 recursions you compile with debuginfo activated and check with gdb:

nim c --debuginfo --lineDir:on recursionlimit.nim

OCaml

When the recursion is a "tail-recursion" there is no limit. Which is important because being a functional programming language, OCaml uses recursion to make loops.

If the recursion is not a tail one, the execution is stopped with the message "Stack overflow": <lang ocaml># let last = ref 0 ;; val last : int ref = {contents = 0}

1. let rec f i =

   last := i;
   i + (f (i+1))
 ;;

val f : int -> int = <fun>

1. f 0 ;;

stack overflow during evaluation (looping recursion?).

1. !last ;;

- : int = 262067</lang>

here we see that the function call stack size is 262067.

<lang ocaml>(* One can build a function from the idea above, catching the exception *)

let rec_limit () =

 let last = ref 0 in
 let rec f i =
   last := i;
   1 + f (i + 1)
 in
 try (f 0)
 with Stack_overflow -> !last

rec_limit ();; 262064

(* Since with have eaten some stack with this function, the result is slightly lower. But now it may be used inside any function to get the available stack space *)</lang>

Oforth

Limit found is 173510 on Windows system. Should be more on Linux system.

<lang Oforth>func: limit { 1 + dup println limit }

0 limit</lang>

ooRexx

Using ooRexx for the program shown under Rexx:

 rexx pgm 1>x1 2>x2

 puts the numbers in x1 and the error messages in x2

...
2785
2786
8 *-*      call self
....
     8 *-*   call self
     3 *-* call self
Error 11 running C:\work.ooRexx\wc\main.4.1.1.release\Win32Rel\StreamClasses.orx line 366:  Control stack full
Error 11.1:  Insufficient control stack space; cannot continue execution

Oz

Translation of: Scheme

Oz supports an unbounded number of tail calls. So the following code can run forever with constant memory use (although the space used to represent Number will slowly increase): <lang oz>declare

 proc {Recurse Number}
    {Show Number}
    {Recurse Number+1}
 end

in

 {Recurse 1}</lang>

With non-tail recursive functions, the number of recursions is only limited by the available memory.

PARI/GP

As per "Recursive functions" in the Pari/GP users's manual. <lang parigp>dive(n) = dive(n+1) dive(0)</lang>

The limit is the underlying C language stack. Deep recursion is detected before the stack is completely exhausted (by checking RLIMIT_STACK) so a gp level error is thrown instead of a segfault.

Pascal

See Delphi

Perl

Maximum recursion depth is memory dependent.

<lang perl>my $x = 0; recurse($x);

sub recurse ($x) {

  print ++$x,"\n";
  recurse($x);

}</lang>

1
2
...
...
10702178
10702179
Out of memory!

Perl 6

Maximum recursion in Perl 6 is implementation dependent and subject to change as development proceeds.

<lang perl6>my $x = 0; recurse;

sub recurse () {

  say ++$x;
  recurse;

}</lang> Using Rakudo 2011.01 this yields:

1
2
...
...
971
972
maximum recursion depth exceeded

This is because the Parrot VM currently imposes a limit of 1000. On the other hand, the niecza implementation has no limit, subject to availability of virtual memory. In any case, future Perl 6 is likely to require tail call elimination in the absence of some declaration to the contrary.

PHP

<lang PHP><?php function a() {

   static $i = 0;
   print ++$i . "\n";
   a();

} a();</lang>

Output:

 1
 2
 3
 [...]
 597354
 597355
 597356
 597357
 597358
 
 Fatal error: Allowed memory size of 134217728 bytes exhausted (tried to allocate 261904 bytes) in [script-location.php] on line 5

PicoLisp

The 64-bit and the 32-bit version behave slightly different. While the 32-bit version imposes no limit on its own, and relies on the 'ulimit' setting of the caller, the 64-bit version segments the available stack (likewise depending on 'ulimit') and allows each (co)routine a maximal stack size as configured by 'stack'.

32-bit version

$ ulimit -s
8192
$ pil +
: (let N 0 (recur (N) (recurse (msg (inc N)))))
...
730395
730396
730397
Segmentation fault

64-bit version

$ ulimit -s
unlimited
$ pil +
: (stack)  # The default stack segment size is 64 MB
-> 64

: (co 'a (yield 7))  # Start a dummy coroutine
-> 7

: (let N 0 (recur (N) (recurse (println (inc N)))))
...
2475
2476
2477
Stack overflow
?

PL/I

<lang PL/I> recurs: proc options (main) reorder; dcl sysprint file; dcl mod builtin;

dcl ri fixed bin(31) init (0);

recursive: proc recursive;

 ri += 1;
 if mod(ri, 1024) = 1 then
   put data(ri);

 call recursive();

end recursive;

call recursive(); end recurs; </lang>

Result (abbreviated):

...
RI=       4894721;
RI=       4895745;
RI=       4896769;
RI=       4897793;
RI=       4898817;

At this stage the program, running on z/OS with a REGION=0M on the EXEC statement (i.e. grab as much storage as you like), ABENDs with a USER COMPLETION CODE=4088 REASON CODE=000003EC

Obviously, if the procedure recursive would have contained local variables, the depth of recursion would be reached much earlier...

PowerShell

Both of these examples will throw an exception when the recursion depth is exceeded, however, the exception cannot be trapped in the script. The exception thrown on a Windows 2008 x64 system is

<lang>The script failed due to call depth overflow. The call depth reached 1001 and the maximum is 1000. System.Management.Automation</lang>

PowerShell v1.0 Example

<lang PowerShell>function Check-Recursion{

  trap [Exception] {
     Write-Host $_.Exception.Message
  }
  Check-Recursion

}

trap [Exception] {

 Write-Host $_.Exception.Message

} Check-Recursion</lang>

PowerShell V2.0+ Example

<lang PowerShell>function Check-Recursion{

   Check-Recursion

}

try { Check-Recursion } catch { Write-Host $_.Exception.Message }</lang>

PureBasic

The recursion limit is primarily determined by the stack size. The stack size can be changed when compiling a program by specifying the new size using '/stack:NewSize' in the linker file.

Procedural

In addition to the stack size the recursion limit for procedures is further limited by the procedure's parameters and local variables which are also stored on the same stack. <lang PureBasic>Procedure Recur(n)

 PrintN(str(n))
 Recur(n+1)

EndProcedure

Recur(1)</lang>

Stack overflow after 86317 recursions on x86 Vista.

Classic

<lang PureBasic>rec:

 PrintN(str(n))
 n+1
 Gosub rec
 Return</lang>
Stack overflow after 258931 recursions on x86 Vista.

Python

<lang python>import sys print sys.getrecursionlimit()</lang> To set it: <lang python>import sys sys.setrecursionlimit(12345)</lang>

R

R's recursion is counted by the number of expressions to be evaluated, rather than the number of function calls. <lang r>#Get the limit options("expressions")

1. Set it

options(expressions = 10000)

1. Test it

recurse <- function(x) {

 print(x)
 recurse(x+1)

} recurse(0)</lang>

Racket

<lang Racket>#lang racket (define (recursion-limit)

 (with-handlers ((exn? (lambda (x) 0)))
   (add1 (recursion-limit))))</lang>

This should theoretically return the recursion limit, as the function can't be tail-optimized and there's an exception handler to return a number when an error is encountered. For this to work one has to give the Racket VM the maximum possible memory limit and wait.

Retro

When run, this will display the address stack depth until it reaches the max depth. Once the address stack is full, Retro will crash.

<lang Retro>: try -6 5 out wait 5 in putn cr try ;</lang>

REXX

recursive procedure

On (IBM's) VM/CMS, the limit of recursion was built-into CMS to stop run-away EXEC programs (this
included EXEC[0], EXEC2, and REXX) being called recursively, it was either 200 or 250 as I recall.
This limit was maybe changed later to allow the user to specify the limit. My memory is really fuzzy about these details. <lang rexx>/*REXX pgm finds the recursion limit: a subroutine that calls itself. */ parse version x; say x; say n=0 call SELF 1 exit /*this statement will never be executed.*/ /*───────────────────────────SELF procedure─────────────────────────────*/ self: procedure expose n n=n+1 say n call self</lang>

Output:

using Regina 3.6 under Windows/XP Pro

REXX-Regina_3.6(MT) 5.00 31 Dec 2011
 .
 .
 .
164423
164424
164425
164426
164427
164428
164429
164430
164431
System resources exhausted

[Your mileage will vary.]

Note that the above recursion limit will be less and it's dependant upon how much virtual memory the program itself uses,
this would include REXX variables and their values, and the program source (as it's kept in virtual memory also),
and the size of the REXX.EXE and REXX.DLL programs, and any other programs executing in the Windows DOS (including
either the CMD.EXE or COMMAND.COM) shell).

Output:

using Personal REXX under Windows/XP Pro

The recursion level wasn't captured, but the last number shown was 240.

REXX/Personal 4.00 21 Mar 1992
 .
 .
 .
    10 +++ call self
    10 +++ call self
    10 +++ call self
    10 +++ call self
    10 +++ call self
    10 +++ call self
    10 +++ call self
    10 +++ call self
    10 +++ call self
     4 +++ call SELF 1
Error 5 on line 10 of D:\SELF.REX: Machine resources exhausted

Output:

using R4 REXX under Windows/XP Pro

REXX-r4 4.00 29 Apr 2012
 .
 .
 .
499
500
501
502
503
504
505
506
507
An unexpected error occurred

Output:

using ROO REXX under Windows/XP Pro

REXX-roo 4.00 28 Jan 2007

 .
 .
 .
374
375
376
377
378
379
380
381
382
An unexpected error occurred

Output:

using ooRexx under Windows/7

See section ooRexx

recursive subroutine

All REXXes were executed under Windows/XP Pro. <lang rexx>/*REXX pgm finds the recursion limit: a subroutine that calls itself. */ parse version x; say x; say n=0 call SELF 2 exit /*this statement will never be executed.*/

/*───────────────────────────SELF subroutine────────────────────────────*/ self: n=n+1 say n call self</lang>

Output:

(paraphrased and edited)

For Regina 3.7,     it was 828,441.
For Regina 3.6,     it was 828,441.
For Regina 3.5,     it was 828,441.
For Regina 3.4,     it was 828,441.
For Regina 3.3,     it was   3,641.
For Personal REXX,  it was     240  (the same).
For R4,             it was     507  (the same).
For ROO,            it was     382  (the same).
For ooRexx          it is    2,480  (the same).

Ruby

<lang ruby>def recurse x

 puts x
 recurse(x+1)

end

recurse(0)</lang>

Output:

Produces a SystemStackError

.
.
.
6074
recurse.rb:3:in `recurse': stack level too deep (SystemStackError)
	from recurse.rb:3:in `recurse'
	from recurse.rb:6

when tracking Stack overflow exceptions ; returns 8732 on my computer :

<lang ruby>def recurse n

 recurse(n+1)

rescue SystemStackError

 n

end

puts recurse(0)</lang>

Run BASIC

<lang runbasic>a = recurTest(1)

function recurTest(n) if n mod 100000 then cls:print n if n > 327000 then [ext]

  n = recurTest(n+1)

[ext] end function</lang>

327000

Rust

<lang rust>fn recurse(n: i32) {

   println!("depth: {}", n);
   recurse(n + 1)

}

fn main() {

   recurse(0);

}</lang>

Output:

...
depth: 18433
depth: 18434
depth: 18435

thread '<main>' has overflowed its stack
An unknown error occurred

To learn more, run the command again with --verbose.

Sather

<lang sather>class MAIN is

 attr r:INT;
 recurse is
   r := r + 1;
   #OUT + r + "\n";
   recurse;
 end;
 main is
   r := 0;
   recurse;
 end;

end;</lang>

Segmentation fault is reached when r is 130560.

Scala

<lang scala>def recurseTest(i:Int):Unit={

  try{
     recurseTest(i+1)
  } catch { case e:java.lang.StackOverflowError => 
     println("Recursion depth on this system is " + i + ".")
  }

} recurseTest(0)</lang>

Output:

depending on the current stack size

Recursion depth on this system is 4869.

If your function is tail-recursive the compiler transforms it into a loop. <lang scala>def recurseTailRec(i:Int):Unit={

  if(i%100000==0) println("Recursion depth is " + i + ".")
  recurseTailRec(i+1)

}</lang>

Scheme

<lang scheme>(define (recurse number)

 (begin (display number) (newline) (recurse (+ number 1))))

(recurse 1)</lang> Implementations of Scheme are required to support an unbounded number of tail calls. Furthermore, implementations are encouraged, but not required, to support exact integers of practically unlimited size.

Sidef

Maximum recursion depth is memory dependent. <lang ruby>func recurse(n) {

  say n;
  recurse(n+1);

}

recurse(0);</lang>

Output:

0
1
2
...
...
357077
357078
357079

Smalltalk

In the Squeak dialect of Smalltalk:

<lang smalltalk> Object subclass: #RecursionTest instanceVariableNames: classVariableNames: poolDictionaries: category: 'RosettaCode' </lang>

Add the following method:

<lang smalltalk> counter: aNumber ^self counter: aNumber + 1 </lang>

Call from the Workspace:

<lang smalltalk> r := RecursionTest new. r counter: 1. </lang>

After some time the following error pops up:

   Warning! Squeak is almost out of memory!
   Low space detection is now disabled. It will be restored when you close or proceed from this error notifier. Don't panic, but do proceed with caution.
   Here are some suggestions:
   If you suspect an infinite recursion (the same methods calling each other again and again), then close this debugger, and fix the problem.
   If you want this computation to finish, then make more space available (read on) and choose "proceed" in this debugger. Here are some ways to make more space available...
   * Close any windows that are not needed.
   * Get rid of some large objects (e.g., images).
   * Leave this window on the screen, choose "save as..." from the screen menu, quit, restart the Squeak VM with a larger memory allocation, then restart the image you just saved, and choose "proceed" in this window.
   If you want to investigate further, choose "debug" in this window.  Do not use the debugger "fullStack" command unless you are certain that the stack is not very deep. (Trying to show the full stack will definitely use up all remaining memory if the low-space problem is caused by an infinite recursion!).

Tcl

<lang tcl>proc recur i {

   puts "This is depth [incr i]"
   catch {recur $i}; # Trap error from going too deep

} recur 0</lang> The tail of the execution trace looks like this:

This is depth 995
This is depth 996
This is depth 997
This is depth 998
This is depth 999

Note that the maximum recursion depth is a tunable parameter, as is shown in this program: <lang tcl># Increase the maximum depth interp recursionlimit {} 1000000 proc recur i {

   if {[catch {recur [incr i]}]} {
       # If we failed to recurse, print how far we got

puts "Got to depth $i"

   }

} recur 0</lang> For Tcl 8.5 on this platform, this prints:

Got to depth 6610

At which point it has exhausted the C stack, a trapped error. Tcl 8.6 uses a stackless execution engine, and can go very deep if required:

Got to depth 999999

TSE SAL

<lang TSE SAL> // library: program: run: recursion: limit <description>will stop at 3616</description> <version>1.0.0.0.3</version> <version control></version control> (filenamemacro=runprrli.s) [kn, ri, su, 25-12-2011 23:12:02] PROC PROCProgramRunRecursionLimit( INTEGER I )

Message( I )
PROCProgramRunRecursionLimit( I + 1 )

END

PROC Main()

PROCProgramRunRecursionLimit( 1 )

END </lang>

TXR

<lang txr>@(do

  (set-sig-handler sig-segv
    (lambda (signal async-p) (throw 'out)))

  (defvar *count* 0)

  (defun recurse ()
    (inc *count*)
    (recurse))

  (catch (recurse)
    (out () (put-line `caught segfault!\nreached depth: @{*count*}`))))</lang>

Output:

$ txr limit-of-recursion.txr 
caught segfault!
reached depth: 2941

UNIX Shell

Works with: Bourne Again SHell

<lang bash>recurse() {

 # since the example runs slowly, the following
 # if-elif avoid unuseful output; the elif was
 # added after a first run ended with a segmentation
 # fault after printing "10000"
 if $(($1 % 5000)) -eq 0 ; then 
     echo $1;
 elif $1 -gt 10000 ; then
     echo $1
 fi
 recurse $(($1 + 1))

}

recurse 0</lang>

The Bash reference manual says No limit is placed on the number of recursive calls, nonetheless a segmentation fault occurs at 13777 (Bash v3.2.19 on 32bit GNU/Linux)

VBScript

Haven't figured out how to see the depth. And this depth is that of calling the O/S rather than calling within.

<lang vb>'mung.vbs option explicit

dim c if wscript.arguments.count = 1 then c = wscript.arguments(0) c = c + 1 else c = 0 end if wscript.echo "[Depth",c & "] Mung until no good." CreateObject("WScript.Shell").Run "cscript Mung.vbs " & c, 1, true wscript.echo "[Depth",c & "] no good."</lang>

Okay, the internal limits version.

<lang vb>'mung.vbs option explicit

sub mung(c) dim n n=c+1 wscript.echo "[Level",n & "] Mung until no good" on error resume next mung n on error goto 0 wscript.echo "[Level",n & "] no good" end sub

mung 0</lang>

Output:

(abbrev.)

[Level 1719] Mung until no good
[Level 1720] Mung until no good
[Level 1721] Mung until no good
[Level 1722] Mung until no good
[Level 1722] no good
[Level 1721] no good
[Level 1720] no good
[Level 1719] no good

x86 Assembly

Works with: nasm

<lang asm> global main

  section .text

main xor eax, eax call recurse ret

recurse add eax, 1 call recurse ret</lang>

I've used gdb and the command print $eax to know when the segmentation fault occurred. The result was 2094783.

zkl

<lang zkl>fcn{self.fcn()}()</lang>

Output:

Stack trace for VM#1 ():
   Cmd.__fcn#1_2 addr:3  args(0) reg(0) R
   <repeats 2096 times>
   Cmd.__constructor addr:3  args(0) reg(0) R
   startup.__constructor addr:2242  args(0) reg(1) ER
   startup.__constructor addr:2178  args(0) reg(22) 
Exception thrown: AssertionError(That is one big stack, infinite recursion?)