Find limit of recursion

From Rosetta Code
Jump to: navigation, search
Task
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.

Contents

[edit] ACL2

(defun recursion-limit (x)
(if (zp x)
0
(prog2$ (cw "~x0~%" x)
(1+ (recursion-limit (1+ x))))))
Output:
(trimmed):
87195
87196
87197
87198
87199
87200
87201

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

[edit] 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;

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

[edit] AutoHotkey

Recurse(0)
 
Recurse(x)
{
TrayTip, Number, %x%
Recurse(x+1)
}

Last visible number is 827.

[edit] AutoIt

;AutoIt Version: 3.2.10.0
$depth=0
recurse($depth)
Func recurse($depth)
ConsoleWrite($depth&@CRLF)
Return recurse($depth+1)
EndFunc

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

[edit] AWK

# syntax: GAWK -f FIND_LIMIT_OF_RECURSION.AWK
#
# version depth messages
# ------------------ ----- --------
# GAWK 3.1.4 2892 none
# XML GAWK 3.1.4 3026 none
# GAWK 4.0 >999999
# MAWK 1.3.3 4976 A stack overflow was encountered at
# address 0x7c91224e.
# TAWK-DOS AWK 5.0c 357 stack overflow
# TAWK-WIN AWKW 5.0c 2477 awk stack overflow
# NAWK 20100523 4351 Segmentation fault (core dumped)
#
BEGIN {
x()
print("done")
}
function x() {
print(++n)
if (n > 999999) { return }
x()
}

[edit] Batch File

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

@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

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.

@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

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

@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

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

@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

[edit] BASIC

[edit] 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:

 
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
 
Output:
(from a 48k Spectrum):

{{{

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

}}}

[edit] BBC BASIC

      PROCrecurse(1)
END
 
DEF PROCrecurse(depth%)
IF depth% MOD 100 = 0 PRINT TAB(0,0) depth%;
PROCrecurse(depth% + 1)
ENDPROC
Output:
from BBC BASIC for Windows with default value of HIMEM:
     37400
No room

[edit] Bracmat

rec=.out$!arg&rec$(!arg+1)

Observed recursion depths:

 Windows XP command prompt: 6588
 Linux: 18276

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

[edit] C

#include <stdio.h>
 
void recurse(unsigned int i)
{
printf("%d\n", i);
recurse(i+1); // 523756
}
 
int main()
{
recurse(0);
return 0;
}

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:

#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;
}

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!

[edit] COBOL

Works with: OpenCOBOL
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.
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

[edit] A more 'canonical' way of doing it

from Richard Plinston on comp.lang.cobol

Works with: OpenCOBOL
       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.
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

[edit] CoffeeScript

 
recurse = ( depth = 0 ) ->
try
recurse depth + 1
catch exception
depth
 
console.log "Recursion depth on this system is #{ do recurse }"
 
Output:
Example on Node.js:
    Recursion depth on this system is 9668

[edit] Common Lisp

 
(defun recurse () (recurse))
(trace recurse)
(recurse)
 
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.

[edit] C#

using System;
class RecursionLimit
{
static void Main(string[] args)
{
Recur(0);
}
 
private static void Recur(int i)
{
Console.WriteLine(i);
Recur(i + 1);
}
}

Through debugging, the highest I achieve is 14250.

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

[edit] Clojure

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

[edit] D

import std.c.stdio;
 
void recurse(in uint i=0) {
printf("%u ", i);
recurse(i + 1);
}
 
void main() {
recurse();
}

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.

[edit] Delphi

Works with: Delphi version 2010 (and probably all other versions)
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.
Output:
Recursion Level is 28781

[edit] Déjà Vu

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


rec-fun n:
!. n
rec-fun ++ n
 
rec-fun 0

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.

[edit] DWScript

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

var level : Integer;
 
procedure Recursive;
begin
Inc(level);
try
Recursive;
except
end;
end;
 
Recursive;
 
Println('Recursion Level is ' + IntToStr(level));

[edit] 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.

[edit] Emacs 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

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.

[edit] 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.

[edit] Forth

: munge ( n -- n' ) 1+ recurse ;
 
: test 0 ['] munge catch if ." Recursion limit at depth " . then ;
 
test \ Default gforth: Recursion limit at depth 3817

Or you can just ask the system:

s" return-stack-cells" environment? ( 0 | potential-depth-of-return-stack -1 )

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

: recur; [ last 2 cells + literal ] @ +bal postpone again ; immediate
 
: test dup if 1+ recur; then drop ." I gave up finding a limit!" ;
 
1 test

[edit] 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
Output:
(snipped):
208914
208915
208916
208917
208918
208919
208920
208921
208922
208923
Segmentation fault (core dumped)


[edit] F#

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

let rec recurse n = 
recurse (n+1)
 
recurse 0

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

let rec recurse n = 
printfn "%d" n
1 + recurse (n+1)
 
recurse 0 |> ignore

[edit] GAP

The limit is around 5000 :

f := function(n)
return f(n+1);
end;
 
# Now loop until an error occurs
f(0);
 
# Error message :
# Entering break read-eval-print loop ...
# you can 'quit;' to quit to outer loop, or
# you may 'return;' to continue
 
n;
# 4998
 
# quit "brk mode" and return to GAP
quit;

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 :

SetRecursionTrapInterval(100000);
# No limit (may crash GAP if recursion is not controlled) :
SetRecursionTrapInterval(0);

[edit] gnuplot

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

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.

[edit] 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."

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)
}

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

[edit] 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,

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

[edit] Groovy

Translation of: Java

Solution:

def recurse;
recurse = {
try {
recurse (it + 1)
} catch (StackOverflowError e) {
return it
}
}
 
recurse(0)
Output:
387

[edit] Icon and Unicon

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

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.

[edit] Inform 7

Home is a room.
 
When play begins: recurse 0.
 
To recurse (N - number):
say "[N].";
recurse N + 1.

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

[edit] 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.

(recur=: verb def 'recur smoutput N=:N+1')N=:0

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").

[edit] 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);
}
}
 
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).

[edit] 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);
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.

[edit] 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

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

Arity-1 Function

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

Results using jq 1.4

 
# Arity 0 - without TCO:
...
23000000 # 1.62 GB
25000000
*** error: can't allocate region
user 0m54.558s
sys 0m2.773s
 
# 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)
# [job cancelled manually after over 5 hours]
 

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).

 
$ time jq -n -f Find_limit_of_recursions.jq
...
10000000 # 360 KB
...
100000000 # 360 KB
# [job cancelled to get a timing]
user 2m0.534s
sys 0m0.329s
 

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.

 
...
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)
# [job cancelled manually after over 9 hours]

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.

[edit] Liberty BASIC

Checks for the case of gosub & for proper subroutine.

 
'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
 
 
'gosub recursion limit- end up on 5767000
[test]
n = n+1
if n mod 1000 = 0 then locate 1,1: print n
gosub [test]
 

[edit]

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.

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

[edit] Lua

 
counter = 0
 
function test()
print("Depth:", counter)
counter = counter + 1
test()
end
 
test()
 

[edit] 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.

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);
}
}
 
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

[edit] Mathematica

The variable $RecursionLimit can be read for its current value or set to different values. eg

$RecursionLimit=10^6

Would set the recursion limit to one million.

[edit] MATLAB / Octave

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

Sample Usage:

>> get(0,'RecursionLimit')
 
ans =
 
500
 
>> set(0,'RecursionLimit',2500)
>> get(0,'RecursionLimit')
 
ans =
 
2500

[edit] 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

[edit] МК-61/52

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

[edit] MUMPS

RECURSE
IF $DATA(DEPTH)=1 SET DEPTH=1+DEPTH
IF $DATA(DEPTH)=0 SET DEPTH=1
WRITE !,DEPTH_" levels down"
DO RECURSE
QUIT
End of the run ...
1918 levels down
1919 levels down
1920 levels down
 DO RECURSE
 ^
<FRAMESTACK>RECURSE+4^ROSETTA
USER 72d0>

[edit] Modula-2

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.

Producing this:

 
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

So the recursion depth is just over half a million.

[edit] 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.

/* 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
 
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 

[edit] Nimrod

proc recurse(i): int =
echo i
recurse(i+1)
echo recurse(0)

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:

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

[edit] 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":

# let last = ref 0 ;;
val last : int ref = {contents = 0}
# let rec f i =
last := i;
i + (f (i+1))
;;
val f : int -> int = <fun>
# f 0 ;;
stack overflow during evaluation (looping recursion?).
# !last ;;
- : int = 262067

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

(* 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 *)

[edit] 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

[edit] 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):

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

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

[edit] PARI/GP

From the documentation:

dive(n) = dive(n+1)
dive(0)

[edit] Pascal

See Delphi

[edit] Perl

Maximum recursion depth is memory dependent.

my $x = 0;
recurse($x);
 
sub recurse ($x) {
print ++$x,"\n";
recurse($x);
}


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

[edit] Perl 6

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

my $x = 0;
recurse;
 
sub recurse () {
say ++$x;
recurse;
}

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.

[edit] 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'.

[edit] 32-bit version

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

[edit] 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
?

[edit] PHP

<?php
function a() {
static $i = 0;
print ++$i . "\n";
a();
}
a();
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

[edit] 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;
 

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...

[edit] 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

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

PowerShell v1.0 Example

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

PowerShell V2.0+ Example

function Check-Recursion{
Check-Recursion
}
 
try { Check-Recursion }
catch { Write-Host $_.Exception.Message }

[edit] 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.

[edit] 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.

Procedure Recur(n)
PrintN(str(n))
Recur(n+1)
EndProcedure
 
Recur(1)
Stack overflow after 86317 recursions on x86 Vista.

[edit] Classic

rec:
PrintN(str(n))
n+1
Gosub rec
Return
Stack overflow after 258931 recursions on x86 Vista.

[edit] Python

import sys
print sys.getrecursionlimit()

To set it:

import sys
sys.setrecursionlimit(12345)

[edit] R

R's recursion is counted by the number of expressions to be evaluated, rather than the number of function calls.

#Get the limit
options("expressions")
 
#Set it
options(expressions = 10000)
 
#Test it
recurse <- function(x)
{
print(x)
recurse(x+1)
 
}
recurse(0)

[edit] Racket

#lang racket
(define (recursion-limit)
(with-handlers ((exn? (lambda (x) 0)))
(add1 (recursion-limit))))

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.

[edit] 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.

: try -6 5 out wait 5 in putn cr try ;

[edit] REXX

[edit] 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.

/*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
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

[edit] recursive subroutine

All REXXes were executed under Windows/XP Pro.

/*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
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).

[edit] Ruby

def recurse x
puts x
recurse(x+1)
end
 
recurse(0)
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 :

def recurse n
recurse(n+1)
rescue SystemStackError
n
end
 
puts recurse(0)

[edit] Run BASIC

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
327000

[edit] Rust

Rust 0.8:

fn recurse(n: int) {
println(fmt!("deep: %?", n));
recurse(n + 1);
}
 
fn main() {
recurse(0);
}

Run:

...
deep: 7892
Segmentation fault

Rust 0.9:

fn recurse(n: int) {
println!("deep: {:d}", n);
recurse(n + 1);
}
 
fn main() {
recurse(0);
}

Run:

...
deep: 12952


There are not many persons who know what wonders are opened to them in the
stories and visions of their youth; for when as children we listen and dream,
we think but half-formed thoughts, and when as men we try to remember, we are
dulled and prosaic with the poison of life. But some of us awake in the night
with strange phantasms of enchanted hills and gardens, of fountains that sing
in the sun, of golden cliffs overhanging murmuring seas, of plains that stretch
down to sleeping cities of bronze and stone, and of shadowy companies of heroes
that ride caparisoned white horses along the edges of thick forests; and then
we know that we have looked back through the ivory gates into that world of
wonder which was ours before we were wise and unhappy.

fatal runtime error:  assertion failed: !ptr.is_null()
Aborted

Run (rust 0.12):

...
deep: 8596

task '<main>' has overflowed its stack

[edit] Sather

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

Segmentation fault is reached when r is 130560.

[edit] 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)
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.

def recurseTailRec(i:Int):Unit={
if(i%100000==0) println("Recursion depth is " + i + ".")
recurseTailRec(i+1)
}

[edit] Scheme

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

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.

[edit] Smalltalk

In the Squeak dialect of Smalltalk:

 
Object subclass: #RecursionTest
instanceVariableNames: ''
classVariableNames: ''
poolDictionaries: ''
category: 'RosettaCode'
 

Add the following method:

 
counter: aNumber
^self counter: aNumber + 1
 

Call from the Workspace:

 
r := RecursionTest new.
r counter: 1.
 

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!).


[edit] Tcl

proc recur i {
puts "This is depth [incr i]"
catch {recur $i}; # Trap error from going too deep
}
recur 0

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:

# 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

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

[edit] 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
 

[edit] 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*}`))))
Output:
$ txr limit-of-recursion.txr 
caught segfault!
reached depth: 2941

[edit] UNIX Shell

Works with: Bourne Again SHell
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

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)

[edit] VBScript

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

'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."

Okay, the internal limits version.

'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
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

[edit] x86 Assembly

Works with: nasm
   global main
 
section .text
 
main
xor eax, eax
call recurse
ret
 
recurse
add eax, 1
call recurse
ret

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

[edit] zkl

fcn{self.fcn()}()
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?)
Personal tools
Namespaces

Variants
Actions
Community
Explore
Misc
Toolbox