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Your task is to create a program or function which randomly errors. Specifically, there must be a nonzero probability of erroring, but also a nonzero probability of running without error.

An error is anything that causes a program to terminate abnormally, such as dividing by zero or using an uninitialized variable. This also includes runtime errors, syntax errors and errors while compiling. Statements which manually throw an error, such as JavaScript's throw are allowed.

This program doesn't need to do anything if it doesn't error, other than exiting gracefully.

This is , so shortest answer per language wins.

Note: For this challenge, "randomly" follows the current consensus (so no using undefined behavior or uninitialized memory for randomness), and the program must be capable of producing both outputs if run or compiled multiple times (so a random number with the same seed is not valid)

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  • 16
    \$\begingroup\$ I think it's a boring challenge. A lot of the answers are doing basically the same thing and there's little room for golfing. Note that downvotes don't necessarily indicate a challenge is ill-specified -- there's close votes for that. \$\endgroup\$
    – xnor
    Apr 10 '20 at 15:43
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    \$\begingroup\$ Separately, I'm not so sure it's clear, since two of the upvoted answers rely on the state of memory when the program starts, despite the question in my earlier comment. \$\endgroup\$
    – xnor
    Apr 10 '20 at 15:45
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    \$\begingroup\$ I don't really think the current definition of random, although settled, is really clear. For example, the python answer and the brainfuck answer below uses some degree of undefined behavior. The c answer, although fits that definition, won't really crash because of the lack of srand. So, I don't really like challenges based on such loose definitions (and when such definition matters a lot). I would change it to upvote if the question is updated, say, to, the program may or may not crash after a rerun/recompile. \$\endgroup\$
    – newbie
    Apr 10 '20 at 15:47
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    \$\begingroup\$ It seems like so many users are posting 1/(random integer) here... \$\endgroup\$
    – Scratch---Cat
    Apr 11 '20 at 1:33
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    \$\begingroup\$ @xnor: agreed, this turns out to be somewhat interesting in some assembly languages, but most high level languages make unpredictable behaviour intentionally difficult other than integer division, or a few other things where 0 is special. \$\endgroup\$
    – Peter Cordes
    Apr 11 '20 at 4:40

57 Answers 57

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2
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Erlang (escript), 22 bytes

Just like the other answers...

f()->1/rand:uniform().

Try it online!

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2
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C (gcc), 28 bytes

f(i){srand(&i);i/=rand()&1;}

Crashes approximately every other time it's run. Abuses ASLR.

Try it online!

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5
  • \$\begingroup\$ Since we ultimately rely on &i for randomness, why not get rid of the rand stuff? f(i){i/=(int)&i<0;} works. \$\endgroup\$
    – Surculose Sputum
    Apr 11 '20 at 7:28
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    \$\begingroup\$ @SurculoseSputum Because then it would violate the rules for "randomly". \$\endgroup\$
    – S.S. Anne
    Apr 11 '20 at 13:46
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    \$\begingroup\$ I just reread the specification - apparently this challenge is much different than I thought... \$\endgroup\$
    – Surculose Sputum
    Apr 11 '20 at 16:04
  • \$\begingroup\$ @SurculoseSputum: I think that's a constant, but such an expression should exist. \$\endgroup\$
    – Joshua
    Apr 24 '20 at 23:00
  • \$\begingroup\$ -1 byte: f(i){i/=rand(srand(&i))&1;} \$\endgroup\$
    – jdt
    Aug 22 at 10:20
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Pyth, 3 bytes

-1 thanks to FryAmTheEggman.

lO2

Try it online!

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1
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    \$\begingroup\$ Using l is shorter than /1, though there still might be something better. \$\endgroup\$
    – FryAmTheEggman
    Apr 10 '20 at 15:22
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Keg, 3 bytes

~~/

Try it online!

This works by pushing two random numbers onto the stack ane dividing them. If the second number is 0, then it errors out.

There is indeed a chance for the second number to be 0, but it is highly unlikely that it will be, as the random range used is quite large. But highly unlikely != zero.

If I were to estimate the odds of erroring out, I'd say it'd be something like 1 in ((int​(​"​9​"​*​3234​) ** 2) + 21) * 2.

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4
  • \$\begingroup\$ Hey, look, it worked: TypeError: integer argument expected, got float \$\endgroup\$
    – S.S. Anne
    Apr 10 '20 at 23:48
  • \$\begingroup\$ With a higher chance of erroring... ~:/ \$\endgroup\$
    – user92069
    Apr 11 '20 at 2:58
  • \$\begingroup\$ @petStorm that's still the same chance. \$\endgroup\$
    – lyxal
    Apr 11 '20 at 3:03
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    \$\begingroup\$ Is "highly unlikely != non-zero" intended to be "highly unlikely != zero"? \$\endgroup\$
    – L. F.
    Apr 11 '20 at 3:39
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Bash, 1/32767 chance of success, 12 bytes

exit $RANDOM
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2
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PowerShell Windows, 10 bytes

ps(random)

get-random returns an int between 0 and 0x7FFFFFFF so it'll eventually match some PID and ps will get the process that are running on the local computer. Maybe...

PowerShell Linux, 11 bytes

gps(random)

TIO does not work because System.UnauthorizedAccessException. Please use your own Powershell. :)

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2
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Ruby, 9 bytes

1/rand(2)

Try it online!

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2
  • 2
    \$\begingroup\$ While very concise, I'm not too sure if this approach works: rand returns a Float, but 1 / 0.0 is Infinity, not an exception. (That's at least in Ruby 2.6, the behaviour might be different in older versions.) \$\endgroup\$
    – Aaron Christiansen
    Apr 11 '20 at 11:23
  • \$\begingroup\$ Damn you're right. \$\endgroup\$
    – Armand Fardeau
    Apr 11 '20 at 17:00
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Taxi, 209 bytes

Go to Heisenberg's:w 1 r 3 r 1 l.Pickup a passenger going to Magic Eight.Pickup a passenger going to Magic Eight.Go to Magic Eight:s 1 r 1 l 3 r.Pickup a passenger going to Cyclone.Go to Taxi Garage:e 2 l 2 r.

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Ungolfed and commented:

[ Heisenberg's produces random integers ]
Go to Heisenberg's:w 1 r 3 r 1 l.
[ Pickup two random integers ]
Pickup a passenger going to Magic Eight.
Pickup a passenger going to Magic Eight.
[ Magic Eight compares two numeric passengers ]
[ It returns the first passenger if it is less than the second and no one if it is not ]
Go to Magic Eight:s 1 r 1 l 3 r.
[ Try to pickup a passenger, which will error if there isn't anyone waiting ]
Pickup a passenger going to Cyclone.
[ Return to the garage to avoid getting the "you're fired" error ]
Go to Taxi Garage:e 2 l 2 r.

Try the ungolfed and commented version online!

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2
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Red - 24 bytes

x: now/time 1 / x/minute
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1
  • \$\begingroup\$ Welcome to the site, I've tried to correct the code formatting in your answer. You can revise it if I have something wrong. Let me know if you are having trouble with the formatting. \$\endgroup\$
    – Grain Ghost ♦
    Apr 15 '20 at 13:13
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Batch, 11 bytes

%random:1=%

Has a probability of 1/65536 Corrected by @Neil, 5/32768 (~0.0015%) to not generate an error.

Edit: Alternatively,

%random:~1%
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2
  • \$\begingroup\$ 5/32768, I think? (1, 11, 111, 1111 and 11111 won't error, and %random% only goes from 0 to 32767.) \$\endgroup\$
    – Neil
    Apr 13 '20 at 13:48
  • \$\begingroup\$ @Neil you're right, will fix that \$\endgroup\$
    – HaxAddict1337
    Apr 13 '20 at 14:44
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Ahead, 4 bytes

50/50 chance of a divide by zero error. X causes the head to go in a random cardinal direction, / is obviously division, and @ ends the program. Because of edge bouncing, the head traveling up or left is the same as it traveling down or right since it's in a corner.

X/
@

Try it online!

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2
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Ruby, 10 bytes

1.div rand

Can't do 1/rand, since 1/0.0 in Ruby is Infinity, not an error.

Try it online!

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2
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Vyxal, 3 bytes

∆Ṙl

Try it Online!

I think this has a \$\frac{1}{2^{19937}}\$ chance of not erroring, but I'm not sure - Vyxal is written in Python, so it's whatever chance python's PRNG has of returning 0.

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2
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Minim, 6 Bytes

[R*S].

Accesses memory at index RandomNumber times MemorySize.

R is between 0 and 1 inclusive, and S is the memory scope size in elements. If R = 1, then the index will be out of bounds, and throws an error.

GitHub Repository

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2
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JavaScript (Node.js), 17 15 14 bytes

new Date%2?a:0

Errors if current millisecond is odd

new Date returns current epoch time
%2 modulo 2, returns 1 if odd, 0 if even
if it's a truthy value (1), a will throw as it's an undefined variable

Try it online!

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6
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    \$\begingroup\$ new Date should work instead of Date.now(), saving 2 bytes. You could also replace the throw 1 with something like a (which causes a ReferenceError). \$\endgroup\$
    – Redwolf Programs
    Aug 20 at 13:25
  • \$\begingroup\$ i never knew new doesn't require (), i completely forgot about reference error lol \$\endgroup\$
    – mekb
    Aug 21 at 5:50
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    \$\begingroup\$ You can save a byte with new Date%2?a:0 \$\endgroup\$
    – Ectogen
    Aug 21 at 7:11
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    \$\begingroup\$ ah yes, ternary, the thing i overuse the most in js \$\endgroup\$
    – mekb
    Aug 21 at 7:25
  • \$\begingroup\$ @Ectogen Or another with new Date%2&&a, but then it's identical to mine \$\endgroup\$
    – Redwolf Programs
    Aug 21 at 17:57
1
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Windows Batch, 1715 bytes

set/a1/%random%
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2
  • 1
    \$\begingroup\$ I don't think you need the x=. \$\endgroup\$
    – Neil
    Apr 10 '20 at 23:34
  • \$\begingroup\$ indeed, -2 bytes thanks to @Neil \$\endgroup\$
    – peter ferrie
    Apr 17 '20 at 22:42
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VBA, 46 42 35 25 13 bytes

Thanks to @Taylor Scott

?1/Int(9*Rnd)

This code only works inside the Immediate Window.

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2
  • 1
    \$\begingroup\$ If you make this an immediate window function, you can get it down to ?1/Int(9*rnd) \$\endgroup\$
    – Taylor Raine
    May 26 '20 at 19:48
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    \$\begingroup\$ @Taylor Scott, I didn't know you can do that. Thanks, mate. \$\endgroup\$
    – 지넬만
    Jul 1 '20 at 1:53
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Befunge-98 (FBBI), 4 bytes

?@
=

Executing = is of the only ways I can get Befunge to actually error (a / won't work there because 0/0 evaulates to 0 in Befunge-98 and asks the user in Befunge-93).

Try it online!

The ? turns the IP in a random direction. If it goes horizontally (1/2 chance), it hits the @ and terminates successfully. If it goes vertically, it hits the = and uses the empty stack (interpreted as the empty string) as the argument. This throws a segfault in FBBI (which implements = as a C system() call).

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1
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05AB1E, 3 bytes

®Ωι

Try it online! Has a \$\frac12\$ chance of erroring.

®Ωι  # full program
  ι  # uninterleave...
 Ω   # random element from...
®    # register_c (starts at -1)...
     # (implicit) split into separate characters...
  ι  # lists which were interleaved together into...
     # (implicit) stack is empty, so bottom of stack...
     # (implicit) converted to list

-1 split into separate characters is ["-", "1"], so it picks a random element from there. Either it picks 1 and uninterleaves 1 list which was interleaved into ["1"] (one list interleaved is the same list, so it returns ["1"]), or it picks -, and tries to uninterleave - lists which were interleaved together into ["-"]. Unlike ï, the built-in integer conversion function, it doesn't handle errors (if it can't be converted to an integer). So, it throws an error.

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1
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Jelly, 4 bytes

9XRy

Try it online!

Errors roughly 55% of the time

How it works

9XRy - Main link. Takes no arguments
9X   - Generate a random number from 1 to 9
  R  - Get the range from 1 to that number
   y - Translate, by pairs

y splits a flat array [a, b, c, d, ..., y, z] into pairs, [[a, b], [c, d], ..., [y, z]], then replaces a with b, c with d etc in the right argument. However, if there are an odd number of elements in the array, it attempts to pop from an empty list when doing the substitution, which errors.

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1
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Japt, 3 bytes

Throws if seconds in a new date object are less than 2 or more than 36 since values outside of that range are not valid for base conversion in Japt.

sKb
s   // Try to convert the input to a string in base
 K  // new Date()
  b // get seconds.

Try it here.

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1
  • 1
    \$\begingroup\$ Nice! I had something similar in sAö. \$\endgroup\$
    – Shaggy
    Aug 16 at 20:10
1
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Java (JDK), 18 bytes

a->1/Math.random()

Try it online!

Well yeah, essentially the answer everyone gave but I didn't see Java yet. For the completeness.

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1
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Japt, 3 bytes

Feck it, may as well post this. Similar in concept to but derived independently from Etheryte's solution. Predicated on the fact that JavaScript's toString method can only handle bases 2-36 so it, therefotre, has a 20% chance of throwing a RangeError (toString() radix argument must be between 2 and 36) - change the A to alter the odds of an error occurring.

sAö

Test it (errors are displayed below the output field)

sAö     :Implicit input of integer U, defaulting to 0
s       :Convert to string in base
 A      :  10
  ö     :  Random integer from range [0,A)
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1
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Perl 5, 8 bytes

1/($$%2)

Here $$ is current PID.

Try it online!

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1
  • \$\begingroup\$ The current PID is not considered random for this challenge. \$\endgroup\$
    – Redwolf Programs
    Aug 22 at 18:55
1
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Perl 5, 10 bytes

1/(time%2)

Try it online!

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0
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(plain) pdfTeX, 34 bytes

\ifcase\pdfuniformdeviate9}\fi\bye
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0
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Commodore BASIC (Commodore C64/128, VIC-20, C16/+4, PET, Ultimate64, THEC64 etc)

0 on6*rnd(.)goto10,20,30,40,50:goto
10 print"working":goto
20 print"working":goto
30 print"working":goto
40 print"working":goto

I've put a prompt in just to show that it does work sometimes too. Essentially, it breaks when the random number generated is rounded to 5 (it generates floating point numbers, but the BASIC interpreter ignores the decimal places).

Commodore C64 BASIC working or not

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1
2

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