Write the shortest code that raises a Segmentation Fault (SIGSEGV) in any programming language.
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52\$\begingroup\$ Wow. Possibly the shortest successful question. \$\endgroup\$– Matthew RohCommented Feb 9, 2017 at 11:42
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6\$\begingroup\$ @MatthewRoh Out of interest, I made this SEDE query. It looks like there are a few with +10 or higher, but this is the first above +40 \$\endgroup\$– rydwolfCommented Nov 21, 2020 at 20:38
85 Answers
C, 5 characters
main;
It's a variable declaration - int
type is implied (feature copied from B language) and 0
is default value. When executed this tries to execute a number (numbers aren't executable), and causes SIGSEGV
.
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5\$\begingroup\$ @Macmade: Actually, it is
0
.static
variables start as0
, andmain;
isstatic
, as I declared it outside function. c-faq.com/decl/initval.html \$\endgroup\$– nullCommented Aug 16, 2013 at 8:20 -
23\$\begingroup\$ last time i played with this thing, i figured out that there's a different reason for the segfault. First of all by calling main you jump to the location of main, not the value, another thing is
main
is an int, it's located in.bss
, usually functions are located in.text
, when the kernel loads the elf program it creates an executable page for.text
and non-executable for.bss
, so by calling main, you jump to a non-executable page, and execution something on a such page is a protection fault. \$\endgroup\$– mniipCommented Dec 6, 2013 at 17:55 -
36\$\begingroup\$ Yep, segfaults in C are pretty much the default :P \$\endgroup\$ Commented May 24, 2014 at 23:23
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1\$\begingroup\$
main __attribute__((section(".text#")))=0xc3;
FTFY (at least it seems to return without crashing on my x86). \$\endgroup\$– jozxyqkCommented Jul 27, 2018 at 4:57 -
3\$\begingroup\$ @jozxyqk Or shorter,
const main=195;
. As interesting it is that it's working, the goal of this code golfing challenge was to make the code segfault, not work :). \$\endgroup\$– nullCommented Jul 27, 2018 at 6:48
Bash, 11
kill -11 $$
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69\$\begingroup\$ Signal 11 in 11 characters. Seems legit. \$\endgroup\$– user344Commented Dec 31, 2013 at 12:39
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19\$\begingroup\$ @nyuszika7h I was going to upvote your comment, but you have 11 upvotes right now, so I'm going to leave it at that. :P \$\endgroup\$– hyperneutrino ♦Commented Nov 25, 2016 at 16:22
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11\$\begingroup\$ @AlexL. other people seem to have spoiled that :( \$\endgroup\$– user36219Commented Jan 21, 2017 at 10:51
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5\$\begingroup\$ @theonlygusti Yeah... That's too bad. :( Oh well, then I can upvote it now. \$\endgroup\$– hyperneutrino ♦Commented Jan 21, 2017 at 14:27
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3\$\begingroup\$ Up to 42 upvotes, no touchee! \$\endgroup\$ Commented Sep 3, 2018 at 15:24
Assembly (Linux, x86-64), 1 byte
RET
This code segfaults.
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8\$\begingroup\$ As an MSDOS .com file, it runs and terminates without error. \$\endgroup\$– J BCommented Dec 26, 2011 at 19:10
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13\$\begingroup\$ My point being: just specifying “assembly” isn't enough to make it segfault. \$\endgroup\$– J BCommented Dec 26, 2011 at 19:56
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65\$\begingroup\$ @JB: On MS DOS, no program will ever produce a segmentation fault. That's because MS DOS runs in real mode where memory protection is nonexistent. \$\endgroup\$– celtschkCommented Feb 4, 2012 at 17:25
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2\$\begingroup\$ @celtschk: You can segfault it anyway like so: mov bx, 1000h ; shr ebx, 4 ; mov eax, [ebx] -> CPU raises the underlying SEGV (AFAIK there's nobody to handle it though). \$\endgroup\$– JoshuaCommented Nov 20, 2016 at 17:18
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2\$\begingroup\$ Whether
ret
by itself crashes depends on whether there's an outer environment to return to. For instance, if this is used as the body ofmain
in a typical C environment, it will not crash (but it will exit with a garbage status), but if it's used as the sole contents of an ELF text segment, it typically will crash because the initial stack frame created by the OS has a null pointer for the return address. \$\endgroup\$– zwolCommented Nov 20, 2016 at 22:59
Python 2, 13
exec'()'*7**6
Windows reports an error code of c00000fd (Stack Overflow) which I would assume is a subtype of segmentation fault.
Thanks to Alex A. and Mego, it is confirmed to cause segmentation faults on Mac and Linux systems as well. Python is the language of choice for portably crashing your programs.
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10
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10\$\begingroup\$
Segmentation fault (core dumped)
on Linux \$\endgroup\$– user45941Commented Nov 2, 2015 at 1:30 -
1\$\begingroup\$ @MegaMan As in take a long time to finish? No, 7**6 is only about 100K so there's no perceptible delay. \$\endgroup\$– feersumCommented Aug 14, 2016 at 2:46
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4\$\begingroup\$ @MaxGasner Try reading the programming language again :) \$\endgroup\$– feersumCommented Aug 6, 2019 at 4:03
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1\$\begingroup\$ Yeah, it doesn't work on 3.8.9 (or at least my build of 3.8.9) on OS X. -- at least this version is clever enough to instead give you a
RecursionError
\$\endgroup\$ Commented Jun 23, 2021 at 18:40
pdfTeX (51)
\def~#1{\meaning}\write0{\expandafter~\string}\bye
This is actually probably a bug, but it is not present in the original TeX, written by Knuth: compiling the code with tex filename.tex
instead of pdftex filename.tex
does not produce a segfault.
LOLCODE, 4 bytes
OBTW
Does not work online, only in the C interpreter.
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33\$\begingroup\$ LOL FANCY CODE M8 8/8 KTHXBYE \$\endgroup\$ Commented Nov 2, 2015 at 15:52
W32 .com executable - 0 bytes
This will seem weird, but on 32 bit Windows systems, creating and executing an empty .com file may cause a segfault, depending on... something. DOS just accepts it (the 8086 having no memory management, there are no meaningful segments to fault), and 64 bit Windows refuses to run it (x86-64 having no v86 mode to run a .com file in).
Python, 33 characters
>>> import ctypes;ctypes.string_at(0)
Segmentation fault
Source: http://bugs.python.org/issue1215#msg143236
Python, 60 characters
>>> import sys;sys.setrecursionlimit(1<<30);f=lambda f:f(f);f(f)
Segmentation fault
Source: http://svn.python.org/view/python/trunk/Lib/test/crashers/recursive_call.py?view=markup
This is the Python version I'm testing on:
Python 2.6.1 (r261:67515, Jun 24 2010, 21:47:49)
[GCC 4.2.1 (Apple Inc. build 5646)] on darwin
In general the Python interpreter is hard to crash, but the above is selective abusiveness...
Bash, 4 bytes
Golfed
. $0
Recursively include the script into itself.
Explained
Recursive "source" (.) operation causes a stack overflow eventually, and as Bash does not integrate with libsigsegv, this results in a SIGSEGV.
Note that this is not a bug, but an expected behavior, as discussed here.
Test
./bang
Segmentation fault (core dumped)
Forth - 3 characters
0 @
(@
is a fetch)
brainfuck (2)
<.
Yes, this is implementation-dependent. SIGSEGV is the likely result from a good compiler.
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6\$\begingroup\$ How is a compiler that segfaults on that "good"? That
<
should either have no effect or wrap around. \$\endgroup\$– user344Commented Jul 4, 2014 at 17:38 -
24\$\begingroup\$ Immediately producing a runtime error on bounds violation is best because it lets the programmer find and fix the bug as fast as possible. Letting the buggy program run for a while and corrupt memory haphazardly before crashing just makes the problem harder to diagnose. Preventing the crash entirely, as you suggest, is worst; the programmer may get the program "working" and then be publicly humiliated when it crashes on standard compilers and interpreters. \$\endgroup\$ Commented Oct 12, 2014 at 23:17
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3\$\begingroup\$ Conversely, catching bounds violations before runtime is not possible in general, nor especially useful in the cases where it is possible. Producing a more descriptive runtime error would be okay, but having the operating system catch it as a segfault is great because it doesn't have any speed cost. (In case it's not clear, the compiler itself doesn't segfault--it produces executables that segfault as soon as they try to access memory out of bounds.) \$\endgroup\$ Commented Oct 12, 2014 at 23:35
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6\$\begingroup\$ Can you provide an implementation that produces this behavior and was created before this challenge was posted? If not, this answer is invalid. \$\endgroup\$– user45941Commented Apr 25, 2016 at 20:10
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1\$\begingroup\$ Bounds checks are implementation specific, so I'm sure there are some that would error on it. Would any SIGSEGV though? I doubt it. There are a large number of programs that depend on the array wrapping to the left though. It can be rather convenient having growable storage on both sides. \$\endgroup\$ Commented Dec 15, 2016 at 20:01
C, 18
main(){raise(11);}
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1\$\begingroup\$ do you need to add #include <signal.h> in the code listing? \$\endgroup\$ Commented Nov 28, 2016 at 8:16
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7\$\begingroup\$ @FlorianCastellane: in C90 and lower, for any function call done without a visible declaration, the compiler implicitly declares it as
int func()
. i.e. a function returningint
, taking unspecified parameters. In this caseraise
is a function returning int, taking an int argument, so this works out (even if the compiler complains). \$\endgroup\$ Commented Nov 28, 2016 at 13:26 -
\$\begingroup\$ @Hasturkun
main(){main();}
\$\endgroup\$ Commented Mar 9, 2020 at 1:16
Perl ( < 5.14 ), 9 chars
/(?{??})/
In 5.14 the regex engine was made reentrant so that it could not be crashed in this way, but 5.12 and earlier will segfault if you try this.
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\$\begingroup\$ I can reproduce this on Perl 5.14 (Debian) and 5.18 (Arch Linux). sprunge.us/RKHT \$\endgroup\$– user344Commented Jan 21, 2014 at 21:51
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1\$\begingroup\$ Reproduced with Perl v5.20.2 (windows) \$\endgroup\$ Commented Mar 11, 2016 at 13:38
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\$\begingroup\$ What about
/(?R)/
on older Perl versions? \$\endgroup\$ Commented Sep 2, 2020 at 13:37
Haskell, 31
foreign import ccall main::IO()
This produces a segfault when compiled with GHC and run. No extension flags are needed, as the Foreign Function Interface is in the Haskell 2010 standard.
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\$\begingroup\$ Awwww. I was gonna post
import Foreign;main=peek nullPtr::IO Int
, but that's 40. \$\endgroup\$ Commented Dec 25, 2020 at 20:02
Python 33
import os
os.kill(os.getpid(),11)
Sending signal 11 (SIGSEGV) in python.
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3\$\begingroup\$ Also 33 characters:
from os import*
andkill(getpid(),11)
\$\endgroup\$– TimtechCommented Jan 8, 2014 at 15:45
C - 11(19) 7(15) 6(14) 1 chars, AT&T x86 assembler - 8(24) chars
C version is:
*(int*)0=0;
The whole program (not quite ISO-compliant, let's assume it's K&R C) is 19 chars long:
main(){*(int*)0=0;}
Assembler variant:
orl $0,0
The whole program is 24 chars long (just for evaluation, since it's not actually assembler):
main(){asm("orl $0,0");}
EDIT:
A couple of C variants. The first one uses zero-initialization of global pointer variable:
*p;main(){*p=0;}
The second one uses infinite recursion:
main(){main();}
The last variant is the shortest one - 7(15) characters.
EDIT 2:
Invented one more variant which is shorter than any of above - 6(14) chars. It assumes that literal strings are put into a read-only segment.
main(){*""=0;}
EDIT 3:
And my last try - 1 character long:
P
Just compile it like that:
cc -o segv -DP="main(){main();}" segv.c
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4\$\begingroup\$ in C isn't main; only 5 charecters \$\endgroup\$ Commented Dec 26, 2011 at 10:50
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3\$\begingroup\$ :Linker doesn't check whether main is function or not .It just pass it to the loader and return sigsegv \$\endgroup\$ Commented Dec 26, 2011 at 12:24
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1\$\begingroup\$ @FUZxxl In this case
main
is a zero-initialized global int variable, so what we get is a result of trying to execute some zero bytes. In x86 it'd be something likeadd %al,(%rax)
which is a perfectly valid instruction which tries to reach memory at address stored in%rax
. Chances of having a good address there are minimal. \$\endgroup\$ Commented Dec 27, 2011 at 20:35 -
9\$\begingroup\$ Of course the last entry can be used for everything, you just have to supply the right compiler arguments. Which should make it the automatic winner of any code golf contest. :-) \$\endgroup\$– celtschkCommented Feb 4, 2012 at 17:20
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6\$\begingroup\$ Usually compiler flags other than ones that choose the language version to use are counted towards the total. \$\endgroup\$ Commented Dec 11, 2014 at 4:32
Perl, 10 / 12 chars
A slightly cheatish solution is to shave one char off Joey Adams' bash trick:
kill 11,$$
However, to get a real segfault in Perl, unpack p
is the obvious solution:
unpack p,1x8
Technically, this isn't guaranteed to segfault, since the address 0x31313131 (or 0x3131313131313131 on 64-bit systems) just might point to valid address space by chance. But the odds are against it. Also, if perl is ever ported to platforms where pointers are longer than 64 bits, the x8
will need to be increased.
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1\$\begingroup\$ What is this
1x8
thing? \$\endgroup\$ Commented Dec 15, 2016 at 10:19 -
1\$\begingroup\$ @HannesKarppila It's a short way to write
"11111111".
\$\endgroup\$ Commented Dec 15, 2016 at 12:18 -
\$\begingroup\$ You can replace
1x8
with$^
, it's long enough in both 32 and 64 bit. \$\endgroup\$– SisyphusCommented Jul 28, 2021 at 8:09
F90 - 39 bytes
real,pointer::p(:)=>null()
p(1)=0.
end
Compilation:
gfortran segv.f90 -o segv
Execution:
./segv
Program received signal SIGSEGV: Segmentation fault - invalid memory reference.
Backtrace for this error:
#0 0x7FF85FCAE777
#1 0x7FF85FCAED7E
#2 0x7FF85F906D3F
#3 0x40068F in MAIN__ at segv.f90:?
Erreur de segmentation (core dumped)
Materials:
gfortran --version
GNU Fortran (Ubuntu 4.8.4-2ubuntu1~14.04.1) 4.8.4
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1
Actually, 17 16 11 10 9 bytes
⌠[]+⌡9!*.
If the above doesn't crash, try increasing the number (multi-digit numbers are specified in Actually with a leading colon)
Crashes the interpreter by exploiting a bug in python involving deeply nested itertools.chain
objects, which actually uses to implement the +
operator.
dc - 7 chars
[dx0]dx
causes a stack overflow
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\$\begingroup\$ Is works, but can you elaborate? Why does it behave that way? \$\endgroup\$ Commented Nov 21, 2016 at 12:17
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3\$\begingroup\$
[dx0]
storesdx0
on the stack, thend
duplicates the top stack element, thenx
pops the top stack element (dx0
) and executes it. Which duplicates the top stack element, and starts executing it... the0
needs to be there to prevent this being a tail call, so they all build up. \$\endgroup\$ Commented Nov 26, 2016 at 7:58 -
\$\begingroup\$ @BenMillwood I wasn't sure if dc had TCO, so to find out that it definitely does is really cool. \$\endgroup\$ Commented May 30 at 23:22
PicoLisp - 4 characters
$ pil
: ('0)
Segmentation fault
This is intended behaviour. As described on their website:
If some programming languages claim to be the "Swiss Army Knife of Programming", then PicoLisp may well be called the "Scalpel of Programming": Sharp, accurate, small and lightweight, but also dangerous in the hand of the inexperienced.
Pyth, 3 characters
j1Z
This would be the part where I explain how I came up with this answer, except I legitimately have no clue. If anyone could explain this for me, I'd be grateful.
Here it is in an online interpreter.
Explanation
j
squares the base and calls itself recursively until the base is at least as large as the number. Since the base is 0, that never happens. With a sufficienly high recursion limit, you get a segfault.
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\$\begingroup\$ Figured something out! From browsing Pyth's source, I found that this code does
j
on1
and0
, which tries to convert1
into base0
. Why that segfaults, I have no idea... \$\endgroup\$ Commented Dec 23, 2016 at 1:24 -
1
-
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\$\begingroup\$ @SeeRhino The Pyth interpreter sets the recursion limit to 100,000. At least on TIO, that's enough for a segfault. \$\endgroup\$– DennisCommented Dec 23, 2016 at 1:38
Python 3.8+, 47 bytes
eval((lambda:0).__code__.replace(co_consts=()))
Not the shortest python entry by far, but this one works without any imports. And as a bonus, this bug is exploitable to get arbitrary code execution, if you're smart about it.
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\$\begingroup\$ Welcome to Code Golf, this is a pretty interesting answer! \$\endgroup\$– rydwolfCommented Dec 9, 2020 at 5:19
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\$\begingroup\$ It's honestly amazing that this language that has such a simple interface when seen from above gives you such incredibly deep access into the language's own runtime internals. \$\endgroup\$ Commented Jan 29, 2021 at 16:14
OCaml, 13 bytes
Obj.magic 0 0
This uses the function Obj.magic
, which unsafely coerces any two types. In this case, it coerces 0 (stored as the immediate value 1, due to the tag bit used by the GC) to a function type (stored as a pointer). Thus, it tries to dereference the address 1, and that will of course segfault.
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1\$\begingroup\$
it coerces 0 (stored as the immediate value 1)
- why is 0 stored as 1? \$\endgroup\$– SkylerCommented Nov 3, 2015 at 14:15 -
1
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1\$\begingroup\$
Obj.magic()0
is one char shorter :) \$\endgroup\$ Commented Nov 26, 2016 at 7:53
19 characters in C
main(a){*(&a-1)=1;}
It corrupts return address value of main function, so it gets a SIGSEGV on return of main
.
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1\$\begingroup\$ It depends on the stack frame layout, so in some architecture it can possibly not fail. \$\endgroup\$ Commented Dec 27, 2011 at 19:41
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\$\begingroup\$ Why not simply
main;
, ormain(){*""=0;}
? \$\endgroup\$ Commented Nov 16, 2020 at 1:23 -
\$\begingroup\$ @Sapphire_Brick
main;
is already given in another answer. \$\endgroup\$– saeednCommented Nov 17, 2020 at 4:08 -
\$\begingroup\$ @saeedn Then why post it at all? This one isn't even the second to shortest! \$\endgroup\$ Commented Nov 17, 2020 at 5:27
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\$\begingroup\$ @Sapphire_Brick At the time I was posting mine,
main;
wasn't posted and I didn't know it works. I was only pointing out that it is already given and no point in changing my answer. Furthermore, people here don't just post only for the shortest, sometimes a different way of solving the problem is also interesting. \$\endgroup\$– saeednCommented Nov 18, 2020 at 17:20
Cython, 14
This often comes in handy for debugging purposes.
a=(<int*>0)[0]
Matlab - Yes it is possible!
In a response to a question of mine, Amro came up with this quirk:
S = struct();
S = setfield(S, {}, 'g', {}, 0)
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\$\begingroup\$ Please give Matlab version -- R2015B (and 2016B also) just throws an error: Error using setfield (line 56) At least one index is required. \$\endgroup\$ Commented Nov 28, 2016 at 8:24
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\$\begingroup\$ @FlorianCastellane Not able to try all versions now, but it has been confirmed to give a segfault in quite some versions, the latest being 2014b and the earliest 2012a. \$\endgroup\$ Commented Nov 29, 2016 at 16:13
C - 14 chars
Be sure to compile an empty file with cc -nostartfiles c.c
Explanation:
What went wrong is that we treated _start as if it were a C function, and tried to return from it. In reality, it's not a function at all. It's just a symbol in the object file which the linker uses to locate the program's entry point. When our program is invoked, it's invoked directly. If we were to look, we would see that the value on the top of the stack was the number 1, which is certainly very un-address-like. In fact, what is on the stack is our program's argc value. After this comes the elements of the argv array, including the terminating NULL element, followed by the elements of envp. And that's all. There is no return address on the stack.
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3\$\begingroup\$ I'm pretty sure you have to score with the additional args \$\endgroup\$– BlueCommented Dec 16, 2016 at 12:36
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\$\begingroup\$ You have to add 14 bytes for the special flag. \$\endgroup\$ Commented Dec 16, 2016 at 12:38
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\$\begingroup\$ @ErikGolferエリックゴルファー -nostartfiles is actually 13 bytes long :) \$\endgroup\$ Commented Dec 16, 2016 at 13:34
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2\$\begingroup\$ @CharlesPaulet I think you have to count the space too. \$\endgroup\$ Commented Dec 16, 2016 at 13:37
Rust, 34 bytes
fn main(){unsafe{*(0 as*mut _)=3}}
Just a null pointer assignment, nothing special here.
As a bonus, 45 44 40 bytes solution not using unsafe
. Arguably a bug in a compiler.
#![no_main]#[no_mangle]static main:i8=0;
Main is usually a function ;).
C# - 62
System.Runtime.InteropServices.Marshal.ReadInt32(IntPtr.Zero);
C# /unsafe, 23 bytes
unsafe{int i=*(int*)0;}
For some reason I don't understand, *(int*)0=0
just throws a NullReferenceException, while this version gives the proper access violation.
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\$\begingroup\$ The
int i=*(int*)0;
returns a NullReferenceException for me. \$\endgroup\$ Commented Dec 30, 2011 at 7:43 -
\$\begingroup\$ You can try to access a negative location, like
*(int*)-1=0
and get an access violation. \$\endgroup\$ Commented Dec 30, 2011 at 7:46 -
\$\begingroup\$ The particular exception is just what the clr wraps it in, and is insignificant. The os itself actually gives the seg fault in all these cases. \$\endgroup\$ Commented Jan 20, 2012 at 17:41
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\$\begingroup\$ The reason why
*(int*)0=0
throws an exception is likely due to optimization. Specifically, to avoid the cost of checking fornull
, the optimizer may remove null checks, but when a segfault occurs it may rethrow it as a properNullReferenceException
. \$\endgroup\$– nullCommented Sep 15, 2018 at 20:55