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Challenges

Bytes to Segfault

+5
−0

Challenge

Cause the currently running program to receive the SIGSEGV signal (on Linux or other *nix systems) as fast as possible. What it does with the signal doesn't matter as long as it receives it.

This is code golf, smallest answer in each language wins. Some languages will have a tougher time than others. Keep in mind that if you depend on uninitalized data, it needs to always result in a segfault; a 0.000001% chance to not work invalidates your solution.

Example program

#include <stddef.h>
#include <stdio.h>

int main() {
    int* ohno = NULL;
    printf("%d", *ohno);
}
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9 answers

+6
−0

C, 5 bytes

main;

This exploits the fact that uninitialised globals live in the .bss section, and that section is not executable. So any attempt to execute code there, regardless of content, will cause a segfault.

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General comments (2 comments)
+3
−0

Python 2, 13 bytes

exec'+1'*5**9

Try it online!

No idea why this works. Something in the Python expression parser?

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General comments (2 comments)
+2
−0

Rust 1.0.0, 58 53 52 37 bytes

fn a(){#[no_mangle]static mmap:u8=0;}
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+2
−0

C, 16 bytes

m(){*(int*)m=0;}

Try it Online!

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General comments (1 comment)
+2
−0

JavaScript (Node.js), 35 32 25 bytes

-7 bytes thanks to @celtschk‭

with(process)kill(pid,11)

Try it online!

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General comments (1 comment)
+1
−0

Bash, 11

kill -11 $$

Try it online!

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+1
−0

C (compliant), 19 bytes

(gcc -std=c18 -pedantic-errors)

int main(){main();}

Godbolt

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+0
−0

Swift 5.4, 17 bytes

func f(){f()};f()

Pretty simple. It just calls itself until it stack overflows. You need to compile and run it, not just do it in the REPL, because the REPL just drops you back into the interpreter once it overflows.

It gives a warning "all paths through this function will call itself", but eh who cares?

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+0
−0

Rust, 47 bytes

fn main(){unsafe{print!("{}",*(0 as*mut i8));}}

Try it online!

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