Minimal C/C++ program that segfaults?

Question

I'm trying to set up the way my server handles core dumps. In order to test it, I'd need a program that always segfaults.

Is there a simple example program that always segfaults?

Coredump or segfault? Usually `abort()` is enough to cause a core dump, and that's actually part of a *real* program. A segfault never is. — Kerrek SB, Sep 13 '12 at 11:01
@ShafikYaghmour : I meant the minimal program, written *either* in C or C++, that segfaults. You can add the C++ tag if you want. — static_rtti, Sep 25 '13 at 13:15

score 10 · Answer 1 · answered Sep 13 '12 at 11:02

10

main() { *(int *)0xdeadbeef = 37; } should do it.

answered Sep 13 '12 at 11:02

nneonneo

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why 37? it should be 42! – arash kordi Sep 25 '13 at 08:08

score 8 · Accepted Answer · edited Jan 17 '22 at 17:09

8

main;

Is portable, and segfault in 5chars.
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.

Source: https://codegolf.stackexchange.com/questions/4399/shortest-code-that-raises-a-sigsegv

edited Jan 17 '22 at 17:09

Sir l33tname

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answered May 25 '14 at 08:52

Thomas Coudray

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Correct, but woefully lacking in any form of explanation as to why it does that (I know why, but not everyone does) – TheSola10 Sep 05 '18 at 15:16

score 4 · Answer 3 · answered Sep 13 '12 at 11:02

4

try this:

long* ptr = 0x0; //-- you can also use other random values and likely you will segfault
printf("%f", *ptr);

answered Sep 13 '12 at 11:02

sergio

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score 2 · Answer 4 · answered Sep 13 '12 at 11:03

2

You can try:

main() {
char *p = NULL;
char c = *p;
}

answered Sep 13 '12 at 11:03

Stephane Rouberol

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This can be eliminated by the compiler's optimizer. You want the pointer to point to a volatile object for this to work. – Alexey Frunze Sep 13 '12 at 11:24
@AlexeyFrunze : just don't turn optimizations on :) – static_rtti Sep 13 '12 at 12:09

Gung Foo · Answer 5 · 2012-09-13T11:24:48.007

0

this should die:

int main() {
    char *die;
    printf("%d",(int *)die * 200);
    return 0;
}

edit:

int main() {
    char *die;
    int killer = 200;
    while(1) {
       printf("%d",(int *)die * killer);
       killer = killer * killer;
    }
    return 0;
}

edited Sep 13 '12 at 11:24

answered Sep 13 '12 at 11:04

Gung Foo

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Why should it? It's UB, so it might, but it could also just print an arbitrary number. – Daniel Fischer Sep 13 '12 at 11:06
Ah, doesn't even compile with gcc. You'd have to explicitly convert the `char*` to an integer for it to compile (wasn't sure whether the compiler permitted an implicit conversion, the standard doesn't, but I think only a diagnostic is required, not termination of the compilation). – Daniel Fischer Sep 13 '12 at 11:15
Now it's still UB since you're using the indeterminate value of the uninitialised pointer, but I don't know of any implementation where that would actually crash. What is the idea how you want to cause a crash? – Daniel Fischer Sep 13 '12 at 11:20
Then you should probably dereference the pointer in some place. – Daniel Fischer Sep 13 '12 at 11:23

Minimal C/C++ program that segfaults?

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