Regexc - Segmentation Fault

Question

I'm relatively new to C and I have a problem with the following:

 regex_t re;
 regmatch_t matches[2];
 int start;
 int end;
 int count = 0;

 int reti = regcomp(&re, "^(?:[a-zA-Z]|[a-zA-Z](?:[a-zA-Z0-9\\-])*[a-zA-Z0-9])(?:\\.(?:[a-zA-Z]|[a-zA-Z](?:[a-zA-Z0-9\\-])*[a-zA-Z0-9]))*$", REG_EXTENDED);

 while (1) {

   printf("Local = %s\n", local);
   reti = regexec(&re, local, 2, matches, 0);

  // More code here

 }

When I run this, I get a segmentation fault.

local was defined as char *local and is being printed out correctly.

I ran the code using GDB and it turns out the issue is arising at the line:

reti = regexec(&re, local, 2, matches, 0);

I can't seem to figure out why.

This is the output from gdb:

Program received signal SIGSEGV, Segmentation fault.
0x00007ffff7af8d42 in regexec () from /lib/x86_64-linux-gnu/libc.so.6
(gdb) backtrace
#0  0x00007ffff7af8d42 in regexec () from /lib/x86_64-linux-gnu/libc.so.6
#1  0x000000000040092d in email_in (str=0x400aa8 "giri@gmail.com") attestdriver.c:70
#2  0x0000000000400780 in main () at testdriver.c:15

Would anyone have any ideas as to what might be the issue?

Thanks for your help.

Can you show us the definition of `regcomp()` and `regexec()`? — nalzok, Apr 03 '16 at 22:34
@sunqingyao https://github.com/lattera/glibc/blob/master/posix/regcomp.c#L465-L519 — Siguza, Apr 03 '16 at 22:38
Did you check your `reti` return value to see that `regcomp` actually returns a successful compilation? — Dolda2000, Apr 03 '16 at 22:45

score 1 · Accepted Answer · edited Jun 20 '20 at 09:12

1

The problem is actually here:

int reti = regcomp(&re, "^(?:[a............

From here:

You need to check the return of regcomp, it will tell you that your regex is invalid.

You will get Invalid preceding regular expression

(?: is not supported by POSIX regexes.

It's throwing this error because:

A common way to get a segfault is to dereference a null pointer

(From here)

The regex cannot initialize because of the illegal (?:

The original error is shown in the return of regcomp, and does not halt execution.

edited Jun 20 '20 at 09:12

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answered Apr 03 '16 at 22:41

Laurel

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And that produces a segfault? Does the POSIX standard allow that? o.0 – Siguza Apr 03 '16 at 22:42
@Siguza: It doesn't segfault on the compilation error, but because an uninitialized `regex_t` is passed to `regexec`. – Dolda2000 Apr 03 '16 at 22:46
@Dolda2000 Oh damn, I confused regexec with regcomp in the stack trace. It makes sense now, thanks. – Siguza Apr 03 '16 at 23:24

Regexc - Segmentation Fault

1 Answers1