Date: Wed, 03 Oct 2018 22:49:17 +0000 From: bugzilla-noreply@freebsd.org To: bugs@FreeBSD.org Subject: [Bug 231926] ldd can't operate on a segfaulting binary Message-ID: <bug-231926-227-47ayDeXKYt@https.bugs.freebsd.org/bugzilla/> In-Reply-To: <bug-231926-227@https.bugs.freebsd.org/bugzilla/> References: <bug-231926-227@https.bugs.freebsd.org/bugzilla/>
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https://bugs.freebsd.org/bugzilla/show_bug.cgi?id=3D231926 Mark Johnston <markj@FreeBSD.org> changed: What |Removed |Added ---------------------------------------------------------------------------- CC| |markj@FreeBSD.org --- Comment #1 from Mark Johnston <markj@FreeBSD.org> --- ldd is very simple: it sets some magic rtld flags and exec()s the specified executable (or uses dlopen(RTLD_TRACE) for shared libs). For an executable, rtld will then print the dependencies and exit without actually calling into the executable. So a bug in the executable's code which causes a segfault should not be triggered when run by ldd, but if the executable itself is corrupted or invalid in some way, rtld may crash. This may or may not be a= bug in rtld, depending on the nature of the corruption. In other words, if the executable is segfaulting before main() gets invoked, then the behaviour you're seeing is probably expected. To say for sure, we need to see exactly where the segfault is occurring. --=20 You are receiving this mail because: You are the assignee for the bug.=
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