Date: Thu, 16 Jun 2016 17:42:01 +0000 From: bugzilla-noreply@freebsd.org To: freebsd-bugs@FreeBSD.org Subject: [Bug 210330] "ar -s" not deterministic by default Message-ID: <bug-210330-8@https.bugs.freebsd.org/bugzilla/>
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https://bugs.freebsd.org/bugzilla/show_bug.cgi?id=3D210330 Bug ID: 210330 Summary: "ar -s" not deterministic by default Product: Base System Version: 11.0-CURRENT Hardware: Any OS: Any Status: New Severity: Affects Only Me Priority: --- Component: bin Assignee: freebsd-bugs@FreeBSD.org Reporter: emaste@freebsd.org ar(1) produces deterministic output by default for the -q/-r options (append/replace), as described in the -D option: -D When used in combination with the -r or -q option, insert 0's instead of the real mtime, uid and gid values and 0644 instead= of file mode from the members named by arguments file .... This ensures that checksums on the resulting archives are reproduci= ble when member contents are identical. This option is enabled by default. If multiple -D and -U options are specified on the c= om=E2=80=90 mand line, the final one takes precedence. It is not documented here, but this is also the case when ar is invoked as ranlib(1). However, 'ar -s <file>' is supposed to be equivalent to ranlib: -s Add an archive symbol table (see ar(5)) to the archive specifi= ed by argument archive. Invoking ar with the -s option alone is equivalent to invoking ranlib. but ar -s does not produce deterministic output by default. In addition to -q/-r, ar -s (AR_S in ar.c) will write an archive symbol tab= le if no mode is specified (-s with no other args), or in combination with -p/-t/-x (which are really read-options). --=20 You are receiving this mail because: You are the assignee for the bug.=
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