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=210330 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 reproducible when member contents are identical. This option is enabled by default. If multiple -D and -U options are specified on the com‐ 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 specified 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 table if no mode is specified (-s with no other args), or in combination with -p/-t/-x (which are really read-options). -- You are receiving this mail because: You are the assignee for the bug.help
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