zstd wrongly uses utime instead of utimensat on BSDs

[asomers]

Describe the bug
zstd and unzstd attempts to set the output file's atime and mtime to match the input file's. This behavior cannot be disabled. On platforms that support it, unzstd uses utimensat. Otherwise, it uses utime. Unfortunately, utime only sets the timestamp's seconds field, not its nanoseconds field. That matters, because zstd wrongly determines which platforms support utimensat. For the BSDs, it simply hardcodes PLATFORM_POSIX_VERSION to a number that's too old (see

zstd/programs/platform.h

Line 77 in 396ef5b

# if (defined(__APPLE__) && defined(__MACH__)) || defined(__SVR4) || defined(_AIX) || defined(__hpux) /* POSIX.1-2001 (SUSv3) conformant */ \

).

Tragically, this means that the output file's mtime is almost always a fraction of a second older than the input file's. So, for example, you can't create a Makefile rule like the below. make will always execute the command.

foo: foo.zst
        unzstd -fo foo foo.zst

To Reproduce

> touch /tmp/foo
> zstd /tmp/foo
> rm /tmp/foo
> touch /tmp/foo.zst
> unzstd /tmp/foo.zst
> stat -f "%N %Fm" /tmp/foo /tmp/foo.zst
/tmp/foo 1694052370.000000000
/tmp/foo.zst 1694052370.336394000

Notice that foo has an older timestamp than foo.zst. It should be equal or greater.

Expected behavior
The output file's timestamps should always be greater than or equal to the input file's.

Desktop (please complete the following information):

• OS: FreeBSD
• Version: 14.0-CURRENT
• Compiler: Clang
• Flags: The defaults. Probably -O2
• Other relevant hardware specs: None
• Build system: FreeBSD's builtin build system