PERockwell

knowitall said:

Ahem, didn't say it wasn't compliant, I said it isn't based on Unix (its based on mach, thats the kernel running).
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macOS is even POSIX (a standard to unify Unix) compliant, but this isn't a guarantee that (heavy) porting of software will work (see one of my earlier posts).

Disagree that macOS isn't based on UNIX. Although it contains a Mach-based kernel, it also contains a BSD subsystem based on 4.4BSD-Lite2 and FreeBSD on top of the kernel that provides OS API services. So it shares UNIX linage back through what most people think as UNIX back to AT&T and Berkeley distributions.

The choice of kernel does not make a UNIX operating system. The Open Group (owner of UNIX) has decided that any operating system that meets the Single Unix Specification standard and passes their UNIX certification program is given a license to be called UNIX.  The standard does not mandate kernel implementation.

macOS not only is POSIX compiiant, but passes the Open Group's Single UNIX Specification (UNIX 03). That's what makes it officially licensed to use the UNIX name. 

Linux is most definitely not UNIX from a kernel and a naming/branding standpoint, although it does strive to be POSIX and Single Unix Specification compatible. It hasn't been (and in all likelihood won't be put) through the certification and licensing process to gain the name. 

Regardless of UNIX/POSIX/Linux designations - there's no guarantee that any "UNIX" software package will compile on every UNIX or UNIX-like system without some "porting" work - not just macOS.  Look at the sources for a lot of open source packages and you'll see a lot of conditionals based on the platform it discovers (not just macOS).