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A comment on How does `cd` work?. Parent comment

Re:How does `cd` work? by aidenn0 on Thursday March 01, 2018 @03:02PM (#16495646)
Also some shells will refuse to run a builtin if there is no executable in the path that matches it.

POSIX actually mandates this behavior for any builtin not on a specific list, though most shells (even dash, which is typically obsessive about complying with POSIX) do not implement this behavior.

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Re:How does `cd` work? by LukeShu on Thursday March 01, 2018 @04:13PM (#16496391)
That didn't sound right to me, but I looked it up, and you're right.

The scripts we're discussing were written in 2002 [1], so the time-appropriate version is POSIX-2001 (Issue 6).

[1]: https://github.com/freebsd/freebsd/commit/55d0b8395514ae4055...

But you don't have to take my word for it: I've paste-bin'ed the entirety of what POSIX-2001 had to say about shell built-ins (in general, I didn't include man-pages for individual built-ins): https://lukeshu.com/dump/posix-2001-builtins.txt

As for whether that's still true today, looking at POSIX-2008 (Issue 7), 2013 edition (I don't have a copy of the 2016 edition handy), none of that has changed.

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Re:How does `cd` work? by aidenn0 on Thursday March 01, 2018 @04:47PM (#16496687)
I only discovered this because I implemented a shell specifically by the specification (just for didactic purposes). I was unable to find a modern shell that acted this way, even with passing the "be more POSIXy" options though.
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