No, I expect the program to do exactly what the manual says.
-h, --human-numeric-sort
compare human readable numbers (e.g., 2K 1G)
Applying -h to the list in my example is expected to be semantically equivalent
to not applying -h:
A = { echo -e "a1\na2" | sort }
B = { echo -e "a1\na2" | sort -h }
~ $ echo -e "a1\na2" | sort
a1
a2
~ $ echo -e "a1\na2" | sort -h
a1
a2
Since A = B, the result of -u must be the same on both sets, by logic. The
program, however, has a mind of its own.
~ $ echo -e "a1\na2" | sort -u
a1
a2
~ $ echo -e "a1\na2" | sort -h -u
a1
-------- Original Message --------
On 2/16/25 23:22, Paul Eggert <egg...@cs.ucla.edu> wrote:
> On 2025-02-16 03:02, Rupert Gallagher wrote:
> > The introduction of the unique operator (-u) returns a wrong answer when
> used with the human sorting operator (-h).
>
> The answer is "wrong" only in the sense that sort's documented and
> implemented behavior is not what you expect.
>
> To fix this mismatch between behavior and expectations, don't use -h. It
> makes sense to not use -h, -h is not intended for uses like that.
>
