I have encountered this in the syntax for case
after calling help case
and getting case WORD in [PATTERN [| PATTERN]...) COMMANDS ;;]... esac
.
There was a question, whether "in" is a bash operator, but answers were answering the script problem of the person rather than saying clearly whether there is or it is an operator at all. I tried to find it in Bash documentation, but since in is a natural language preposition, the search provides lots of hits. Trying to find it as an operator, I was not successful.
foo
in Bash?" by invokingtype foo
in Bash (ortype -a foo
if you want). This works for manyfoo
s, includingin
; trytype in
. – Kamil Maciorowski Apr 18 '23 at 14:49