I have the following line:
printf "---- %.55s\n" "$1 --------------"
When I run this under bash I get the following error:
printf: --: invalid option printf: usage: printf [-v var] format [arguments]
This is because (I think) printf interprets the string "----" as the beginning of an argument. I have tried to escape this using "\" like so:
printf "\---- %.55s\n" "$1 --------------"
The following is the result:
\---- content of $1 -------
As you can see the "\" became part of the output which is not what I want.
So the question is, how can i persuade printf that I am not trying to pass a long argument?