I’m trying to use printf to format some pretty output in a bash script
e.g.:
----------------------- | This is some output | -----------------------
But I’ve stumbled over some behavior I don’t understand.
$ printf "--"
gives me the error:
printf: usage: printf [-v var] format [arguments]
and
$ printf "-stuff"
results in
-bash: printf: -s: invalid option
So apparently printf thinks I’m trying to pass some arguments while I’m not.
Meanwhile, completely by accident, I’ve found this workaround:
$ printf -- "--- this works now ----n"
gives me
--- this works now ----
Can anyone explain this behavior?
Answers:
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Method 1
The -- is used to tell the program that whatever follows should not be interpreted as a command line option to printf.
Thus the printf "--" you tried basically ended up as “printf with no arguments” and therefore failed.
Method 2
-- is being interpreted as an option (in this case, to signify that there are no more options).
A format string should always be included when using printf to prevent bad interpretation. For your particular case:
printf '%sn' '-----------------------'
Method 3
There are differences between printf builtin and /usr/bin/printf, the second one do “what you mean” without these annoying errors.
printf "-----" # error printf -- "-----" # ok /usr/bin/printf "-----" # ok /usr/bin/printf -- "-----" # ok
Method 4
POSIX provides the option of using octal encoding.
printf "--"
can become:
printf "55-"
This is a portable way to avoid the ambiguity of whether -- is being used as a format string or as a marker for the end of options.
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0