bash printf literal verbatim string

问题:

To make my code portable, I try to use printf rather than echo. But then

printf "-dogs-cats" 

returns an error. A workaround in the present case is:

printf "-";printf "dogs-cats"

But is there a general, portable command (or an option with printf) that will print an arbitrary string as a literal/verbatim, not try to interpret the string as a format?

I work in BSD UNIX (on a Mac) but my objective is code that would work in other UNIX flavors as well.

回答1:

Just use -- after printf to let it know that no more arguments are to come and to consider the string as so:

$ printf -- "-dogs-cats" 
-dogs-cats                    # no new line after this

This is a *NIX-trick that can be used for many other commands. As Bash Reference Manual → 4 Shell Builtin Commands says:

Unless otherwise noted, each builtin command documented as accepting options preceded by ‘-’ accepts ‘--’ to signify the end of the options. The :, true, false, and test builtins do not accept options and do not treat ‘--’ specially. The exit, logout, return, break, continue, let, and shift builtins accept and process arguments beginning with ‘-’ without requiring ‘--’. Other builtins that accept arguments but are not specified as accepting options interpret arguments beginning with ‘-’ as invalid options and require ‘--’ to prevent this interpretation.

---

Note why this happens:

$ printf "-dogs-cats" 
bash: printf: -d: invalid option
printf: usage: printf [-v var] format [arguments]

This makes printf understand the first part of the string, -d, as an argument.

回答2:

Use a format specification:

printf '%s' "-dogs-cats"