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"
来源：https://stackoverflow.com/questions/40423203/bash-printf-literal-verbatim-string