I can't seem to use the Bash “-c” option with arguments after the “-c” option string

Question

The man page for Bash says, regarding the -c option:

-c string If the -c option is present, then comma

Answer 1

martin is right about the interpolation: you need to use single quotes. But note that if you're trying to pass arguments to a command that is being executed within the string, you need to forward them on explicitly. For example, if you have a script foo.sh like:

#!/bin/bash
echo 0:$0
echo 1:$1
echo 2:$2

Then you should call it like this:

$ bash -c './foo.sh ${1+"$@"}' foo "bar baz"
0:./foo.sh
1:bar baz
2:

Or more generally bash -c '${0} ${1+"$@"}' [argument]...

Not like this:

$ bash -c ./foo.sh foo "bar baz"
0:./foo.sh
1:
2:

Nor like this:

$ bash -c './foo.sh $@' foo "bar baz"
0:./foo.sh
1:bar
2:baz

This means you can pass in arguments to sub-processes without embedding them in the command string, and without worrying about escaping them.