You're interpreting the man page wrong. Firstly, the part about
-- signalling the end of options is irrelevant to what you're trying to do. The
-c overrides the rest of the command line from that point on, so that it's no longer going through bash's option handling at all, meaning that the
-- would be passed through to the command, not handled by bash as an end of options marker.
The second mistake is that extra arguments are assigned as positional parameters to the shell process that's launched, not passed as arguments to the command. So, what you're trying to do could be done as one of:
/bin/bash -c 'echo "$0" "$1"' foo bar
/bin/bash -c 'echo "$@"' bash foo bar
In the first case, passing echo the parameters
$1 explicitly, and in the second case, using
"$@" to expand as normal as "all positional parameters except $0". Note that in that case we have to pass something to be used as
$0 as well; I've chosen "bash" since that's what
$0 would normally be, but anything else would work.
As for the reason it's done this way, instead of just passing any arguments you give directly to the command you list: note that the documentation says "commands are read from string", plural. In other words, this scheme allows you to do:
/bin/bash -c 'mkdir -p -- "$1" && cd -P -- "$1" && touch -- "$2"' bash dir file
But, note that a better way to meet your original goal might be to use
env rather than
/usr/bin/env -- "ls" "-l"
If you don't need any of the features that a shell is providing, there's no reason to use it - using
env in this case will be faster, simpler, and less typing. And you don't have to think as hard to make sure it will safely handle filenames containing shell metacharacters or whitespace.