What's the purpose of the argument at the end of "bash -c command argument"?

Question

From man bash :

If the -c option is present, then commands are read from
the first non-option argument command_string.  If there
are arguments after the command_string, the first argument
is  assigned  to $0 and any remaining arguments are assigned
to the positional parameters. The assignment to $0 sets the
name of the shell, which is used in warning and error messages.

I don't understand the purpose of the $0 assignment.

I was exploring ways to pass arguments to multiple commands with xargs. The following solution works just fine (inspired from here), but I fail to understand why an argument is needed at the end :

[ Edit : following chepner's answer, the argument doesn't need to be empty, it can be anything indeed, but it has to exist ].

$ cat test
abc
def
ghi

$ cat test | xargs -n 1 /bin/bash -c 'echo "1-$@"; echo "2-$@";' 'dummyArgument'
1-abc
2-abc
1-def
2-def
1-ghi
2-ghi

Obviously the 'dummyArgument' is necessary for bash -c to be able to interpret $@, (it can even be empty '') because without it I get the following result :

$ cat test | xargs -n 1 /bin/bash -c 'echo "1-$@"; echo "2-$@";'
1-
2-
1-
2-
1-
2-

But how does it work ? Why do I need that 'dummyArgument' ?

Answer 1

I don't understand the purpose of the $0 assignment.

Its purpose is to let you give your inline script a name, potentially for decorating diagnostic messages. There is nothing interesting about it.

$ cat foo.sh
echo my name is $0
$
$ bash foo.sh a b c
my name is foo.sh
$
$ bash -c 'echo my name is $0' foo a b c
my name is foo