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

Because '$0' and '$1' in your string is replaced with a variable #0 and #1 respectively.

Try :

bash -c "echo arg 0: \$0, arg 1: \$1" arg0 arg1

In this code $ of both are escape so base see it as a string $ and not get replaced.

The result of this command is:

arg 0: arg0, arg 1: arg1

Hope this helps.