How to use positional parameters with “bash -c” command?

Question

I\'d like to know the best way of using positional parameters when using the command bash -c.

The man pages indicates for the -c option tha

Answer 1

The parameter $0 is special in that it does not participate in shift (for example).

This works as expected:

$ bash -c 'shift; printf "%s %s %s\n" $1 $2 $3' _ par1 par2 par3 par4
par2 par3 par4

This does not:

$ bash -c 'shift; printf "%s %s %s\n" $0 $1 $2' par1 par2 par3 par4
par1 par3 par4

This has been defined in POSIX (simplified):

sh -c command_string [command_name [argument...]]

The following additional options shall be supported:

-c

Read commands from the command_string operand. Set the value of special parameter 0 (see Special Parameters) from the value of the command_name operand and the positional parameters ($1, $2, and so on) in sequence from the remaining argument operands.

So, use this:

sh -c 'commands' command_name arg1 arg2 arg3 ....