[Help-bash] Clarify the meaning of 's' in $- and the meaning of -s when invoking bash

[zoicas]

Hello all,

It seems to me that the 's' flag is supposed to be shown in $- when
bash reads  the standard input. Such  a case occur when  bash is invoked
with the -s option. Example:

    $ echo 'echo $-' | bash -s
    hBs

However,  bash reads  the standard  input also  when there  are no  more
arguments  after option  processing.  Example

   $ echo 'echo $-' | bash
   hB

Can somebody explain what is the difference between running bash with
or without '-s'?

Thank you in advance
Cristian.


btw:

  Another case when bash reads the standard input is when simply
  started with the command

   $ bash