Why is $0 not -bash in an interactive login shell?

Question

I was reading about bash interactive and on interactive logins and found that running bash with -l option will start it as login shell but after doing so, when I ran command echo $0 to verify, it still shows as non login shell.

[root@localhost ~]# bash -l
[root@localhost ~]# echo $0
bash

It should show - as prefix to shell name. Please confirm if I am going wrong.

Answer 1

To have an - as the first character of the command name is just one way to signal a login shell, there are other signals. In fact, the correct way to detect that the present shell is a login shell is to ask the shell itself.

In the bash manual:

A login shell is one whose first character of argument zero is a -, or one started with the --login option.

In bash:

shopt -p login_shell

Will print -u if the shell is not login, and -s if it is.

Lets test it:

$ ln -s $(which bash) ./-bash       # make a local copy of bash
$ PATH=$PATH:.                      # Big security problem, don't use it.
$ -bash                             # start the local copy of bash
$ echo $0                           # Its name start with an -
-bash                               
$ shopt -p login_shell              # Is it a login shell?
shopt -s login_shell                # Yes!, the answer is -s

$ exit                              # leave the login shell
$ PATH=${PATH%:.}                   # Remove the local pwd from the path.
$ rm ./-bash                        # Remove the local copy of bash

In fact, that $0 has a - doesn't have to mean that the shell is a login shell:

$  bash -c 'echo $0; shopt -p login_shell' -bash 1 2 3
-bash
shopt -u login_shell

The other way (from the manual) to get a login shell, is just ask for it:

$ bash --login                 # ask for a login shell
$ echo $0                      # What is its name?
bash
$ shopt -p login_shell         # Is it a login shell?
shopt -s login_shell           # Yes, it is!

$ exit                         # leave the login shell.