Does exiting a login shell necessarily log out of the OS?












-5














Bash manual says




logout [n]



Exit a login shell, returning a status of n to the shell’s parent.




On a virtual console of Ubuntu, I first log in, and in the login shell, run:



$ pstree -paus $$

systemd,1 --system --deserialize 19

  `-login,30488 -p --  

      `-bash,31728,t

          `-pstree,31774 -paus 31728


and then run logout, and all the processes starting from login,30488 till below disappear, more than just "return to the shell's parent" which is thelogin process.



In a login bash shell , does bash builtin commandlogout log out of the OS, not just exit the shell?



Does exiting a login shell (not necessarily by logout, but also any other way, such as bash builtin exit among others) necessarily lead to logging out of the OS?



May I also ask what logging out of OS means?



Thanks.




Extended to: https://stackoverflow.com/questions/53887813/how-does-the-login-program-in-linux-invoke-a-login-shell






bash logout







share|improve this question





























    -5














    Bash manual says




    logout [n]



    Exit a login shell, returning a status of n to the shell’s parent.




    On a virtual console of Ubuntu, I first log in, and in the login shell, run:



    $ pstree -paus $$
    
systemd,1 --system --deserialize 19
    
  `-login,30488 -p --  
    
      `-bash,31728,t
    
          `-pstree,31774 -paus 31728


    and then run logout, and all the processes starting from login,30488 till below disappear, more than just "return to the shell's parent" which is thelogin process.



    In a login bash shell , does bash builtin commandlogout log out of the OS, not just exit the shell?



    Does exiting a login shell (not necessarily by logout, but also any other way, such as bash builtin exit among others) necessarily lead to logging out of the OS?



    May I also ask what logging out of OS means?



    Thanks.




    Extended to: https://stackoverflow.com/questions/53887813/how-does-the-login-program-in-linux-invoke-a-login-shell






    bash logout







    share|improve this question



























      -5












      -5








      -5







      Bash manual says




      logout [n]



      Exit a login shell, returning a status of n to the shell’s parent.




      On a virtual console of Ubuntu, I first log in, and in the login shell, run:



      $ pstree -paus $$
      
systemd,1 --system --deserialize 19
      
  `-login,30488 -p --  
      
      `-bash,31728,t
      
          `-pstree,31774 -paus 31728


      and then run logout, and all the processes starting from login,30488 till below disappear, more than just "return to the shell's parent" which is thelogin process.



      In a login bash shell , does bash builtin commandlogout log out of the OS, not just exit the shell?



      Does exiting a login shell (not necessarily by logout, but also any other way, such as bash builtin exit among others) necessarily lead to logging out of the OS?



      May I also ask what logging out of OS means?



      Thanks.




      Extended to: https://stackoverflow.com/questions/53887813/how-does-the-login-program-in-linux-invoke-a-login-shell






      bash logout







      share|improve this question















      Bash manual says




      logout [n]



      Exit a login shell, returning a status of n to the shell’s parent.




      On a virtual console of Ubuntu, I first log in, and in the login shell, run:



      $ pstree -paus $$
      
systemd,1 --system --deserialize 19
      
  `-login,30488 -p --  
      
      `-bash,31728,t
      
          `-pstree,31774 -paus 31728


      and then run logout, and all the processes starting from login,30488 till below disappear, more than just "return to the shell's parent" which is thelogin process.



      In a login bash shell , does bash builtin commandlogout log out of the OS, not just exit the shell?



      Does exiting a login shell (not necessarily by logout, but also any other way, such as bash builtin exit among others) necessarily lead to logging out of the OS?



      May I also ask what logging out of OS means?



      Thanks.




      Extended to: https://stackoverflow.com/questions/53887813/how-does-the-login-program-in-linux-invoke-a-login-shell






      bash logout




      bash logout






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 14 hours ago

























      asked yesterday









      Tim

      25.6k74245449




      25.6k74245449






















          1 Answer
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          active

          oldest

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          2














          In bash speak, "login shell" means a shell invoked with the -l flag, or where the first character of argument 0 begins with a -. (See man bash INVOCATION section).



          In your example, you can see process 31728 is called -bash, so begins with a - and so was invoked as a login shell.



          logout simply exits a login shell.



          So if you run bash -l and then logout you'll find yourself back at the calling shell.



          $ echo $$
          
32145
          
$ bash -l
          
bash-4.2$ logout
          
$ echo $$
          
32145


          Now the login program (process 30488 in your example) waits for the child shell to exit, then performs some cleanup actions (eg wtmp, utmp) then exits. That is why you no longer see this process after running logout.






          share|improve this answer























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            In bash speak, "login shell" means a shell invoked with the -l flag, or where the first character of argument 0 begins with a -. (See man bash INVOCATION section).



            In your example, you can see process 31728 is called -bash, so begins with a - and so was invoked as a login shell.



            logout simply exits a login shell.



            So if you run bash -l and then logout you'll find yourself back at the calling shell.



            $ echo $$
            
32145
            
$ bash -l
            
bash-4.2$ logout
            
$ echo $$
            
32145


            Now the login program (process 30488 in your example) waits for the child shell to exit, then performs some cleanup actions (eg wtmp, utmp) then exits. That is why you no longer see this process after running logout.






            share|improve this answer




























              2














              In bash speak, "login shell" means a shell invoked with the -l flag, or where the first character of argument 0 begins with a -. (See man bash INVOCATION section).



              In your example, you can see process 31728 is called -bash, so begins with a - and so was invoked as a login shell.



              logout simply exits a login shell.



              So if you run bash -l and then logout you'll find yourself back at the calling shell.



              $ echo $$
              
32145
              
$ bash -l
              
bash-4.2$ logout
              
$ echo $$
              
32145


              Now the login program (process 30488 in your example) waits for the child shell to exit, then performs some cleanup actions (eg wtmp, utmp) then exits. That is why you no longer see this process after running logout.






              share|improve this answer


























                2












                2








                2






                In bash speak, "login shell" means a shell invoked with the -l flag, or where the first character of argument 0 begins with a -. (See man bash INVOCATION section).



                In your example, you can see process 31728 is called -bash, so begins with a - and so was invoked as a login shell.



                logout simply exits a login shell.



                So if you run bash -l and then logout you'll find yourself back at the calling shell.



                $ echo $$
                
32145
                
$ bash -l
                
bash-4.2$ logout
                
$ echo $$
                
32145


                Now the login program (process 30488 in your example) waits for the child shell to exit, then performs some cleanup actions (eg wtmp, utmp) then exits. That is why you no longer see this process after running logout.






                share|improve this answer














                In bash speak, "login shell" means a shell invoked with the -l flag, or where the first character of argument 0 begins with a -. (See man bash INVOCATION section).



                In your example, you can see process 31728 is called -bash, so begins with a - and so was invoked as a login shell.



                logout simply exits a login shell.



                So if you run bash -l and then logout you'll find yourself back at the calling shell.



                $ echo $$
                
32145
                
$ bash -l
                
bash-4.2$ logout
                
$ echo $$
                
32145


                Now the login program (process 30488 in your example) waits for the child shell to exit, then performs some cleanup actions (eg wtmp, utmp) then exits. That is why you no longer see this process after running logout.







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited yesterday









                Stephen Kitt

                163k24362440




                163k24362440










                answered yesterday









                Stephen Harris

                24.4k24477




                24.4k24477






























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