.zsh/functions/logout

   1 # -*- mode:sh -*-
   2
   3 # Process the exceptions which allow logging out while still having
   4 # attached jobs.
   5
   6 # This is weird . . $1 seems to contain "logout" because the "logout"
   7 # command has been aliased to "noglob logout" - and $0 is the noglob.
   8 # I can't believe that this is what is meant to happen.
   9 if [[ $1 = "logout" ]]
  10 then
  11         shift
  12 fi
  13 # If the checkjobs option is unset, just plain exit.  (This option
  14 # emerged in zsh 3.1.8.)
  15 if [[ ! -o checkjobs ]]
  16 then
  17     reallyexit ${1+"$@"}
  18 fi
  19 # Did I type logout twice in a row?  Note check for history number.
  20 if [[ $history[$#history] = (exit|(logout|lo)(|\ *)) || $lastlogout = $#history ]]; then
  21     reallyexit ${1+"$@"}
  22 fi
  23
  24 # Did I type jobs, then logout?
  25 if [[ $history[$#history] = (j(|obs)(|\ *)) ]]; then
  26     reallyexit ${1+"$@"}
  27 fi
  28
  29 lastlogout=$#history
  30
  31 # Emulate the code that protects against logout if I have
  32 # attached jobs.
  33 if [[ $#jobstates -gt 0 ]]; then
  34     # The first listed state is good enough to be a hint of my job states.
  35     echo "You have ${jobstates[${${(k)jobstates}[1]}]%%:*} jobs."
  36     return
  37 fi
  38
  39 # If we got here we're really exiting.
  40 reallyexit ${1+"$@"}