#!/bin/bash
env
This spits out a bunch of environment variables. $0, $@, $_, etc. are not output. Is there a way to dump them out, similar to how env
or set
dumps the rest of the environment? I don't mean anything like echo $0
or something, I want to see all of the possible special variables in their current state without directly calling all of them.
Built-in variables are included in the output to the set
command. ($0
and $@
aren't variables at all, even special ones, so they aren't included -- but $_
, $BASH_ALIASES
, $BASH_SOURCE
, $BASH_VERSINFO
, $DIRSTACK
, $PPID
, and all the other things that are built-in variables are present).
$0
, $*
, $@
, etc. are not built-in variables; they are special parameters instead. Semantics are quite different (you can use declare -p
to print a variable's value, but not that of a special parameter; many built-in variables lose their special behavior if reassigned, whereas special parameters can never be the target of an assignment; etc).
http://wiki.bash-hackers.org/syntax/shellvars covers both built-in variables and special parameters.
If your goal is to generate a log of current shell state, I suggest the following:
(set; set -x; : "$0" "$@") &>trace.log
set
dumps the things that are actually built-in variables (including $_
), and the set -x
log of running : "$0" "$@"
will contain enough information to reproduce all special parameters which are based on your positional parameters ("$*"
, "$@"
, etc); whereas the output from set
will include all other state.