bash - declaring a variable without assigning a value

Question

I'm trying to figure out how to declare a variable without assigning a value to it. According to the bash doc, that should be ok:

declare [-aAfFgilnrtux] [-p] [name[=value] ...]

Declare variables and/or give them attributes.

The "=value" bit is optional, but using "declare var" without an assignment doesn't seem to do anything.

#!/bin/bash

function check_toto_set() {
  if [ -z "${toto+x}" ] ; then
    echo toto not defined!
  else
    echo toto=$toto
    echo toto defined, unsetting
    unset toto
  fi
}

function set_toto() {
  declare -g toto
}

function set_toto_with_value() {
  declare -g toto=somevalue
}

check_toto_set
toto=something
check_toto_set
declare toto
check_toto_set
set_toto
check_toto_set
set_toto_with_value
check_toto_set

Basically I would expect to have "toto not defined!" just for the first "check_toto_set", and all the others should find toto being declared, even if empty but the ouput is:

toto not defined!
toto=something
toto defined, unsetting
toto not defined!
toto not defined!
toto=somevalue
toto defined, unsetting

I'm using bash 4.3.46 on ubuntu

echo $BASH_VERSION
4.3.46(1)-release

So am I misunderstanding something about declare, or am I testing if a variable is set the wrong way? (I'm using info from How to check if a variable is set in Bash? for that)

BTW, unset effectively undeclares a variable; it doesn't just remove the value. — Charles Duffy
– Charles Duffy, Commented Mar 9, 2017 at 22:01

Charles Duffy · Accepted Answer · 2017-03-09 22:00:05Z

3

You're testing if the variable is set (even to an empty value). That's different from whether it has been declared.

To determine whether it has been declared, you can use declare -p:

varstat() {
  if declare -p "$1" >/dev/null 2>&1; then
    if [[ ${!1+x} ]]; then
      echo "set"
    else
      echo "declared but unset"
    fi
  else
    echo "undeclared"
  fi
}

export -f varstat

bash -c 'varstat toto'                 # output: "undeclared"
bash -c 'declare toto; varstat toto'   # output: "declared but unset"
bash -c 'declare toto=; varstat toto'  # output: "set"

edited Mar 9, 2017 at 22:00

answered Mar 9, 2017 at 21:54

Charles Duffy

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4 Comments

XSen Over a year ago

This should work but somehow when i tried it worked in BASH 3 but not BASH4

Charles Duffy Over a year ago

@XSen, exported functions don't work across version boundaries (well, not that particular version boundary; the format was reworked due to shellshock). Thus, the shell in which you run export -f varstat needs to be the same major version as the one invoked by bash -c.

XSen Over a year ago

I'm just testing with "declare toto ; declare -p toto" in the shell, with no function. Somehow this doesn't work with 4.3.46(1)-release shipped with ubuntu LTS 16.04. But I've downloaded the sourcecode for bash 4.4.0 from gnu.org, built it and tried it in there, and it works.... so there is some funny stuff going in on with the packaged versions (didn't work at work on a redhat 4.1.x either...)

Charles Duffy Over a year ago

shrug. declareing isn't really very useful if you aren't adding flags (declare -x, declare -i, etc) -- I'm not immensely surprised to see a corner case. (Since variables in shell are global by default, inside a function it's necessary for declare foo to cause a change in state -- by marking it local -- but if declare -g foo is a noop, it has no real impact in terms of documented semantics).

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