I've read what's listed in Bibliography regarding unset, declare, local and "Shell Functions".
My version of Bash is
GNU bash, version 5.1.16(1)-release (x86_64-pc-linux-gnu)
var='outer'
declare -p var
unset var
declare -p var
function foo {
echo \""I'm in"\"
local var='inner'
declare -p var
unset var
declare -p var
}
foo
(the strange quoting around I'm in is there just to preserve syntax highlithing in the following block quote.) This prints
declare -- var="outer"
bash: declare: var: not found
"I'm in"
declare -- var="inner"
declare -- var
There is a difference in unsetting a global variable and unsetting a local variable inside a function. In the former case, the variable is removed. In the latter, the variable stays declared (but unset).
Is there a way to completely remove a local variable inside a function (before the function returns)? That is that the output would be
declare -- var="outer"
bash: declare: var: not found
"I'm in"
declare -- var="inner"
bash: declare: var: not found
That would be useful to test if a variable is non existing inside a function, like in
function foo {
local var
while
declare -p var
do
echo "$var"
((var++))
[[ "$var" -eq 4 ]] \
&& unset var
done
}
This code loops indefinitely, printing
declare -- var="1"
1
declare -- var="2"
2
declare -- var="3"
3
declare -- var
declare -- var="1"
1
declare -- var="2"
2
declare -- var="3"
3
declare -- var
[omissis]
Is there anything wrong in wanting to checkchecking if a variable exists using declare -p?
The only mention I found in the Bash reference manual about the difference between unsetting a global variable and a local one is
If a variable at the current local scope is unset, it will remain so (appearing as unset) until it is reset in that scope or until the function returns. (ref. Bash reference manual - sec. "Shell Functions")
where the word appearing is the only clue.
Bibliography
- Bash reference manual sec. "4 Shell Builtin Commands"
- What does unset do?