Using a function locally to checks blank strings - bash

Question

Cannot understand why the function myfunc fails. I put a string in src and dst, and I get the message
src is blank
dst is blank
Here is the code
zblank ()
{
local s="$1"
declare -i f=1 num=0
[[ $s =~ ^[[:space:]]*$ ]] && num=1
return $num
}
myfunc
{
src="angio"
dst="angio"
(
function myzblank() { zblank "$#"; }
myzblank "$src" && printf '%s\n' "src is blank"
myzblank "$dst" && printf '%s\n' "dst is blank"
)
}

Answer 1

When you use return $num, the result is interpreted as an exit status.
Exit status 0 means success.
Every nonzero exit status means failure.
Thus, when you set num=1 only if [[ $s =~ ^[[:space:]]*$ ]], you're reporting success unless that happens, so the && branch is executed.
If you want zblank to return success only if the variable you pass it is blank, it should instead look like:
zblank() { [[ $1 =~ ^[[:space:]]*$ ]]; }
You don't need return: By default the return value of a function is that of the last command it ran.
You don't need $s: it doesn't make things any terser or clearer than $1.
You don't need declare -i: When a variable is used in an arithmetic context its value is treated as numeric even if it was a string previously.
And if you wanted it to return success only in the case where $1 is not all-spaces, you could write it as:
znonblank() { ! [[ $1 =~ ^[[:space:]]*$ ]]; }
...addition of a leading ! being the only change needed.
All that said -- you don't need bash extensions like [[ ]] to do this at all. POSIX sh is powerful enough for what you're doing here:
zblank() {
case $1 in
*[![:space:]]*) return 1;; # At least one non-space character exists
*) return 0;; # No non-spaces exist; success.
esac
}