Hello lejeczek,
On Wed, 19 Apr 2023 08:10:16 +0200 lejeczek peljasz@yahoo.co.uk wrote:
On 19/04/2023 08:04, wwp wrote:
Hello lejeczek,
On Wed, 19 Apr 2023 07:50:29 +0200 lejeczek via CentOS centos@centos.org wrote:
Hi guys.
I cannot wrap my hear around this:
-> $ unset _Val; test -z ${_Val}; echo $? 0 -> $ unset _Val; test -n ${_Val}; echo $? 0 -> $ _Val=some; test -n ${_Val}; echo $? 0
What is this!? How should two different, opposite tests give the same result Is there some bash option which affects that and if so, then what would be the purpose of such nonsense?
Surround ${_Val} with double quotes (as you should) and things will be different:
$ unset _Val; test -n "${_Val}"; echo $? 1
Now you get it? :-)
I don't know, am not sure, I remembered it differently, did not think enclosing quotes were necessary(always?) for that were {}
{} does not prevent this (at least not in bash):
$ FOO="a b"
$ test -z $FOO bash: test: a: binary operator expected
$ test -z ${FOO} bash: test: a: binary operator expected
Because after $FOO or ${FOO} variable expansion, bash parsed: test -z a b 'b' is unexpected, from a grammar point of view.
Quoting is expected, here: $ test -z "$FOO" <no error>
When FOO is unset, apparently it's a different matter, where you end up with $?=0 in all unquoted -n/-z cases, interestingly. I could not find this specific case in the bash documentation. That may not be portable to other shells, BTW. I only use {} when necessary (because of what bash allows to do between {}, plenty!, or when inserting $FOO into a literal string that may lead the parser to take the whole string for a variable name: echo $FOObar != echo ${FOO}bar).
Regards,