In bash, are [[ $variable ]] and [[ -n $variable ]] completely equivalent? It appears to be the case judging by the output below, but I see both forms of usage prevalent in shell scripts.
$ z="abra"
$ [[ $z ]]
$ echo $?
0
$ [[ -n $z ]]
$ echo $?
0
$ z=""
$ [[ $z ]]
$ echo $?
1
$ [[ -n $z ]]
$ echo $?
1
$ unset z
$ [[ $z ]]
$ echo $?
1
$ [[ -n $z ]]
$ echo $?
1
[ "$var" ] is equivalent to [ -n "$var" ] in bash and most shells nowadays. In other older shells, they're meant to be equivalent, but suffer from different bugs for some special values of "$var" like =, ( or !.
I find [ -n "$var" ] more legible and is the pendant of [ -z "$var" ].
[[ -n $var ]] is the same as [[ $var ]] in all the shells where that non-standard ksh syntax is implemented.
test "x$var" != x would be the most reliable if you want to be portable to very old shells.
According to Test for non-zero length string in bash: [ -n “$var” ] or [ “$var” ], yes, they are equivalent.
They are equivalent even quoting the name of the variable.
Important to notice: the name of the question I cite refers only to [, but the answer considers both [ and [[.
[[]] is not the equivalent of [] :)
[[ -n $(echo -ne "\0") ]]; echo $?and[ -n $(echo -ne "\0") ]; echo $?differ[ -n ], the same[ -n -n ]. In shells other than zsh, command (even builtin) arguments or shell variables can't contain NUL characters.