For Bash versions prior to "GNU bash, Version 4.2" are there any equivalent alternatives for the -v option of the test command? For example:
shopt -os nounset
test -v foobar && echo foo || echo bar
# Output: bar
foobar=
test -v foobar && echo foo || echo bar
# Output: foo
Portable to all POSIX shells:
if [ -n "${foobar+1}" ]; then
echo "foobar is defined"
else
echo "foobar is not defined"
fi
Make that ${foobar:+1} if you want to treat foobar the same way whether it is empty or not defined. You can also use ${foobar-} to get an empty string when foobar is undefined and the value of foobar otherwise (or put any other default value after the -).
In ksh, if foobar is declared but not defined, as in typeset -a foobar, then ${foobar+1} expands to the empty string.
Zsh doesn't have variables that are declared but not set: typeset -a foobar creates an empty array.
In bash, arrays behave in a different and surprising way. ${a+1} only expands to 1 if a is a non-empty array, e.g.
typeset -a a; echo ${a+1} # prints nothing
e=(); echo ${e+1} # prints nothing!
f=(''); echo ${f+1} # prints 1
The same principle applies to associative arrays: array variables are treated as defined if they have a non-empty set of indices.
A different, bash-specific way of testing whether a variable of any type has been defined is to check whether it's listed in ${!PREFIX*}. This reports empty arrays as defined, unlike ${foobar+1}, but reports declared-but-unassigned variables (unset foobar; typeset -a foobar) as undefined.
case " ${!foobar*} " in
*" foobar "*) echo "foobar is defined";;
*) echo "foobar is not defined";;
esac
This is equivalent to testing the return value of typeset -p foobar or declare -p foobar, except that typeset -p foobar fails on declared-but-unassigned variables.
In bash, like in ksh, set -o nounset; typeset -a foobar; echo $foobar triggers an error in the attempt to expand the undefined variable foobar. Unlike in ksh, set -o nounset; foobar=(); echo $foobar (or echo "${foobar[@]}") also triggers an error.
Note that in all situations described here, ${foobar+1} expands to the empty string if and only if $foobar would cause an error under set -o nounset.
echo "${foobar:+1}" doesn't print 1 if declare -a foobar was previously issued and thus foobar is an indexed array. declare -p foobar correctly reports declare -a foobar='()'. Does "${foobar:+1}" only work for non-array variables?
${foobar+1} (without the :, I inverted two examples in my original answer) is correct for arrays in bash if your definition of “defined” is “would $foobar work under set -o nounset”. If your definition is different, bash is a bit weird. See my updated answer.
0 index nor a key is defined as it is true for a=(), ${a+1} correctly returns nothing.
defined operator. Test could DO that; it can't be difficult (um ....)
To sum up with Gilles' answer I made up my following rules:
[[ -v foobar ]] for variables in Bash version >= 4.2.declare -p foobar &>/dev/null for array variables in Bash version < 4.2.(( ${foo[0]+1} )) or (( ${bar[foo]+1} )) for subscripts of indexed (-a) and keyed (-A) arrays (declare), respectively. Options 1 and 2 don't work here.I use the same technique for all variables in bash, and it works, e.g.:
[ ${foobar} ] && echo "foobar is set" || echo "foobar is unset"
outputs:
foobar is unset
whilst
foobar=( "val" "val2" )
[ ${foobar} ] && echo "foobar is set" || echo "foobar is unset"
outputs:
foobar is set
foobar="" will then report that foobar is unset. No wait, I take that back. Really only tests if first element is empty or not, so it seems to be only a good idea if you know the variable is NOT an array, and you only care about emptiness, not definedness.
-visn't an option totest, but an operator for conditional expressions.optionto a commandtest -v, anoperatorto aconditional expressionand aunary test primaryfor[ ]. Don't mix English language with shell definitions.