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    What about arrays? In Bash version "GNU bash, version 4.1.10(4)-release (i686-pc-cygwin)" 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? Commented Nov 27, 2012 at 9:29
  • @TimFriske ${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. Commented Nov 27, 2012 at 10:56
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    Regarding the topic "In bash, arrays behave in a different and surprising way." the behavior can be explained from the bash(1) manpages, section "Arrays". It states that "Referencing an array variable without a subscript is equivalent to referencing the array with a subscript of 0.". Thus if neither a 0 index nor a key is defined as it is true for a=(), ${a+1} correctly returns nothing. Commented Nov 27, 2012 at 21:57
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    @TimFriske I know that the bash implementation conforms to its documentation. But treating an empty array like an undefined variable is really strange design. Commented Nov 27, 2012 at 22:19
  • I prefer the result from: How to check if a variable is set in Bash? --> The Right Way ... I strikes a chord with how I thing this ought to work. Dare I say, Windows has a defined operator. Test could DO that; it can't be difficult (um ....) Commented Apr 11, 2016 at 14:48