7

Does declare -a A create an empty array A in bash, or does it just set an attribute in case A is assigned to later?

Consider this code:

set -u
declare -a A
echo ${#A[*]}
echo ${A[*]}
A=()
echo ${#A[*]}
echo ${A[*]}
A=(1 2)
echo ${#A[*]}
echo ${A[*]}

What should be the expected output?

In Bash 4.3.48(1) I get bash: A: unbound variable when querying the number of elements after declare. I also get that error when accessing all the elements. I know that later versions of Bash treat this differently. Still I'd like to know whether declare actually defines a variable (to be empty).

  • Well, you could try declare -a P; Q=(), then declare -p P Q or set | grep -w '^[PQ]' to show bash's idea of your variables. Does this satisfy you or are you looking for something deeper? – mosvy May 28 at 10:53
13

That depends whether the corresponding variable has already been declared in the current scope (top-level aka global or current function) before.

If it hasn't been declared in the current scope (and beware that in the top-level scope, the variable may have declared (and assigned) by importing it from the environment), then it declares it (makes it local to the function when in function scope), assigning it a type, but doesn't initialise it, not even to an empty list (declare -p a shows declare -a a, not declare -a a=() as it would if you had declared and/or assigned it with a=()).

If it had already been declared in the current scope (for instance because it was imported as a scalar variable from the environment when in the global scope), then declare -a a would try to convert it to an array.

If it was previously a scalar, then it becomes a ([0]=value-of-the-variable) array. If it was already an array, it is left untouched. If it was an associative array, it fails with a cannot convert associative to indexed array error.

Note that declare a would not convert an array or hash to scalar. bash would not be able to convert a hash/array to scalar anyway. You can use declare +aA a to force a scalar (that would fail with an error if the variable was previously a hash/array in the current scope).

In your case, the variable was probably not already declared in the current scope, so it ended up declared but not assigned which explains why trying to expand it fails under set -u.

That distinction between two declared and assigned/set states of a variable is not specific to bash. In POSIX sh, you can also export a variable or make it readonly without giving it a value.

$ sh -uc 'unset -v var; readonly var; : "$var"'
sh: 1: var: parameter not set

Note that unset both unsets and undeclares the variable. In bash, mksh and yash it may restore the variable from an outer scope.

In zsh, except in sh emulation, using typeset on a variable declares and sets it to an empty value if it was not already set or was set but from a different type (scalar vs array vs associative array).

  • So my conclusion is that when A is undeclared before declare -a A, then A still is no array (because otherwise it would be an empty array with 0 elements), but undefined. So what's the purpose of declare -a when I can create arrays by assignment (A=(), A[0]=...)? – U. Windl May 28 at 11:26
  • 1
    @U.Windl: The declare -a /declare -A are just to differentiate if you want to use the array with name following as an indexed or an associative array. Neither of them set/initialize values – Inian May 28 at 11:29
  • 2
    @U.Windl, it still declares it as a array so that for instance a=foo would do a[0]=foo and declare -p a would show it as an array. But the main usage of declare in in function to make the function local to the function. It's like for export, it doesn't assign it but remembers the export attribute in case the variable is assigned later. – Stéphane Chazelas May 28 at 11:35

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