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
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
(=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.
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
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
unset both unsets and undeclares the variable. In
yash it may restore the variable from an outer scope.
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).