On Wed, Jan 07, 2026 at 10:15:55 +0800, 王伟 wrote:
> local all_var_info=${ local;}
> declare -p all_var_info
> test_case6 ret----------
> declare -- all_var_info=""
The man page says:
This type of command substitution superficially resembles executing an
unnamed shell function: local variables are created as when a shell
function is executing, and the return builtin forces command to com-
plete; however, the rest of the execution environment, including the
positional parameters, is shared with the caller.
Since bash provides no way to print the local variables of your caller,
the curly-brace command substitution is simply not what you want here.
Continue using $( ) instead, since that does what you need. (Or write
local's output to a temp file and read it back in. That also works.)
