Emanuele Torre writes:
> `declare -f "something="' fails with the following error:
>
> $ declare -f 'a=x'
> bash: declare: cannot use `-f' to make functions
> That error is not very useful. Bash makes `declare -f' fail with that
> error when an argument looks like an assignment.
It's an
On Dez 04 2022, Dale R. Worley wrote:
> In default mode, you actually can do
> $ function a=b { printf hi\\n; }
> though you can't execute it:
> $ a=b foo
> bash: foo: command not found
You just have to quote any part of the function name upto the equal sign
to stop if from being inte
Dale, thanks for explaining.
So basically the behavior is as documented (not a bug), but the design
decision was poor:
declare a
a=b
has a different semantic as
declare a=b
which I consider to be bad.
Ulrich
>>> "Dale R. Worley" schrieb am 02.12.2022 um 17:39 in
Nachricht <878rjpahfz..
>>> Martin D Kealey schrieb am 03.12.2022 um 02:26 in
Nachricht
:
...
> Found in the real code (intended to trigger a bug):
>> declare ERRORS=0 ARGS=$(getopt -o "$S_OPTS" -l "$L_OPTS" -n "$0" --
>> "$@")
>>
> if [ $? -ne 0 ]; then
>> usage
>> fi
>>
>
> That is a well-known a
