>>> 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
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..
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
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
