Ok, I accept your points, but please read on and decide after that.
> to do anything but assign a value to `var'. Very few people, when asked,
> would say that it were more intuitive to cause a variable named `bar' to
> spring into existence with the value 7. If you want nameref behavior, you
> have to explicitly declare it.
But why is nameref behavior in the evaluation part activated by default
and not for the assignments? I think that one should
either make nameref behavior the default case for both,
assignment and evaluation, or the non-default case for both.
In the latter case, where namerefs are off for both things, the lines (1) and
(2)
1) var=bar; bar=3; echo ((var))
2) var=bar; ((var=3)) or var=bar; bar=0; ((var++))
should lead to an error message!
However the current status quo is, that bash happily does (1) as you’d
expect but gives out no error in (2) and overwrites (1).
> That is variable indirection. It has superficially similar effects to what
> we are discussing, which is arithmetic expansion; however, it has nothing
> to do with the behavior of (( )) or $(( )).
Again, I never wrote, that, from a technical point of view, it has something to
do
with (()) or $(()). Merely the visible behavior is similar in the cases I
showed.
> Am 24.04.2017 um 01:43 schrieb Chet Ramey <chet.ra...@case.edu>:
>
> On 4/23/17 4:25 PM, Florian Mayer wrote:
>>> That's not a reasonable expectation.
>> Why not? Why is it not reasonable to expect an intuitive
>> result from (())? The most intuitive thing, in my opinion,
>> would be to use nameref for side effects by default, because in order
>> to get a value from an id, (()) already follows namerefs.
>
> The thing that makes that result intuitive for you is your opinion about
> how things should work. That's fine. I don't happen to share your opinion
> of what is "intuitive" in this case.
>
> However, it is unreasonable to expect
>
> var=bar
> (( var=7 ))
>
> to do anything but assign a value to `var'. Very few people, when asked,
> would say that it were more intuitive to cause a variable named `bar' to
> spring into existence with the value 7. If you want nameref behavior, you
> have to explicitly declare it.
>
>
>>> It's not indirection, and I am not sure why you show the completely
>> I was mentioning that, not because I use „a mental model“ that falsely
>> unifies
>> both things, but because I wanted to point out that there exists something
>> with a similar behavior.
>>
>> And if ${!<varid>} does not portray some kind of indirection, what do you
>> call it then?
>
> That is variable indirection. It has superficially similar effects to what
> we are discussing, which is arithmetic expansion; however, it has nothing
> to do with the behavior of (( )) or $(( )).
