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.
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 $(( )).
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