Re: $(()): "?:": false "assignment to non-variable"

[Steffen Nurpmeso]

Andreas Schwab wrote in
 <mvmpmiblsvn.fsf@suse.de>:
 |On Jul 09 2022, Steffen Nurpmeso wrote:
 |>   $ bash -c 'I=3; echo "$((1?(I*=I):I+=I))";echo $I'
 |
 |The third operand of ?: cannot contain an assignment expression, thus,
 |like in C, this is parsed as `(1?(I*=I):I)+=I'.
 ..
 --End of <mvmpmiblsvn.fsf@suse.de>

Chet Ramey wrote in
 <52011ddb-824b-73d0-f1c8-c295ffa1fe59@case.edu>:
 |On 7/9/22 5:58 PM, Steffen Nurpmeso wrote:
 ...
 |>    $ bash -c 'I=3; echo "$((1?(I*=I):(I+=I)))";echo $I'
 |
 |The parens override precedence.
 |
 |>    $ bash -c 'I=3; echo "$((1?(I*=I):I+=I))";echo $I'
 |>    bash: line 1: 1?(I*=I):I+=I: attempted assignment to non-variable \
 |>    (error token is "+=I")
 |
 |The normal rules of precedence apply, and the conditional expression on the
 |rhs of the `:' can't contain an assignment, since the assignment operator
 |has higher precedence.
 |
 |
 |>    $ bash -c 'I=3; echo "$((1?I*=I:(I+=I)))";echo $I'
 |
 |The parens override precedence.

Sorry for the noise then.
And thanks for the explanation.  (Assignment has lower.)

--steffen
|
|Der Kragenbaer,                The moon bear,
|der holt sich munter           he cheerfully and one by one
|einen nach dem anderen runter  wa.ks himself off
|(By Robert Gernhardt)