Re: declare XXX=$(false);echo $?

[David]

On Fri, 2 Dec 2022 at 21:29, Ulrich Windl
<Ulrich.Windl@rz.uni-regensburg.de> wrote:

> Surprisingly "declare XXX=$(false);echo $?" outputs "0" (not "1")
> There is no indication in the manual page that "declare" ignores
> the exit code of commands being executed to set values.

The above is not surprising at all.

'declare' is a builtin command. It succeeded.

$ help declare | tail -n 3
    Exit Status:
    Returns success unless an invalid option is supplied or a variable
    assignment error occurs.

'man bash' explains this comprehensively.

    The return value is 0 unless an invalid option is encountered, an
    attempt is made to define a function using ``-f foo=bar'', an attempt
    is made to assign a value to a readonly variable, an attempt is made to
    as‐ sign a value to an array variable without using the compound as‐
    signment syntax (see Arrays above), one of the names  is  not  a valid
    shell variable name, an attempt is made to turn off read‐ only status
    for a readonly variable, an attempt is made to  turn off array status
    for an array variable, or an attempt is made to display a non-existent
    function with -f.

Also, $(false) probably does not produce the value
of $XXX that you expect. Try running 'echo $(false)'.