Re: making a variable local counts as a statement as per $?

On Tuesday, February 24, 2026, Luca Saiu <[email protected]> wrote:

> Hello.
>
> I was surprised to discover that the following two lines behave
> differently:
>
>
>   f () { a=$(false); echo $?; }; f         # This prints 1 as expected
>
>   f () { local a=$(false); echo $?; }; f   # This prints 0
>
> The way I interpret this is that making the variable local “counts as a
> statement”, and overwrites the value of $? with 0, since making the
> variable local succeeds.  Is this correct?  Is this the intended
> semantics?  I certainly find it counterintuitive, and cannot see it
> being documented.


local is a command like everything else, `echo a=$(false)' returns 0 as
well. You can always do

local a
a=$(false)

instead.


>
> Thanks, and thanks for your precious work on Bash.
>
> --
> Luca Saiu      https://ageinghacker.net
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>
> I support everyone's freedom of mocking any opinion or belief, no
> matter how deeply held, with open disrespect and the same unrelented
> enthusiasm of a toddler who has just learned the word "poo".
>


-- 
Oğuz