Re: sh: "local" assignment from command loses exit status

Eric van Gyzen <eric_at_vangyzen.net> wrote:

> On 11/06/2014 12:30, Fabian Keil wrote:
> > Eric van Gyzen <eric_at_vangyzen.net> wrote:
> >
> >> In sh, if I use a single statement to declare a local variable and
> >> assign the output of a command to it, the exit status of that command is
> >> lost.  For example:
> >>
> >>     should_return_false() {
> >>         local var1=`false`
> >>     }
> >>
> >> The function should return non-zero, but it returns zero.
> > The function should return the return code of the last command.
> > In your example, the last command is "local".
> 
> Fair enough.  What about errexit?  The shell ran a command whose exit
> status was not tested, that status was failure, yet the shell did not exit.

That's unrelated to the "local", though. Compare:

fk_at_r500 ~ $sh -e -c 'true `false; echo "Not reached"`; echo Reached'
Reached

While it's not obvious from the man page[1], my assumption is that this
is intentional behaviour. The return code of the command substitution
subshell can't be checked in the parent shell, as $? belongs to the
command that gets the output as argument (in your case "local"),
thus counting this as an untested return code would be inconvenient.

Fabian

[1] It could be argued that the behaviour is documented as
"-e ... tends to behave in unexpected ways", though ...