Date: Fri, 7 Nov 2014 14:42:30 +0100 From: Fabian Keil <freebsd-listen@fabiankeil.de> To: Eric van Gyzen <eric@vangyzen.net> Cc: Freebsd current <freebsd-current@freebsd.org> Subject: Re: sh: "local" assignment from command loses exit status Message-ID: <63396928.4e2735a2@fabiankeil.de> In-Reply-To: <545BB08D.4020402@vangyzen.net> References: <545BADDE.5050605@vangyzen.net> <74dd7d12.793ea654@fabiankeil.de> <545BB08D.4020402@vangyzen.net>
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Eric van Gyzen <eric@vangyzen.net> wrote:
> On 11/06/2014 12:30, Fabian Keil wrote:
> > Eric van Gyzen <eric@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=3D`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".
>=20
> 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 exi=
t.
That's unrelated to the "local", though. Compare:
fk@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 ...
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