Date: Sun, 19 Jul 2009 16:32:28 -0400 From: Glen Barber <glen.j.barber@gmail.com> To: =?ISO-8859-1?Q?Romain_Tarti=E8re?= <romain@blogreen.org> Cc: stable@freebsd.org Subject: Re: Value of $? lost in the beginning of a function. Message-ID: <4ad871310907191332r2f933a33l36c121903bc0742f@mail.gmail.com> In-Reply-To: <20090719202638.GA85228@blogreen.org> References: <20090719202638.GA85228@blogreen.org>
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2009/7/19 Romain Tartière <romain@blogreen.org>:
> Hi!
>
> Simple test case:
>
> ----8<----------------------
> #!/bin/sh
> foo()
> {
> echo "\$?=$? \$1=$1"
> }
> false
> foo $?
> ----8<----------------------
>
> % sh foo.sh
> $?=0 $1=1
> % zsh foo.sh
> $?=1 $1=1
> % bash foo.sh
> $?=1 $1=1
>
> As you can see, the value of $? is « lost » when FreeBSD sh enters a
> function. Is this supposed to behave this way?
>
Hi.
I'm no expert at shell scripting, but my first presumption is that
since you have '#!/bin/sh' at the beginning of the script, it is
creating a new subshell, and overwriting the value. What happens if
you replace '#!/bin/sh' with '#!/usr/local/bin/zsh' ?
--
Glen Barber
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