2009/7/19 Romain Tartière <rom...@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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