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Michal Privoznik
28482f2630
In shell, the following function doesn't echo '1' but '0':
func() {
local var=$(false)
echo $?
}
This is because '$?' does not refer to 'false' but 'local'. The
bash_builtins(1) manpage explains it well. And it also mentions
other commands behaving the same: export, declare and readonly.
Since it is really easy to miss this pattern, introduce a
syntax-check rule. Mind you, the following pattern (which passes
the rule) does check for the subshell exit code:
func() {
local var
var=$(false)
echo $?
}
Signed-off-by: Michal Privoznik <mprivozn@redhat.com>
Reviewed-by: Ján Tomko <jtomko@redhat.com>