I have a export function looking like:
mybuild()
{
(
set -e
make
echo "build okay"
)
}
I wish to use this function this way:
mybuild && do_other_stuff
But whatever (success or failure) make returns "build okay" is always printed
and do_other_stuff always gets executed, which is my expectation.
I have found below descriptions on bash manual.
-e Exit immediately if a pipeline (which may consist of a
single simple command), a subshell command enclosed in parentheses, or one of
the
commands executed as part of a command list enclosed by
braces (see SHELL GRAMMAR above) exits with a non-zero status. The shell does
not
exit if the command that fails is part of the command
list immediately following a while or until keyword, part of the test following
the if
or elif reserved words, part of any command executed in a
&& or || list except the command following the final && or ||, any command in
a
pipeline but the last, or if the command's return
value is being inverted with !. A trap on ERR, if set, is executed before the
shell
exits. This option applies to the shell environment and
each subshell environment separately (see COMMAND EXECUTION ENVIRONMENT above),
and
may cause subshells to exit before executing all the
commands in the subshell.
To my understanding, the purpose of these limitations is to prevent the shell
exiting if a command fails in conditional testing context such as if, elif,
while, isn't it?
My confusion is why the && operator disables `-e' effects in the subshell. It
does not make sense to continue executing remaining commands since I wish the
subshell to stop on error.
Or I am wrong and the results match the original designs?
BR
