If I understand why you want to do it can't work without using a file or fifo.On Thu, May 3, 2012 at 12:16 AM, Dan Douglas <address@hidden> wrote:On Wednesday, May 02, 2012 11:21:03 PM Peng Yu wrote:On Wed, May 2, 2012 at 11:12 PM, Dan Douglas <address@hidden> wrote:On Wednesday, May 02, 2012 10:37:04 PM Peng Yu wrote:Hi, (export XXX=a) echo $XXX In the above code, the echo command will print nothing. I'm wondering if there is a way make the variable set in () accessible outside.Any time there's a subshell involved, you're effectively dealing with a separate independent process with its own shell execution environment. There's no way to make anything inside directly accessible to the outside. The only ways to interact with a subshell are the same as with any other external command - mostly that's via standard streams. Lots of examples here: http://mywiki.wooledge.org/BashFAQ/002 Also remember that export is a POSIX special builtin, and Bash (thankfully, IMO) doesn't follow the spec on this by default. ~ $ ( bash -c 'xxx=a export a; echo "$xxx"' ) ~ $ ( bash --posix -c 'xxx=a export a; echo "$xxx"' ) a The above isn't related to a subshell issue. Is there a specific problem you're trying to solve?I need pass something computed in () to the enclosing environment. The walkaround that I found was to use a tempfile.Normally people only suffer this problem in unavoidable cases like when using pipes or process substitutions. Ideally the explicit subshell should be avoided if at all possible, e.g. ,using a function and local variables for encapsulation.http://lists.gnu.org/archive/html/help-bash/2012-02/msg00000.html See the example of holdcpu.sh in the above mail (holdcpu.sh was mistyped to holdcup.sh there.)There are many, many other workarounds depending on the situtation. Just reading the vars through a pipe usually works. If there's absolutely no alternative than to move around variables with temporary files, a hack I commonly use (and wouldn't commonly advertise to others) is to write to an empty herestring. This shows both: shopt -s lastpipe; { { x=5 let x+=y=5; declare -p x; printf %s "$y" >/dev/stdin; } | . /dev/stdin; declare -p x; IFS= read -rd -- y; declare -p y; } <<<'' As I recall, looking at this in strace, bash doesn't actually fsync, and up to a certain size this is pretty fast (even if TMPDIR doesn't happen to be on a tmpfs) -- Dan Douglas
your not trying to get information from sub process as much as from a concurrent co-process. In fact if it understand it correctly from multiple co-processes.
I played with something similar a while ago, but its not too pretty.
https://github.com/dethrophes/Experimental-Bash-Module-System/blob/master/bash/JobQueue.sh
I mainly just did it to bet a better feel for bash, I think I'd do it a lot differently now, but it might give you some ideas.
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