On Wed, Jan 27, 2010 at 03:07:40PM +0200, Pierre Gaston wrote: > On Wed, Jan 27, 2010 at 2:49 PM, Sharuzzaman Ahmat Raslan > <sharuzza...@gmail.com> wrote: > > Somehow, the backtick for foo() execute the function, echoing the correct > > output, but fails to set the variable $gang to the correct value. Because of > > that, the function bar() did not echoing anything because the variable $gang > > is null. > > `` and $() introduce subshells. The parent shell cannot be modified by > the subshells.
For hints on working around this, see http://mywiki.wooledge.org/BashFAQ/084 The bad news is that it's pretty much impossible to capture stdout from a function *and* set global variables at the same time. You need to choose one or the other, or use some file system object to store output.
