Re: weird behavior of set -e

[Linda Walsh]

mhenn wrote:

> Am 24.06.2011 11:51, schrieb Harald Dunkel:
> > Hi folks,
> >
> > A colleague pointed me to this problem: If I run
> >         ( set -e; ( false; echo x ) )
> > in bash 4.1.5, then there is no screen output, as
> > expected. If I change this to
> >         ( set -e; ( false; echo x ) || echo y )
> > then I get "x" instead of "y". How comes?
> > Any helpful comment would be highly appreciated.
> >
> > Harri
>
> Hey Harald,
> Thus, "false" is executed, but the command sequence is not cancelled
> because of the "||". Then, after the ";" "echo x" is executed, which
> gives a 0 exit status, which leads to "echo y" never to be executed
> because of evaluation shortcut of "||". For this, see also:
>         
>         man bash | less -p 'The \&\&'

---

        Has this behavior changed in more recent versions?

The above description seems to be rather non-intuitive, given that ( ) should
mean it is evaluated in a sub-shell process, -- thus whether or not it aborts
after 'false', should be independent of the context.

If it was (set -e; { false ; echo x;} || echo y )
Then I could see {} being a type of parenthesis that allows evaluation
linearly, in the same process and would honor such dependencies.

Or is it the case that the "false;echo x", isn't executed in it's own separate 
process?

Of course who knows what it does now with the seeming concerted effort to kill
any use or usefulness for the -e switch...(sigh)...

perl has -w and use strict;  If you don't want -e to do it, maybe the 
"language"** should offer alternatives.

**-- I suppose that's the rub -- is it a language as in programming language,
or is it just supposed to be for the most primitive scripting needs.
Adding integers, arrays and hashes sure makes it seem more like a language than
the latter.