[lug] How does bash "set -e" work?

[Jeffrey S. Haemer]

Rob,

Great puzzle. I saw this this morning, but couldn't dive into it because I
went to the Tanner all afternoon, then out to dinner and a movie. I was
certain someone else would have gotten to it before I returned, but maybe
you were all in the same bind. :-)

Much of the example is red herring. It's enough just to type, at a command
prompt, this pair:

$ (set -e; false; echo hello); echo $?

 1

$ (set -e; false; echo subshell) || false; echo $?
> subshell

0


In the first, the subshell fails, in the second, it succeeds.

The " || " is the culprit.

All the shells I have handy -- bash (with or without "--posix"), zsh, dash,
and the busybox shell (another ash variant) -- act this way. That's enough
for me to decide it's "correct" -- existing practice -- so my next question
is "What's the man page say about 'set -e'?"

*The shell does not exit  if  the  command that fails is [...] part of any
> command executed in a && or  ||  list except  the  command  following  the
> final && or ||*


Thus, if we swap the order, the subshell should fail, and exit after its
"false."

$ false || (set -e; false; echo subshell); echo $?
1

Sho' nuff.

This looks like an un-intuitive manifestation of an actual, useful feature.
By reflex, I start my shell scripts with *#!/bin/bash -eux* which catches a
variety of coding errors I often make.

I want to be able to write code like this:


#!/bin/bash -eux
[ -d foo ] || mkdir foo


without having the whole script die at the "[ -d foo ]" if the directory
doesn't exist.

Does that really mean it should it ignore failures when the "set -e"
happens *within* a subshell in that list? *A priori*? Not obvious, at least
to me. But *de facto*?  Looks like it's the traditional, if
slightly-cryptically-documented, behavior.

HTH.

-- 
Jeffrey Haemer <jeffrey.haemer at gmail.com>
720-837-8908 [cell], http://seejeffrun.blogspot.com [blog],
http://www.youtube.com/user/goyishekop [vlog]
*פרייהייט? דאס איז יאַנג דינען וואָרט.*
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