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Re: set -e yet again (Re: saving bash....)

From: Greg Wooledge
Subject: Re: set -e yet again (Re: saving bash....)
Date: Fri, 12 Aug 2011 15:42:38 -0400
User-agent: Mutt/

On Fri, Aug 12, 2011 at 12:19:59PM -0700, Linda Walsh wrote:
> Under -e, it would fail on the 'let' statement

This is one of the cases I mention on http://mywiki.wooledge.org/BashFAQ/105

> "-e" -- in a ****WELL DESIGNED PROG***, where errors are caught,
> shouldn't cause a otherwise working program to fail.

In my opinion, a well-designed program will not use set -e.  We're
going to have to "agree to disagree" about this apparently.

> The statements:
>   ((a=0))
> and
>   let a=0
> are both assignment statements (not 'commands')

I was going to just let this drop, until I saw this part.  This is
simply incorrect, and can't be ignored.  ((a=0)) is *not* an assignment.
It is an arithmetic command.  More specifically, it is documented in
the bash manual under Compound Commands, with this description:

           The expression is evaluated according to the rules described
           below under ARITHMETIC EVALUATION.  If the value of the
           expression is non-zero, the return status is 0; otherwise the
           return status is 1.  This is exactly equivalent to let

So, it is a command -- a compound command.  It does what the manual
says it does.  It is not, however, an *assignment*, which is a very
specific, different, thing.  Assignments are defined here:


a=$((0)) is an assignment.  ((a=0)) is not.

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