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

[Greg Wooledge]

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:

      ((expression))
           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
           "expression".

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:

http://pubs.opengroup.org/onlinepubs/009695399/basedefs/xbd_chap04.html#tag_04_21

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