Shell script silently exits. Why?

[Ralf Friedl]

Arnout Vandecappelle wrote:
>   I guess you have a 'set -e' somewhere in your script?
>
>   let returns the value it computed, so if $foo is different from 0, it will
> return non-0. This causes the shell to exit.
>
>   It's POSIX behaviour.
Bash "let" returns 1 if there is a sytax error within the expression or 
the expression is zero, 0 otherwise. So it is basically a "not" 
operator, the opposite of what you describe.
I'm not sure there is a good reason for "let" to return an exit status 
depending on the value assigned, whether 0 is returned for a zero or a 
non-zero value assigned.

But bash does exit when let returns 1 and "-e" is set, so that might be 
the reason.

Regarding POSIX behaviour, the list of shell built-ins at 
http://pubs.opengroup.org/onlinepubs/007904875/idx/sbi.html doesn't even 
mention "let", and "let" can only implemented as a built-in as it must 
assign a value to a shell variable.