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.
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