On 12/09/2010 10:52 AM, Dominic Raferd wrote: > Description: > > $ val=0; let val++; echo $val,$?; unset val > 1,1
Not a bug. > > see the error code 1. Setting any other start value (except undefined) > for val does not produce this error, the problem occurs for let val++ > and let val-- if the start value is 0. let intentionally returns status 1 if the value was 0; and status > 1 if there was an error. Why? So you can do loops such as: countdown=10 while let countdown--; do ... ; done > Why does this happen? Is it 'by design'? Yes. The same as for 'expr' which is standardized by POSIX to have the same behavior. -- Eric Blake ebl...@redhat.com +1-801-349-2682 Libvirt virtualization library http://libvirt.org
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