On 06/09/2014 07:11 AM, Thibault, Daniel wrote: > Description: > The string comparison operators -n and -z are designed to be mutually > complementary. ! -z should be interchangeable with -n and ! -n should be > interchangeable with -z. But such is not the case. Consider these lines:
The bug is not in bash, but in your script. > > $ if [ -z `pgrep pname` ]; then echo "not r" ; else echo "r" ; fi Insufficient quoting. If `pgrep pname` produces no output, then you are running "[ -z ]", which has entirely different semantics (always true, because the single-argument string -z is not empty). You MEANT to write "[ -z "`pgrep pname`" ]"; (or modernize your script, and use "[ -z "$(pgrep pname)" ]", or use a bash-ism, and use "[[ -z $(pgrep pname) ]]") > $ if [ ! -z `pgrep pname` ]; then echo "r" ; else echo "not r" ; fi > $ if [ -n `pgrep pname` ]; then echo "r" ; else echo "not r" ; fi > $ if [ ! -n `pgrep pname` ]; then echo "not r" ; else echo "r" ; fi Same problem of insufficient quoting here. Just as [ -z ] is always true, so is [ -n ]; this behavior is required by POSIX. > Turns out this is how the script needs to be written to work correctly: > > $ if [ -z "`pgrep pname`" ]; then echo "not r" ; else echo "r" ; fi > $ if [ ! -z "`pgrep pname`" ]; then echo "r" ; else echo "not r" ; fi > $ if [ -n "`pgrep pname`" ]; then echo "r" ; else echo "not r" ; fi > $ if [ ! -n "`pgrep pname`" ]; then echo "not r" ; else echo "r" ; fi Glad you figured out your quoting bug. -- Eric Blake eblake redhat com +1-919-301-3266 Libvirt virtualization library http://libvirt.org
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