Inconsistent string comparison operators n and z

[Thibault, Daniel]

Configuration Information [Automatically generated, do not change]:
Machine: x86_64
OS: linux-gnu
Compiler: gcc
Compilation CFLAGS:  -DPROGRAM='bash' -DCONF_HOSTTYPE='x86_64' 
-DCONF_OSTYPE='linux-gnu' -DCONF_MACHTYPE='x86_64-pc-linux-gnu' 
-DCONF_VENDOR='pc' -DLOCALEDIR='/usr/share/locale' -DPACKAGE='bash' -DSHELL 
-DHAVE_CONFIG_H   -I.  -I../bash -I../bash/include -I../bash/lib  
-D_FORTIFY_SOURCE=2 -g -O2 -fstack-protector --param=ssp-buffer-size=4 -Wformat 
-Wformat-security -Werror=format-security -Wall
uname output: Linux sds-dut-vb 3.9.3 #1 SMP Mon Mar 24 18:48:39 EDT 2014 x86_64 
x86_64 x86_64 GNU/Linux
Machine Type: x86_64-pc-linux-gnu

Bash Version: 4.2
Patch Level: 25
Release Status: release

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:

$ 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

They should be equivalent but are not: -z correctly detects the process's
presence or absence, while -n returns true even when the process is not
running.

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

Repeat-By:
     See the examples above.