It is useful to understand how the shell behaves. The common bash shell is intended to be a superset of the standard Unix shell, which is described in this standards document http://pubs.opengroup.org/onlinepubs/9699919799/utilities/V3_chap02.html -- though by web search, you will be able to find other explanations of the unix shell which may be easier to read and understand. Anyhow, we are interested in how simple commands are parsed and executed, in order to understand why your original alternative did not work. http://pubs.opengroup.org/onlinepubs/9699919799/utilities/V3_chap02.html#tag_18_09_01 Notice the following statement about a simple command that has both a variable assignment and a command name: If no command name results, or if the command name is a special built-in or function, variable assignments shall affect the current execution environment. Otherwise, the variable assignments shall be exported for the execution environment of the command and *shall not affect the current execution environment* except as a side-effect of the expansions performed in step 4. In this case it is unspecified: (emphasis mine) This requirement of the standard is why your original command did not work as you desired. Here are some simple uses of 'echo' that show several cases of how shell variables and environment variables behave -- the last one is particularly surprising to someone not familiar with the rules of the Unix shell: $ unset unsetvar $ echo "${unsetvar-unset}" unset $ var="x y z" $ echo "${var-unset}" x y z $ unset ENVVAR $ ENVVAR="x y z" echo "${ENVVAR-unset}" unset $ ENVVAR="x y z"; export ENVVAR $ ENVVAR="a b c" echo "${ENVVAR-unset}" x y z Jeff
