On Wed, Feb 27, 2008 at 11:46:13AM -0600, Les Mikesell alleged: > Stephen Harris wrote: > > >>Yes, but I'm looking for what happens before and after. Why does > >>unset foo > >>foo=bar >$foo > >>do something you might expect, but > >>unset foo > >>foo=bar echo $foo >$foo > >>doesn't? > > > >What would you expect the last to do? "foo=bar echo $foo" only sets "foo" > >inside the scope of the "echo" statement, so in the scope of ">$foo" the > >variable is unset. In the first case there's no command so the shell is > >evaluating left-to-right and it works. > > How does the shell know that there's no command without evaluating, and > if it evaluates left to right, why isn't the result the same? Actually > I found the answer to that, which is that > name=val command > is processed like > (name=val; command) > which also explains why the value isn't left lingering for the next > operation. If was like '(name=val; command)' then 'foo=bar echo $foo' would do what you want. 'name=val' is very different from 'name=val command'. They aren't parsed the same way and follow different rules. The first is a variable assignment. The second is a command execution with a supplied env list. 'name=val command' is not "evaluated from left-to-right". After all variable substitution, expansions, and word splitting is done, then the command is executed with the supplied env var. 'name=val >$name' is a weird case that I can't explain. It is not related to 'name=val command' because there is no command. It acts like two statements when syntacticaly it is one. It think it is a bug. Avoid it. -- Garrick Staples, GNU/Linux HPCC SysAdmin University of Southern California Please avoid sending me Word or PowerPoint attachments. See http://www.gnu.org/philosophy/no-word-attachments.html
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