BASH question

[Dave Mason]

I wasn't aware of this new-fangled syntax:

    while ...; do
	...
    done < <(ls -l)

but it does appear to work.  If you want something that works with older
shells, you can do something like:

	ls -l | (
	   while ...; do
		 x=...
	   done
	   ... $x ...
	   )

where everything in the () runs in a separate shell with stdin set to
the pipe.

../Dave
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