On Thu, Mar 11, 2004 at 07:58:15AM -0700, brook@biology.nmsu.edu wrote:
> I am trying to write a shell script that will construct a list of
> words while inside a loop.  Here is the logic I would have expected to
> work:
> 
>      #!/bin/sh
>      LIST=""
>      cat - | while true
       ^^^^^^^^^^^^^^^^^^
>      do
> 	 read WORD JUNK
> 	 if [ $? -ne 0 ]; then break; fi
> 	 LIST="$LIST $WORD"
> 	 echo "LIST:  $LIST"
>      done
>      echo "final LIST:  $LIST"
> 
> The problem is that the final value of $LIST is empty, though within
> the loop the value grows as expected.  That is, the assignment inside
> the loop seems to have no effect on the outermost $LIST.  Adding
> export does not seem to matter.
> 
> I must be missing something obvious about shell variables.  How can
> this be done?

Using a pipe here means you're launching the while loop in a sub-shell.
Hence, $LIST inside the loop is not the same as the external $LIST, and
get lost when the sub-shell exits.

The 'cat -' here is really useless, just remove it and 'read' will take
its input from the standard input anyway.

If you really need a cat here, you'll have to make the sub-shell return
a value on the standard output, and assign it, or something like that.

Quentin Garnier.