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Brian,
	With regard to John Ramsdell's question about parameter
expansion:  The syntax
	${param:?[value]}
is supposed to substitute "param" if it is not null, otherwise it
should write "value" to stderr.  If value is omitted, then sh writes
	param: parameter null or not set

This is paraphrased from UNIX Shell Programming, by Kochan and Wood.
As you can see from the transcript below, sh will also accept
${param?[val]} (e.g. without the internal ":").  sh only writes the
"value" string if the variable "param" is not set.  If the variable is
set, sh attempts to execute the contents of param as a command.

Script started on Fri Jan  5 14:41:09 1990
bash$ sh
$ set
ARCH=sun3
BASH=/bin/sh
..
..many other variables...
..
$ echo $abc

$ ${abc?}
abc: parameter null or not set
$ ${abc:?}
abc: parameter null or not set
$ ${abc?"What a crock"}
abc: What a crock
$ ${abc?ls}
abc: ls
$ abc=
$ ${abc?}
$ ${abc?ls}
$ abc=zip
$ ${abc:?}
zip: not found
$ ${abc:?"What a crock, II"}
zip: not found
$ : ${abc:?}
$ : ${abc:?"What now?"}
$ exit

script done on Fri Jan  5 14:43:23 1990

Bash, on the other hand, doesn't seem to handle this.  If abc is not
set, the syntax
	${abc:?}
will exit the shell, and 
	${abc:?"Help"} will echo Help and *then* exit the shell.  In
neither case should bash exit the current shell.  If abc=ls, then bash
behaves correctly, e.g. executes ls, and remains in the current shell.

I verified this behavior in both 1.03 and 1.04 on a Sun-3 running
SunOS 4.0.3

Thanks, 
Jeff Barratt