Subject: Re: report

From: Rob Schreiber <[email-masked]>
Date: Mon, 3 Aug 92 10:14:17 PDT

Mary Zosel reports this from the meeting:

	Pijush's example ...
	too surprising that C can be mapped where we have

	common A(100), B(100), C(100)  Dim X(100) EQV (X, A(50)) in routine 1
	common Q(50), R(150), C(100) Dim X(100) EQV (X,R) in routine 2

According to the draft I sent before the meeting, the common block is SEQUENTIAL,
becouse it fails to have the same set of distribution components in all instances.
In the first, there are two distribution components, of lengths 200, 100.
In the second, there are three, of lengths 50, 150, 100.   SO, C cannot
be distributed, since it is a sequential variable by virtue of its membership
in a sequential common block.

The rule is that a distribution component contains ALL of each and every variable
that is a part of it, so all of A and B are included in the component formed
when X is equivalenced to A(50) (first instance above).

	BUT if we just have
	common A(100), B(100), C(100)
	common Q(50), R(150), C(100) 
	in two different routines - C cannot be mapped, becaused it isn't
	saved by the messy equivalence.

Right, it can't be mapped.   There is no way to save this one, sorry.
Next case.

Rob