Check double write read order
authorMathieu Desnoyers <mathieu.desnoyers@polymtl.ca>
Mon, 6 Apr 2009 20:30:04 +0000 (16:30 -0400)
committerMathieu Desnoyers <mathieu.desnoyers@polymtl.ca>
Mon, 6 Apr 2009 20:30:04 +0000 (16:30 -0400)
commit821586101518e2cfa82b1c33f9e78d75f43c6dc6
tree1b920fced12edbcec89bb6e278a436194427deba
parent8baf2c9506f8e4bbac913c7ea2a8b2e9a3cd0932
Check double write read order

Check the following scenario :

> > P.S.:  Regarding very weak memory-ordering models, I recall that in our
> > earlier discussions we considered this scenario:

> >     CPU 0                   CPU 1
> >     -----                   -----
> >     x = 1;                  y = 1;
> >     mb();                   mb();
> >     read y                  read x

> > with the question being whether it was possible for both reads to
> > obtain the value 0.  With sufficiently weak ordering it is possible,
> > but it turned out that there are places in the Linux kernel (I don't
> > remember where) that essentially assume this cannot happen.  Have you
> > gotten far enough in your outline of the semi-formal model to address
> > this sort of question?
formal-model/ooomem-double-update/DEFINES [new file with mode: 0644]
formal-model/ooomem-double-update/Makefile [new file with mode: 0644]
formal-model/ooomem-double-update/mem.sh [new file with mode: 0644]
formal-model/ooomem-double-update/mem.spin [new file with mode: 0644]
formal-model/ooomem-double-update/read_order.ltl [new file with mode: 0644]
formal-model/ooomem-double-update/read_order_no_rmb.define [new file with mode: 0644]
formal-model/ooomem-double-update/read_order_no_wmb.define [new file with mode: 0644]
formal-model/ooomem-double-update/references.txt [new file with mode: 0644]
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