Summary:
It turns out keeping attributes (such as invalidation facts) separate
from the memory is a bad idea and leads to loss of precision and false
positives, as seen in the new test (which previously generated a
report).
Allow me to illustrate on this example, which is a stylised version of
the issue in the added test: previously we'd have:
```
state1 = { x = 1; invalids={} }
state2 = { x = 2; invalids ={1} }
join(state1, state2) = { x = {1, 2}; invalids={{1, 2}} }
```
So even though none of the states said that `x` pointed to an invalid
location, the join state says it does because `1` and `2` have been
glommed together. The fact `x=1` from `state1` and the fact "1 is
invalid" from `state2` conspire together and `x` is now invalid even
though it shouldn't.
Instead, if we record attributes as part of the memory we get that `x`
is still valid after the join:
```
state1 = { x = (1, {}) }
state2 = { x = (2, {}) }
join(state1, state2) = { x = ({1, 2}, {}) }
```
Reviewed By: mbouaziz
Differential Revision: D12958130
fbshipit-source-id: 53dc81cc7
Summary:
Getting this right will be long and complex so for now the easiest is to
underreport and only consider as invalid the addresses we know to be invalid on
both sides of a join. In fact the condition for an address to be invalid after
a join is more complex than this: it is invalid only if *all* the addresses in
its equivalence class as discovered by the join are invalid.
Reviewed By: skcho
Differential Revision: D12823925
fbshipit-source-id: 2ca109356
Summary:
Instead of the non-sensical piecewise join we had until now write
a proper one. Hopefully the comments explain what it does. Main one:
```
(* high-level idea: maintain some union-find data structure to identify locations in one heap
with locations in the other heap. Build the initial join state as follows:
- equate all locations that correspond to identical variables in both stacks, eg joining
stacks {x=1} and {x=2} adds "1=2" to the unification.
- add all addresses reachable from stack variables to the join state heap
This gives us an abstract state that is the union of both abstract states, but more states
can still be made equal. For instance, if 1 points to 3 in the first heap and 2 points to 4
in the second heap and we deduced "1 = 2" from the stacks already (as in the example just
above) then we can deduce "3 = 4". Proceed in this fashion until no more equalities are
discovered, and return the abstract state where a canonical representative has been chosen
consistently for each equivalence class (this is what the union-find data structure gives
us). *)
```
Reviewed By: mbouaziz
Differential Revision: D10483978
fbshipit-source-id: f6ffd7528
Jules Villard