xazax.hun added inline comments.
================
Comment at: clang/lib/Analysis/FlowSensitive/TypeErasedDataflowAnalysis.cpp:49
+ // in tests.
+ std::set<const CFGBlock *> Preds;
Preds.insert(Block.pred_begin(), Block.pred_end());
----------------
sgatev wrote:
> xazax.hun wrote:
> > xazax.hun wrote:
> > > Are we sure that the memory addresses of CFGBlocks are stable enough for
> > > a deterministic order? Alternatively, we could use the block ids for the
> > > ordering.
> > Also, could you describe where the flakiness is originated from? Naively,
> > I'd expect that the order in which we process the predecessors should not
> > change the results of the analysis.
> You're right, using block ids for ordering is better. I updated the code.
>
> > Also, could you describe where the flakiness is originated from?
>
> Say we have a block `B1` with predecessors `B2` and `B3`. Let the environment
> of `B2` after evaluating all of its statements is `B2Env = { Expr1 -> Loc1 }`
> and the environment of `B3` after evaluating all of its statement is `B3Env =
> { Expr2 -> Loc2 }` where `ExprX -> LocX` refers to a particular mapping of
> storage locations to expressions.
>
> What we want for the input environment of `B1` is `{}` because `B2Env` and
> `B3Env` do not contain common assignments of storage locations to
> expressions. What we got before this patch is either `B2Env.join(B3Env) = {
> Expr1 -> Loc1 }` or `B3Env.join(B2Env) = { Expr2 -> Loc2 }`.
>
> Without deterministic ordering of predecessors the test that I'm introducing
> in this patch is flaky.
> What we got before this patch is either B2Env.join(B3Env) = { Expr1 -> Loc1 }
> or B3Env.join(B2Env) = { Expr2 -> Loc2 }.
I think I'm still missing something. With this patch, wouldn't both
B2Env.join(B3Env) and B3Env.join(B2Env) produce the empty environment? If that
is the case, do we still care about a deterministic order?
Repository:
rG LLVM Github Monorepo
CHANGES SINCE LAST ACTION
https://reviews.llvm.org/D117754/new/
https://reviews.llvm.org/D117754
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