[PATCH] D123773: [clang][analyzer][ctu] Make CTU a two phase analysis

[Gábor Horváth via Phabricator via cfe-commits]

xazax.hun added inline comments.


================
Comment at: clang/lib/StaticAnalyzer/Core/ExprEngineCallAndReturn.cpp:446
+    }
+    const bool BState = State->contains<CTUDispatchBifurcationSet>(D);
+    if (!BState) { // This is the first time we see this foreign function.
----------------
xazax.hun wrote:
> martong wrote:
> > xazax.hun wrote:
> > > So if we see the same foreign function called in multiple contexts, we will only queue one of the contexts for the CTU. Is this the intended design? 
> > > 
> > > So if I see:
> > > ```
> > > foreign(true);
> > > foreign(false);
> > > ```
> > > 
> > > The new CTU will only evaluate `foreign(true)` but not `foreign(false)`. 
> > This is intentional.
> > ```
> > foreign(true);
> > foreign(false);
> > ```
> > The new CTU will evaluate the following paths in the exploded graph:
> > ```
> > foreign(true); // conservative evaluated
> > foreign(false); // conservative evaluated
> > foreign(true); // inlined
> > foreign(false); // inlined
> > ```
> > The point is to keep bifurcation to a minimum and avoid the exponential blow up.
> > So, we will never have a path like this:
> > ```
> > foreign(true); // conservative evaluated
> > foreign(false); // inlined
> > ```
> > 
> > Actually, this is the same strategy that we use during the dynamic dispatch of C++ virtual calls. See `DynamicDispatchBifurcationMap`.
> > 
> > The conservative evaluation happens in the first phase, the inlining in the second phase (assuming the phase1 inlining option is set to none).
> > The new CTU will evaluate the following paths in the exploded graph:
> > ```
> > foreign(true); // conservative evaluated
> > foreign(false); // conservative evaluated
> > foreign(true); // inlined
> > foreign(false); // inlined
> > ```
> 
> When we encounter `foreign(true)`, we would add the decl to `CTUDispatchBifurcationSet`. So the second time we see a call to the function `foreign(false);`, we will just do the conservative evaluation and will not add the call to the CTU worklist. So how will `foreign(false);` be inlined in the second pass? Do I miss something? 
> 
Oh, I think I now understand. Do you expect `foreign(false);` to be inlined after we return from `foreign(true);` in the second pass? Sorry for the confusion, in that case it looks good to me.


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https://reviews.llvm.org/D123773