I have a Function pass, called firstPass, which does some analysis and populates:
A a;
where
typedef std::map< std::string, B* > A;
class firstPass : public FunctionPass {
A a;
}
typedef std::vector< C* > D;
class B {
D d;
}
class C {
// some class packing information about basic blocks;
}
Hence I have a map of vectors traversed by std::string.
I wrote associated destructors for these classes. This pass works successfully on its own.
I have another Function pass, called secondPass, needing this structure of type A to make some transformations. I used
bool secondPass::doInitialization(Module &M) {
errs() << "now running secondPass\n";
a = getAnalysis<firstPass>().getA();
return false;
}
void secondPass::getAnalysisUsage(AnalysisUsage &AU) const {
AU.addRequired<firstPass>();
AU.setPreservesAll();
}
The whole code compiles fine, but I get a segmentation fault when printing this structure at the end of my first pass only if I call my second pass (since B* is null).
To be clear:
opt -load ./libCustomLLVMPasses.so -passA < someCode.bc
prints in doFinalization() and exits successfully
opt -load ./libCustomLLVMPasses.so -passA -passB < someCode.bc
gives a segmentation fault.
How should I wrap this data structure and pass it to the second pass without issues? I tried std::unique_ptr instead of raw ones but I couldn't make it work. I'm not sure if this is the correct approach anyway, so any help will be appreciated.
EDIT: I solved the problem of seg. fault. It was basically me calling getAnalysis in doInitialization(). I wrote a ModulePass to combine my firstPass and secondPass whose runOnModule is shown below.
bool MPass::runOnModule(Module &M) {
for(Function& F : M) {
errs() << "F: " << F.getName() << "\n";
if(!F.getName().equals("main") && !F.isDeclaration())
getAnalysis<firstPass>(F);
}
StringRef main = StringRef("main");
A& a = getAnalysis<firstPass>(*(M.getFunction(main))).getA();
return false;
}
This also gave me to control the order of the functions processed. Now I can get the output of a pass but cannot use it as an input to another pass. I think this shows that the passes in llvm are self-contained.