A question about alias analysis


I got a strange result when I used alias analysis DSAA, can you tell what is

1. The following is the primary body of my pass "fps.cpp":

  AliasAnalysis *AA = &getAnalysis<AliasAnalysis>();
  AliasSetTracker AST(*AA);
  for (Module::iterator fi = M.begin(), fe = M.end(); fi != fe; ++fi )
    for (Function::iterator bi = fi->begin(), be = fi->end(); bi != be; ++bi)

  for (AliasSetTracker::iterator I = AST.begin(), E = AST.end(); I != E; ++I) {
  AliasSet &AS = *I;
  AS.print (std::cerr);

2. The follwoing is the test program "hello.cpp":

   int *a, *b;
   int main() {
      int t;
      a = &t;
      b = &t;
      printf ("%d, %d", a, b);

3. I compiled the test program with "llvm-gcc hello.cpp -o hello", and made alias
   analysis with:
          opt -ds-aa -load ../Debug/lib/fps.so -FPS < hello.bc > /dev/null

   opt printed that:
      AliasSet[XXXX,1] must alis, Mod Pointers: (sbyte** %a, 4)
      AliasSet[YYYY,1] must alis, Mod Pointers: (sbyte** %b, 4)
   I donot know why "a" and "b" are not alias? They donot point to a same memory
object? But, when I donot use DSAA, that is, "opt -load ../Debug/lib/fps.so -FPS <
hello.bc > /dev/null", "a" and "b" are alias!



This is telling you that &a != &b.