CS553 Colorado State University
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Interprocedural Analysis
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12/8/09

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What is Datalog?
    
    -> slides 10-13 in John Whaley's PLDI Tutorial slides.
    http://suif.stanford.edu/~jwhaley/PLDITutorial.ppt

    -> graph example with bddbddb
        path.dtl and edge.tuples
    
    
    -> academic family tree example with bddbddb
        family.dtl and child.tuples
        
        -> then add rules from slide 14
            family-extend.dtl
            
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Reaching definitions in Datalog
    - how does the example in slide 15 match the book?
    It doesn't but I like the formulation in the slides better.
    
    -> have students construct the statementAt, assign, and follows tuples 
    files for our typical data-flow example
        They need a mapping for each domain they are dealing with,
        which maps integers to names.
        
        They need a .tuples files that encodes the relationships in the
        program.
    
    -> work on checking the output
    
    
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Relationship between datalog and BDDs
    -> slides 34-46
    

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How to specify Andersens and Steensgaard in Datalog

    ----------------
    Andersens
        Program
            a = &b
            c = a
            a = &d
            e = a
            
        Constraints
            ptsto(a) = {b,d}
            ptsto(c) superseteq ptsto(a)
            ptsto(e) superseteq ptsto(a)

        Answer
            ptsto(c) = ptsto(a) = {b,d}
            ptsto(e) = ptsto(a) = {b,d}


        Specifying the analysis in datalog
        Different than book because we are encoding a stack-only pointer analysis.
        
            output
                ptsto(v,w) means stack variable v can point at stack variable w
            input
                addressof(a,b) means a contains the address of b, a=&b
                ptrassign(p,q) means that p has been assigned the pointer value q, p=q
                dereflhs(p,q) means *p=q
                derefrhs(p,q) means p=*q
            
            rules to solve with Andersens
                ptsto(a,b) :- addressof(a,b).
                ptsto(p,x) :- ptrassign(p,q), ptsto(q,x).
                ptsto(y,x) :- dereflhs(p,q), ptsto(q,x), ptsto(p,y).
                ptsto(p,y) :- derefrhs(p,q), ptsto(q,x), ptsto(x,y).

# Goes with example on slide 12 of lecture22-aliasanalysis.ppt.
# p is 1, q is 2, x is 3, a is 4, b is 5, d is 6, 
# f is 7, c is 8, e is 9, g is 10
# t1 is 11, t2 is 12

int g;

int** foo(int **p, **q)
{
   int **x;
   
   x = p;
   . . .
   x = q;
   return x;
}

int main()
{
   int **a, *b, *d, *f,
         c, e, t1, t2;

   a = foo(&b, &f);
   t1 = &c;
   *a = t1;
   a = foo(&d, &g);
   t2 = &e;
   *a = t2;
}

Answer should be 
    p -> {b,d}          (1,5), (1,6)
    q -> {f,g}          (2,7), (2,10)
    t1-> {c}            (11,8)
    t2-> {e}            (12,9)
    x -> {b,d,f,g}      (3,5), (3,6), (3,7), (3,10)
    a -> {b,d,f,g}      (4,5), (4,6), (4,7), (4,10)
    b -> {c,e}          (5,8), (5,9)
    d -> {c,e}          (6,8), (6,9)
    f -> {c,e}          (7,8), (7,9)
    g -> {c,e}          (10,8), (10,9)
    

    ----------------
    Steensgaard

            More efficiently solved with a union/find data structure instead
            of using datalog and BDDs, but here is the datalog specification

            extending Andersen rules to solve for Steensgaard
                ptsto(a,b) :- addressof(a,b).
                ptsto(p,x) :- ptrassign(p,q), ptsto(q,x).
                ptsto(y,x) :- dereflhs(p,q), ptsto(q,x), ptsto(p,y).
                ptsto(p,y) :- derefrhs(p,q), ptsto(q,x), ptsto(x,y).
      
                # extension is modification of last three rules ablve
                ptsto(q,x) :- ptrassign(p,q), ptsto(p,x).
                ptsto(q,x) :- dereflhs(p,q), ptsto(p,y), ptsto(y,x).
                ptsto(q,y) :- derefrhs(p,q), ptsto(p,y), ptsto(q,x).
                
                
# Goes with example on slide 12 of lecture22-aliasanalysis.ppt.
# p is 1, q is 2, x is 3, a is 4, b is 5, d is 6, 
# f is 7, c is 8, e is 9, g is 10
      
Answer is
    p -> {b,d,f,g}      (1,5), (1,6), (1,7), (1,10)
    q -> {f,g,b,d}      (2,7), (2,10), (2,6), (2,5)
    t1-> {c}            (11,8), (11,9)
    t2-> {e}            (12,9), (12,8)
    x -> {b,d,f,g}      (3,5), (3,6), (3,7), (3,10)
    a -> {b,d,f,g}      (4,5), (4,6), (4,7), (4,10)
    b -> {c,e}          (5,8), (5,9)
    d -> {c,e}          (6,8), (6,9)
    f -> {c,e}          (7,8), (7,9)
    g -> {c,e}          (10,8), (10,9)


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mstrout@cs.colostate.edu, 12/8/09