1. Back to lattice (D, v, ?, >, t, u ) = (2 A, ¶, A, ;, Å, [ ) where A = { x ! N | x 2 Vars Æ N 2 Z } What’s the problem with this lattice? Lattice is infinitely high, which means we can’t guarantee termination

2. Better lattice Suppose we only had one variable

3. Better lattice Suppose we only had one variable D = { ?, > } [ Z 8 i 2 Z. ? v i Æ i v > height = 3

4. For all variables Two possibilities Option 1: Tuple of lattices Given lattices (D 1, v 1, ? 1, > 1, t 1, u 1 )... (D n, v n, ? n, > n, t n, u n ) create: tuple lattice D n =

5. For all variables Two possibilities Option 1: Tuple of lattices Given lattices (D 1, v 1, ? 1, > 1, t 1, u 1 )... (D n, v n, ? n, > n, t n, u n ) create: tuple lattice D n = ((D 1 £... £ D n ), v, ?, >, t, u ) where ? = ( ? 1,..., ? n ) > = ( > 1,..., > n ) (a 1,..., a n ) t (b 1,..., b n ) = (a 1 t 1 b 1,..., a n t n b n ) (a 1,..., a n ) u (b 1,..., b n ) = (a 1 u 1 b 1,..., a n u n b n ) height = height(D 1 ) +... + height(D n )

6. For all variables Option 2: Map from variables to single lattice Given lattice (D, v 1, ? 1, > 1, t 1, u 1 ) and a set V, create: map lattice V ! D = (V ! D, v, ?, >, t, u )

7. Back to example x := y op z in out F x := y op z (in) =

8. Back to example x := y op z in out F x := y op z (in) = in [ x ! in(y) op in(z) ] where a op b =

9. General approach to domain design Simple lattices: –boolean logic lattice –powerset lattice –incomparable set: set of incomparable values, plus top and bottom (eg const prop lattice) –two point lattice: just top and bottom Use combinators to create more complicated lattices –tuple lattice constructor –map lattice constructor

10. May vs Must Has to do with definition of computed info Set of x ! y must-point-to pairs –if we compute x ! y, then, then during program execution, x must point to y Set of x ! y may-point-to pairs –if during program execution, it is possible for x to point to y, then we must compute x ! y

11. May vs must MayMust most optimistic (bottom) most conservative (top) safe merge

12. May vs must MayMust most optimistic (bottom) empty setfull set most conservative (top) full setempty set safeoverly bigoverly small merge [Å

13. Direction of analysis Although constraints are not directional, flow functions are All flow functions we have seen so far are in the forward direction In some cases, the constraints are of the form in = F(out) These are called backward problems. Example: live variables –compute the set of variables that may be live

14. Example: live variables Set D = Lattice: (D, v, ?, >, t, u ) =

15. Example: live variables Set D = 2 Vars Lattice: (D, v, ?, >, t, u ) = (2 Vars, µ, ;,Vars, [, Å ) x := y op z in out F x := y op z (out) =

16. Example: live variables Set D = 2 Vars Lattice: (D, v, ?, >, t, u ) = (2 Vars, µ, ;,Vars, [, Å ) x := y op z in out F x := y op z (out) = out – { x } [ { y, z}

17. Example: live variables x := 5 y := x + 2 x := x + 1 y := x + 10... y...

18. Example: live variables x := 5 y := x + 2 x := x + 1 y := x + 10... y... How can we remove the x := x + 1 stmt?

19. Revisiting assignment x := y op z in out F x := y op z (out) = out – { x } [ { y, z}

20. Revisiting assignment x := y op z in out F x := y op z (out) = out – { x } [ { y, z}

21. Theory of backward analyses Can formalize backward analyses in two ways Option 1: reverse flow graph, and then run forward problem Option 2: re-develop the theory, but in the backward direction