Optimizing Compilers CISC 673 Spring 2011 Yet More Data flow analysis

1. Optimizing CompilersCISC 673Spring 2011Yet More Data flow analysis John Cavazos University of Delaware

2. Overview • Additional Data Flow Problem • Constant Propagation • Generalization of Data Flow

3. Constant Propagation • Along every path to point p, variable v has same “known” value

4. Constant Propagation • Specialize computation at p based on v’s value

5. … -2 -1 0 1 2 ... The Lattice of Constant Prop Constant Propagation Lattice Hasse diagram • Lattice is a poset with (V, ≤) • V meet (or join) operator • ≤ ordering operator

6. Two Special Elements in Lattice • ⊤ (called top) • Not a Constant • ⊥ (called bottom) • Unknown

7. Some Lattice Theory • Generalized Meet Operator ⊔ • Used at “merge” points

8. Some Lattice Theory (cont’d) • Meet Operator (⊔) for Avail Expressions • Avail Expression (w/ sets):  • Avail Expression (w/ Bit Vectors): ⋀ • Produces Least Upper Bound (LUB) • X ⊔ Y = least element Z s.t. • X ≤ Z and Y ≤ Z

9. Iterative Data Flow Analysis • Initialize non-entry nodes to identify element • Identity element for meet function • Remember last time: X ⋀ 1 = X

10. Iterative Data Flow Analysis (cont’d) • If node function is monotone: • Each re-evaluation of node moves up the lattice, if it moves at all • If height of lattice is finite, must terminate

11. Top ⊤ versus Bottom ⊥ • ⊤ means definitely NOT a constant • Inputs are ⊤ • ⊥ means Unknown or Undefined • Could be constant but we do not know!

12. … -2 -1 0 1 2 ... The Lattice of Constant Prop More Lattice Theory: Ordering Op • ≤ ordering operator • ⊥ ≤ anything • ∀x ⊥ ⊔ x = x • anything ≤ ⊤ • ∀x x ⊔ ⊤= ⊤ • ∀x x ⊔ x = x

13. Relate to Constant Prop?! • i and j are integer values • i ⊔ i = i • i ⊔ j = ⊤ if i ≠ j

14. What is X in Block 3? 1. 2. 3.

15. What is X in Block 3? 1. 2. 3. x is ⊤

16. Constant Propagation Problem • Variable v has constant value c at point p, iff ∀ paths from entry to p, the most recently assigned value for v is c

17. Constant Prop Example Meet Operator = ⊔ Identity= ? X ⊔ ? = X

18. Constant Prop Example Meet Operator = ⊔ Identify= ⊥ X ⊔ ⊥ = X Initial Condition?

19. Constant Prop Example Meet Operator = ⊔ Identify= ⊥ X ⊔ ⊥ = X Initial Condition Boundary Condition? XYZ ⊥⊥⊥ ⊥⊥⊥ ⊥⊥⊥ ⊥⊥⊥

20. Constant Prop Example ⊥⊥⊥ Meet Operator = ⊔ Identify= ⊥ X ⊔ ⊥ = X Initial Condition Boundary Condition XYZ 1. ⊥⊥⊥ 2. 3. ⊥⊥⊥ ⊥⊥⊥ 4. ⊥⊥⊥

21. Constant Prop Example ⊥⊥⊥ XYZ out1= ??? 1. ⊥⊥⊥ 2. 3. ⊥⊥⊥ ⊥⊥⊥ 4. ⊥⊥⊥

22. Constant Prop Example ⊥⊥⊥ XYZ out1= 1⊥⊥ out2= ??? 1. 1⊥⊥ ⊥⊥⊥ 2. 3. ⊥⊥⊥ ⊥⊥⊥ 4. ⊥⊥⊥

23. Constant Prop Example ⊥⊥⊥ XYZ out1= 1⊥⊥ out2= 023 out3= ??? 1. 1⊥⊥ ⊥⊥⊥ 2. 3. ⊥⊥⊥ ⊥⊥⊥ 023 4. ⊥⊥⊥

24. Constant Prop Example ⊥⊥⊥ XYZ out1= 1⊥⊥ out2= 023 out3= 12⊥ out4= ??? 1. 1⊥⊥ ⊥⊥⊥ 2. 3. ⊥⊥⊥ ⊥⊥⊥ 12⊥ 023 4. ⊥⊥⊥

25. Constant Prop Example ⊥⊥⊥ XYZ out1= 1⊥⊥ out2= 023 out3= 12⊥ out4= ⊤23 1. 1⊥⊥ ⊥⊥⊥ 2. 3. ⊥⊥⊥ ⊥⊥⊥ Does this make sense? 12⊥ 023 4. ⊥⊥⊥ ⊤23

26. Next Time • Static-Single Assignment (SSA Form) • Read Efficiently computing static single assignment form and the control dependence graph, Cytron et al. http://portal.acm.org/citation.cfm?id=115320

27. Next Time • Intelligent Compilation