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Efficiently Computing SSA

Efficiently Computing SSA. Static Single Assignment form. Each name is defined exactly once Each use refers to exactly one name Why use SSA form?. Joins. What to do when two different values meet on the Control Flow Graph? Φ -Functions

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Efficiently Computing SSA

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  1. Efficiently Computing SSA

  2. Static Single Assignment form • Each name is defined exactly once • Each use refers to exactly one name • Why use SSA form?

  3. Joins • What to do when two different values meet on the Control Flow Graph? • Φ-Functions • A Φ-function is a special kind of copy that selects one of its parameters

  4. Sparse Representation • Only want to propagate facts where relevant • Don’t care about the rest • Use the SSA Graph

  5. SSA Graph • Add edges from definitions to uses

  6. Example i = j = k = l = 1 repeat if (p) then begin j = i if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 print (i,j,k,l) repeat if R then l = l + 4 until S i = i + 6 until T

  7. i = j = k = l = 1 repeat if (p) then begin j = i if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 print (i,j,k,l) repeat if R then l = l + 4 until S i = i + 6 until T Example Number existing defns i1 = j = k = l = 1 repeat if (p) then begin j = i if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 print (i,j,k,l) repeat if R then l = l + 4 until S i2 = i + 6 until T

  8. i = j = k = l = 1 repeat if (p) then begin j = i if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 print (i,j,k,l) repeat if R then l = l + 4 until S i = i + 6 until T Example Add ϕ-functions where needed i1 = j = k = l = 1 repeat i3 = ϕ() if (p) then begin j = i if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 print (i,j,k,l) repeat if R then l = l + 4 until S i2 = i + 6 until T

  9. i = j = k = l = 1 repeat if (p) then begin j = i if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 print (i,j,k,l) repeat if R then l = l + 4 until S i = i + 6 until T Example Fill in the use numbers --- Then repeat for other variables i1 = j = k = l = 1 repeat i3 = ϕ(i1,i2) if (p) then begin j = i3 if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 print (i3,j,k,l) repeat if R then l = l + 4 until S i2 = i3 + 6 until T

  10. i = j = k = l = 1 repeat if (p) then begin j = i if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 print (i,j,k,l) repeat if R then l = l + 4 until S i = i + 6 until T Example i1 = j1 = k1 = l1 = 1 repeat i3 = f(i1,i2) j2 = f(j1,j4) k2 = f(k5,k1) l2 = f(l9,l1) if (p) then begin j3 = i2 if Q then l3 = 2 else l4 = 3 l5 = f(l3,l4) k3 = k2 + 1 end else k4 = k2 + 2 j4 = f(j3,j2) k5 = f(k3,k4) l6 = f(l2,l5) print (i3,j4,k5,l6) repeat l7 = f(l9,l6) if R then l8 = l7 + 4 l9 = f(l7,l8) until S i2 = i3 + 6 until T

  11. i = j = k = l = 1 repeat if (p) then begin j = i if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 print (i,j,k,l) repeat if R then l = l + 4 until S i = i + 6 until T Example - CFG Entry 1 2 3 4 5 7 6 8 9 10 11 Exit 12

  12. Constructing SSA (Naively) • Insert Φ-functions at every join for every name • Solve Reaching Definitions • Rename each use to the definition that reaches it

  13. Constructing SSA (Naively) • Why is the previous bad? • Too many ϕ-functions! • Can we do better?

  14. Efficiently Constructing SSA • Perform Control-flow analysis • Insert ϕ-Functions • Rename Values • However – where do we put the ϕ-Functions?

  15. Formalizing ϕ placement • We need a Φfunction at node Z if • Two non-null CFG paths that both define v • Such that both paths start at two distinct nodes and end at Z

  16. Dominance Frontiers Illustration Dominated by X Dominance Frontier of X (Not Dominated by X)

  17. Dominance Frontiers • If z is the first node we encounter on the path from x which x does not strictly dominate, z is in the dominance frontier of x • For this to happen, there is some path from node x to z, x …  y  z where (x SDOM y) but not (x SDOM z).

  18. Example {1} {7} • DF(1) = • DF(2) = • DF(3) = • DF(4) = • DF(5) = • DF(6) = • DF(7) = {6} {6} {1, 7} {7} ∅

  19. Computing Dominance Frontiers • Two components to DF(X): • DFlocal(X) = {Y∊ succ(X) | X> Y} • Any child of X not (strictly) dominated by X is in DF(X) • Let Z be such that idom(Z) = X • idom(Z) is the parent of Z in the dominator tree • DFup(Z) = {Y∊ DF(Z) | X>Y} • Nodes from DF(Z) that are not strictly dominated by X are also in DF(X)

  20. Algorithm • Let SDOM(X) = {Y | X>Y} • For each Z such that idom(Z) = X do • DF(X) = DFlocal(X) ∪ (DF(Z) - SDOM(X)) • I.e., DF(X) = DFlocal(X) ∪ DFup(Z)

  21. SSA f-placement For each variable M in the program • Set Worklist = DM, the set of CFG nodes that contain assignments to variable M. • While items left in the Worklist • remove some node X from the Worklist • for all W in DF(X), if W has never been in M's Worklist, add a f-term at W. Place W into the Worklist.

  22. i = j = k = l = 1 repeat if (p) then begin j = i if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 print (i,j,k,l) repeat if R then l = l + 4 until S i = i + 6 until T CFG for example – Variable k Entry 1 2 3 1 2 4 5 7 7 3 8 6 4 5 6 9 8 11 10 9 10 12 11 Exit 12

  23. 1 dominates the entire graph Entry 1 2 3 • Worklist = {1,6,7} – location of k’s assignments • Compute DF(1) = {2,3,4,5,6,7,8,9,10,11,12,Exit } – {2,3,4,5,6,7,8,9,10,11,12,Exit} = { } Remember that DF(x) = (SUCC(DOM-1(x)) – SDOM-1(x) where SUCC(x) is set of successors of x in the CFG) 4 5 7 6 8 9 10 11 Exit 12

  24. Entry 1 2 3 • Worklist = {6,7} • Compute DF(6) = {8} – {} = {8} • put f in node 8 Remember that DF(x) = (SUCC(DOM-1(x)) – SDOM-1(x) where SUCC(x) is set of successors of x in the CFG) 4 5 7 node 6 only dominates itself 6 f 8 9 10 11 Exit 12

  25. Entry 1 2 3 • Worklist = {7,8} • Compute DF(7) = {8} – {} = {8} • node 8 already in worklist Remember that DF(x) = (SUCC(DOM-1(x)) – SDOM-1(x) where SUCC(x) is set of successors of x in the CFG) 4 5 7 node 7 only dominates itself 6 f 8 9 10 11 Exit 12

  26. Entry 1 f 2 3 • Worklist = {8} • Compute DF(8) = {2,9,10,11,12, exit} – {9,10,11,12,exit} = {2} • put f in node 2 Remember that DF(x) = (SUCC(DOM-1(x)) – SDOM-1(x) where SUCC(x) is set of successors of x in the CFG) 4 5 7 node 8 dominates {8,9,10,11,12,exit} 6 f 8 9 10 11 Exit 12

  27. Entry 1 f 2 3 • Worklist = {2} • Compute DF(2) = {2-12, exit} – {3-12,exit} = {2} Remember that DF(x) = (SUCC(DOM-1(x)) – SDOM-1(x) where SUCC(x) is set of successors of x in the CFG) 4 5 7 node 2 dominates {2-12,exit} 6 f 8 9 10 11 Exit 12

  28. i = j = k = l = 1 repeat if (p) then begin j = i if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 print (i,j,k,l) repeat if R then l = l + 4 until S i = i + 6 until T CFG for example – Variable k Entry k = 1 1 2 k = f(k,k) 3 4 5 7 k =k + 2 6 k =k + 1 8 k = f(k,k) 9 10 11 12 Exit

  29. Linking it all together Initially, for each variable V C(V) = 0 /* counter */ S(V) = Empty /* stack for each variable*/ Other important information: RHS(A) = set of variables used on right side of assignment LHS(A) = set of variables assigned to on left side WhichPred(X,Y) = j where Y is the jth predecessor of X in CFG Children(X) = children in the IDOM tree ‘Search(entry)’ will do renaming of all variables in program.

  30. top(S(v)) will hold the ‘current’ # for variable v Search (X: CFG node) for each statement A in X If A is an ordinary assignment statement For each v in RHS(A) replace the use of v with vi, where i = top(S(v)) For each v in LHS(A) i = C(v); replace v by vi push i on S(v); C(v)++ end end For each Y in Succ(X) do J = WhichPred(X,Y) /* predecessors have unique # wrt successor */ For each f-function F in Y Replace j-th operand v in RHS(F) by vi where i = Top(S(v)) End end For each Y in Children() call Search() end For each assignment A in X For each v in LHS(A) pop S(v) end end end New assignment – update the counter and push

  31. Search (X: CFG node) for each statement A in X If A is an ordinary assignment statement For each v in RHS(A) replace the use of v with vi, where i = top(S(v)) For each v in LHS(A) i = C(v); replace v by vi push i on S(v); C(v)++ end end For each Y in Succ(X) do j = WhichPred(X,Y) /* predecessors have unique # wrt successor */ For each f-function F in Y Replace j-th operand v in RHS(F) by vi where i = Top(S(v)) End end For each Y in Children() call Search() end For each assignment A in X For each v in LHS(A) pop S(v) end end end Look at successors of this node in CFG Resolve any ϕ functions in successors

  32. Search (X: CFG node) for each statement A in X If A is an ordinary assignment statement For each v in RHS(A) replace the use of v with vi, where i = top(S(v)) For each v in LHS(A) i = C(v); replace v by vi push i on S(v); C(v)++ end end For each Y in Succ(X) do J = WhichPred(X,Y) /* predecessors have unique # wrt successor */ For each f-function F in Y Replace j-th operand v in RHS(F) by vi where i = Top(S(v)) End end For each D in Children(X) call Search(D) end For each assignment A in X For each v in LHS(A) pop S(v) end end end Look at all nodes dominated by X

  33. Search (X: CFG node) for each statement A in X If A is an ordinary assignment statement For each v in RHS(A) replace the use of v with vi, where i = top(S(v)) For each v in LHS(A) i = C(v); replace v by vi push i on S(v); C(v)++ end end For each Y in Succ(X) do J = WhichPred(X,Y) /* predecessors have unique # wrt successor */ For each f-function F in Y Replace j-th operand v in RHS(F) by vi where i = Top(S(v)) End end For each D in Children(X) call Search(D) end For each assignment A in X For each v in LHS(A) pop S(v) end end end Pop the var # pushed earlier

  34. 1 1 2 2 3 3 7 8 4 5 7 4 5 6 9 10 11 6 12 8 9 10 11 12

  35. Just deal with k in this example Search(1) C(k) = 1 S(k) = {0} Children(1) = {2} k0 = 1 1 2 k = f(k0,k) 3 4 5 7 1 k =k + 2 6 2 k =k + 1 3 7 8 8 k = f(k,k) 9 4 5 6 9 10 10 11 12 11 Exit 12

  36. Search(2) C(k) = 2 S(k) = {0,1} None of the successors have f functions Children(2) = {3,7,8} k0 = 1 1 2 k1 = f(k0,k) 3 4 5 7 1 k =k + 2 6 2 k =k + 1 3 7 8 8 k = f(k,k) 9 4 5 6 9 10 10 11 12 11 Exit 12

  37. Search(3) C(k) = 2 S(k) = {0,1} Children(3) = {4,5,6} Search(4) Children(4) = {} Search(5) Children(5) = {} Search(6) C(k) = 3 S(k) = {0,1,2} Children(6) = {} Pop(S(k)) = {0,1} k0 = 1 1 2 k1 = f(k0,k) 3 4 5 7 k =k + 2 6 1 k2 =k1 + 1 8 k = f(k2,k) 2 9 3 7 8 10 4 5 6 9 11 10 11 Exit 12 12

  38. Search(7) C(k) = 4 S(k) = {0,1,3} Children(7) = { } Pop(S(k)) = {0,1} k0 = 1 1 2 k1 = f(k0,k) 3 4 5 7 1 k3 =k1 + 2 6 2 k2 =k1 + 1 8 3 7 8 k = f(k2,k3) 9 4 5 6 9 10 10 11 12 11 Exit 12

  39. Search(8) C(k) = 5 S(k) = {0,1,4} Children(8) = {9} Search(9) Children(9) = {10,11} Search(10) Search(11) k0 = 1 1 2 k1 = f(k0,k) 3 4 5 7 k3 =k1 + 2 1 6 k2 =k1 + 1 2 8 k4 = f(k2,k3) 3 7 8 9 4 5 6 9 10 10 11 11 Exit 12 12

  40. Search(12) C(k) = 5 S(k) = {0,1,4} k0 = 1 1 2 k1 = f(k0,k4) 3 4 5 7 k3 =k1 + 2 1 6 2 k2 =k1 + 1 8 k4 = f(k2,k3) 3 7 8 9 4 5 6 9 10 10 11 11 12 Exit 12

  41. i = j= k = l = 1 Repeat if (p) then begin j = i if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 print (i,j,k,l) repeat if R then l = l + 4 until S i = i + 6 until T i = j = k1 = l = 1 Repeat k2 = f(k1,k5) if (p) then begin j = i if Q then l = 2 else l = 3 k3 = k2 + 1 end else k4 = k2 + 2 k5 = f(k3,k4) print (i,j,k5,l) repeat if R then l = l + 4 until S i = i + 6 until T

  42. i = j = k = l = 1 repeat if (p) then begin j = i if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 print (i,j,k,l) repeat if R then l = l + 4 until S i = i + 6 until T Consider L Entry l = 1 2 3 4 5 7 l = l = 6 print l 8 9 10 l = l + 11 Exit 12

  43. i = j = k = l = 1 repeat if (p) then begin j = i if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 print (i,j,k,l) repeat if R then l = l + 4 until S i = i + 6 until T Computing the DF Entry l = 1 l = f 2 3 1 4 5 7 2 l = l = 7 3 8 6 l = f l = f 4 5 6 9 print l 8 9 11 10 l = f 10 12 l = l + l = f 11 Exit 12

  44. i = j= k = l = 1 repeat if (p) then begin j = i if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 print (i,j,k,l) repeat if R then l = l + 4 until S i = i + 6 until T i = j = k = l = 1 repeat l = f(l,l) if (p) then begin j = i if Q then l = 2 else l = 3 l = f(l,l) k = k + 1 end else k = k + 2 l = f(l,l) print (i,j,k,l) repeat l = f(l,l) if R then l = l + 4 l = f(l,l) until S i = i + 6 until T

  45. i = j = k = l = 1 repeat if (p) then begin j = i if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 print (i,j,k,l) repeat if R then l = l + 4 until S i = i + 6 until T Adding the numbers:search(1) Entry l1 = 1 l = f(l1) 2 3 1 4 5 7 2 l = l = 7 3 8 6 l = f l = f 4 5 6 9 print l 8 9 11 10 l = f 10 12 l = l + l = f 11 Exit 12

  46. i = j = k = l = 1 repeat if (p) then begin j = i if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 print (i,j,k,l) repeat if R then l = l + 4 until S i = i + 6 until T Adding the numbers:search(2) Entry l1 = 1 l2 = f(l1) 2 3 1 4 5 7 2 l = l = 7 3 8 6 l = f l = f 4 5 6 9 print l 8 9 11 10 l = f 10 12 l = l + l = f 11 Exit 12

  47. i = j = k = l = 1 repeat if (p) then begin j = i if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 print (i,j,k,l) repeat if R then l = l + 4 until S i = i + 6 until T Adding the numbers:search(3) Entry l1 = 1 l2 = f(l1) 2 3 1 4 5 7 2 l = l = 7 3 8 6 l = f l = f 4 5 6 9 print l 8 9 11 10 l = f 10 12 l = l + l = f 11 Exit 12

  48. i = j = k = l = 1 repeat if (p) then begin j = i if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 print (i,j,k,l) repeat if R then l = l + 4 until S i = i + 6 until T Adding the numbers:search(4) Entry l1 = 1 l2 = f(l1) 2 3 1 4 5 7 2 l = l3 = 7 3 8 6 l = f(l3) l = f 4 5 6 9 print l 8 9 11 10 l = f 10 12 l = l + l = f 11 Exit 12

  49. i = j = k = l = 1 repeat if (p) then begin j = i if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 print (i,j,k,l) repeat if R then l = l + 4 until S i = i + 6 until T Adding the numbers:search(5) Entry l1 = 1 l2 = f(l1) 2 3 1 4 5 7 2 l4 = l3 = 7 3 8 6 l = f(l3,l4) l = f 4 5 6 9 print l 8 9 11 10 l = f 10 12 l = l + l = f 11 Exit 12

  50. i = j = k = l = 1 repeat if (p) then begin j = i if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 print (i,j,k,l) repeat if R then l = l + 4 until S i = i + 6 until T Adding the numbers:search(6) Entry l1 = 1 l2 = f(l1) 2 3 1 4 5 7 2 l4 = l3 = 7 3 8 6 l5 = f(l3,l4) l = f(l5 4 5 6 9 print l 8 9 11 10 l = f 10 12 l = l + l = f 11 Exit 12

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