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SSA

SSA. Agenda. SSA Introduction Converting to SSA Converting out of SSA SSA Example. SSA Introduction. SSA is a intermediate representation. SSA was developed by Ron Cytron, Jeanne Ferrante, Barry Rosen, Mark Wegman, and Ken Zadeck, researchers at IBM in the 1980s. Benefits

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SSA

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  1. SSA

  2. Agenda • SSA Introduction • Converting to SSA • Converting out of SSA • SSA Example

  3. SSA Introduction • SSA is a intermediate representation. • SSA was developed by Ron Cytron, Jeanne Ferrante, Barry Rosen, Mark Wegman, and Ken Zadeck, researchers at IBM in the 1980s. • Benefits • constant propagation • sparse conditional constant propagation • dead code elimination • global value numbering • partial redundancy elimination • strength reduction • register allocation

  4. SSA Introduction • Static Single Assignment Form • Each variable has only one reaching definition. • When two definitions merge, a Ф function is introduced to with a new definition of the variable. • A Ф operand represents the reaching definition from the corresponding predecessor.

  5. SSA Introduction a3= a= a1= a= = a1+5 = a+5 = a+5 a4= Ф(a1,a3) a2= a= = a4+5 = a2+5 = a+5

  6. SSA Introduction • SSA Condition • If two nonnull paths X →+ Z and Y →+ Z converge at a node, and nodes X and Y contain assignments to V (in the original program), then a trivial -function V = (V,…,V) has been inserted at Z (in the new program) • Each mention of V in the original program or in an inserted -function has been replaced by a mention of a new variable Vi, leaving the new program in SSA form • Along any control flow path, consider any use of a variable V (in the original program) and the corresponding use of Vi (in the new program). Then V and Vi has same value.

  7. SSA Introduction • Where to insert -functions? • By condition 1, a node Z needs a -function for V because Z is a convergence point for two nonnull paths X →+ Z and Y →+ Z that start at nodes X and Y already containing assignments to V.

  8. SSA Introduction • What about arrays? • Treating A[i] as a variable would be awkward, both because an assignment to A[i] may or may not change the value of A[j] and because the value of A[i] could be changed by assigning to i rather than to A[i]. • Treat entire array as a scalar variable. = A[i] A[j] = V = A[k] = R(A,i) A = W(A,j,V) = R(A,k) = R(A8,i7) A9 = W(A8,j6,V5) = R(A9,k4)

  9. SSA Introduction • W operator may introduce unnecessary liveness for A. Introduce HW (HiddenW). repeat i2 = Ф(i1,i3) A2= HW(i2,i2) i3 = i2 +1 until i3>10 repeat A[i] = i i = i +1 until i>10 repeat i2 = Ф(i1,i3) A1 = Ф(A0,A2) A2= W(A1,i2,i2) i3 = i2 +1 until i3>10

  10. Converting to SSA • Big picture, translation to SSA form is done in 3 steps • The dominance frontier mapping is constructed form the control flow graph. • Using the dominance frontiers, the locations of the -functions for each variable in the original program are determined. • The variables are renamed by replacing each mention of an original variable V by an appropriate mention of a new variable Vi

  11. CFG • A control flow graph G = (V, E) • Set V contains distinguished nodes ENTRY and EXIT • every node is reachable from ENTRY • EXIT is reachable from every node in G. • ENTRY has no predecessors • EXIT has no successors. • Notation: predecessor, successor, path

  12. Dominance Relation • If X appears on every path from ENTRY to Y, then X dominates Y. • Dominance relation is both reflexive and transitive. • idom(Y): immediate dominator of Y • Dominator Tree • ENTRY is the root • Any node Y other than ENTRY has idom(Y) as its parent • Notation: parent, child, ancestor, descendant

  13. Dominator Tree Example ENTRY ENTRY a EXIT a c b b d c d CFG DT EXIT

  14. Dominator Tree Example

  15. Dominance Frontier • Dominance Frontier DF(X) for node X • Set of nodes Y • X dominates a predecessor of Y • X does not strictly dominate Y • Equation 1: DF(X) = {Y|(P∈Pred(Y))(XP and XY )} • Equation 2: DF(X) = DFlocal(X)∪ ∪Z∈Children(X)DFup(z) • DFlocal(X) = {Y∈Succ(x)|idom(Y)≠X} • DFup(X) = {Y∈DF(Z)|idom(Z) ≠X}

  16. Dominance Frontier • How to proof equation 1 and equation 2 are correct? • Easy to See that => is correct • Still have to show everything in DF(X) has been accounted for. • Suppose Y ∈DF(X) and UY be the edge that X dominate U but doesn’t strictly dominate Y. • If U == X, then Y ∈DFlocal(X) • If U ≠X, then there exists a path from X to U in Dominator Tree which implies there exists a child Z of X dominate U. Z doesn’t strictly dominate Y because X doesn’t strictly dominate Y. So Y ∈DFup(X).

  17. Ф-function and Dominance Frontier • Intuition behind dominance frontier • Y ∈DF(X) means: • Y has multiple predecessors • X dominate one of them, say U, U inherits everything defined in X • Reaching definition of Y are from U and other predecessors • So Y is exactly the place where Ф-function is needed

  18. Control Dependences and Dominance Frontier • A CFG node Y is control dependent on a CFG node X if both the following hold: • There is nonnull path p: X →+ Y such that Y postdominate every node after X. • The node Y doesn’t strictly postdominate the node X • If X appears on every path from Y to Exit, then X postdominate Y.

  19. Control Dependences and Dominance Frontier • In other words, there is some edge from X that definitely causes Y to execute, and there is also some path from X that avoids executing Y.

  20. RCFG • The reverse control flow graph RCFG has the same nodes as CFG, but has edge Y →X for each edge X→Y in CFG. • Entry and Exit are also reversed. • The postdominator relation in CFG is dominator relation in RCFG. • Let X and Y be nodes in CFG. Then Y is control dependent on X in CFG iff X∈DF(Y) in RCFG.

  21. SSA Construction– Place Ф Functions • For each variable V • Add all nodes with assignments to V to worklist W • While X in W do • For each Y in DF(X) do • If no Ф added in Y then • Place (V = Ф (V,…,V)) at Y • If Y has not been added before, add Y to W.

  22. SSA Construction– Rename Variables • Rename from the ENTRY node recursively • For node X • For each assignment (V = …) in X • Rename any use of V with the TOS of rename stack • Push the new name Vi on rename stack • i = i + 1 • Rename all the Ф operands through successor edges • Recursively rename for all child nodes in the dominator tree • For each assignment (V = …) in X • Pop Vi in X from the rename stack

  23. Rename Example TOS a= a1= a1 a0 a1+5 Rename expr a= Ф(a1,a) a = a+5 a+5

  24. Converting out of SSA • Eventually, a program must be executed. • The Ф-function have precise semantics, but they are generally not represented in existing target machines.

  25. Converting Out of SSA • Naively, a k-input Ф-function at entrance to node X can be replaced by k ordinary assignments, one at the end of each control predecessor of X. • Inefficient object code can be generated.

  26. Dead Code Elimination • Where does dead code come from? • Assignments without any use • Dead code elimination method • Initially all statements are marked dead • Some statements need to be marked live because of certain conditions • Mark these statements can cause others to be marked live. • After worklist is empty, truly dead code can be removed

  27. Allocation By Coloring • It might seem possible to map all occurrence of Vi back to V and delete all Ф-functions. • After certain optimization, this assumption might not be true.

  28. Allocation By Coloring • Use graph coloring techniques to allocate storage for variables. • In the case above,V1 and V2 are alive simultaneously, they have different color, and will be converted back to two different variables.

  29. SSA Example • Dead Code Elimination Intuition • Because there is only one definition for each variable, if the list of uses of the variable is empty, the definition is dead. • When a statement v x  y is eliminated because v is dead, this statement must be removed from the list of uses of x and y. This might cause those definitions to become dead.

  30. SSA Example • Simple Constant Propagation Intuition • If there is a statement v  c, where c is a constant, then all uses of v can be replaced for c. • A  function of the form v  (c1, c2, …, cn) where all ci are identical can be replaced for v  c. • Using a work list algorithm in a program in SSA form,we can perform constant propagation in linear time.

  31. j  i k  k+1 j  k k  k+2 SSA Example B1 i  1 j  1 k 0 i=1; j=1; k=0; while(k<100) { if(j<20) { j=i; k=k+1; } else { j=k; k=k+2; } } return j; } B2 if k<100 B4 B3 return j if j<20 B6 B5 B7

  32. j  i k3  k+1 j  k k5  k+2 SSA Example B1 i  1 j  1 k1 0 i=1; j=1; k=0; while(k<100) { if(j<20) { j=i; k=k+1; } else { j=k; k=k+2; } } return j; } B2 if k<100 B4 B3 return j if j<20 B6 B5 B7

  33. j  i k3  k+1 j  k k5  k+2 SSA Example B1 i  1 j  1 k1 0 i=1; j=1; k=0; while(k<100) { if(j<20) { j=i; k=k+1; } else { j=k; k=k+2; } } return j; } B2 if k<100 B4 B3 return j if j<20 B6 B5 B7 k4  (k3,k5)

  34. j  i k3  k+1 j  k k5  k+2 SSA Example B1 i  1 j  1 k1 0 i=1; j=1; k=0; while(k<100) { if(j<20) { j=i; k=k+1; } else { j=k; k=k+2; } } return j; } B2 k2  (k4,k1) if k<100 B4 B3 return j if j<20 B6 B5 B7 k4  (k3,k5)

  35. j  i k3  k2+1 j  k k5  k2+2 SSA Example B1 i  1 j  1 k1 0 i=1; j=1; k=0; while(k<100) { if(j<20) { j=i; k=k+1; } else { j=k; k=k+2; } } return j; } B2 k2  (k4,k1) if k2<100 B4 B3 return j if j<20 B6 B5 B7 k4  (k3,k5)

  36. B1 i1  1 j1  1 k1 0 B2 j2  (j4,j1) k2  (k4,k1) if k2<100 B4 B3 return j2 if j2<20 B6 B5 j3  i1 k3  k2+1 j5  k2 k5  k2+2 B7 j4  (j3,j5) k4  (k3,k5) SSA Example i=1; j=1; k=0; while(k<100) { if(j<20) { j=i; k=k+1; } else { j=k; k=k+2; } } return j; }

  37. B1 B1 i1  1 j1  1 k1 0 i1  1 j1  1 k1 0 B2 B2 j2  (j4,j1) k2  (k4,k1) if k2<100 j2  (j4,1) k2  (k4,0) if k2<100 B4 B4 B3 B3 return j2 return j2 if j2<20 if j2<20 B6 B6 B5 B5 j3  1 k3  k2+1 j3  i1 k3  k2+1 j5  k2 k5  k2+2 j5  k2 k5  k2+2 B7 B7 j4  (j3,j5) k4  (k3,k5) j4  (j3,j5) k4  (k3,k5) SSA Example: Constant Propagation

  38. B1 i1  1 j1  1 k1 0 B2 B2 j2  (j4,1) k2  (k4,0) if k2<100 j2  (j4,1) k2  (k4,0) if k2<100 B4 B4 B3 B3 return j2 if j2<20 return j2 if j2<20 B6 B6 B5 B5 j3  1 k3  k2+1 j3  1 k3  k2+1 j5  k2 k5  k2+2 j5  k2 k5  k2+2 B7 j4  (j3,j5) k4  (k3,k5) B7 j4  (j3,j5) k4  (k3,k5) SSA Example: Dead Code Elimination

  39. B2 B2 j2  (j4,1) k2  (k4,0) if k2<100 j2  (j4,1) k2  (k4,0) if k2<100 B4 B4 B3 B3 return j2 return j2 if j2<20 if j2<20 B6 B6 B5 B5 j3  1 k3  k2+1 j3  1 k3  k2+1 j5  k2 k5  k2+2 j5  k2 k5  k2+2 B7 B7 j4  (1,j5) k4  (k3,k5) j4  (j3,j5) k4  (k3,k5) SSA Example: Constant Propagation and Dead Code Elimination

  40. k3  k2+1 j5  k2 k5  k2+2 SSA Example: One Step Further But block 6 is never executed! How can we find this out, and simplify the program? B2 j2  (j4,1) k2  (k4,0) if k2<100 B4 B3 SSA conditional constant propagation finds the least fixed point for the program and allows further elimination of dead code. return j2 if j2<20 B6 B5 B7 j4  (1,j5) k4  (k3,k5)

  41. B2 j2  (j4,1) k2  (k4,0) if k2<100 B4 return j2 B5 k3  k2+1 k3  k2+1 j5  k2 k5  k2+2 B7 j4  (1) k4  (k3) SSA Example: Dead Code Elimination B2 j2  (j4,1) k2  (k4,0) if k2<100 B4 B3 return j2 if j2<20 B6 B7 j4  (1,j5) k4  (k3,k5)

  42. B2 B2 j2  (j4,1) k2  (k4,0) if k2<100 j2  (j4,1) k2  (k4,0) if k2<100 B4 return j2 return j2 B5 B5 k3  k2+1 k3  k2+1 B7 B7 j4  (1) k4  (k3) j4  1 k4  k3 SSA Example: Single Argument  Function Elimination B4

  43. B2 B2 j2  (j4,1) k2  (k4,0) if k2<100 j2  (1,1) k2  (k3,0) if k2<100 B4 return j2 return j2 B5 B5 k3  k2+1 k3  k2+1 B7 B7 j4  1 k4  k3 j4  1 k4  k3 SSA Example: Constant and Copy Propagation B4

  44. B2 j2  (1,1) k2  (k3,0) if k2<100 B2 j2  (1,1) k2  (k3,0) if k2<100 B4 return j2 B4 return j2 B5 k3  k2+1 B5 k3  k2+1 B7 j4  1 k4  k3 SSA Example: More Dead Code

  45. B2 B2 j2  (1,1) k2  (k3,0) if k2<100 j2  1 k2  (k3,0) if k2<100 B4 B4 return j2 return j2 B5 B5 k3  k2+1 k3  k2+1 SSA Example: More  Function Simplification

  46. B2 B2 j2  1 k2  (k3,0) if k2<100 j2  1 k2  (k3,0) if k2<100 B4 B4 return j2 return 1 B5 B5 k3  k2+1 k3  k2+1 SSA Example: More Constant Propagation

  47. B2 j2  1 k2  (k3,0) if k2<100 B4 return 1 B4 return 1 B5 k3  k2+1 SSA Example: Ultimate Dead Code Elimination

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