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Static Single Assignment

Static Single Assignment. CS 540. Efficient Representations for Reachability. Efficiency is measured in terms of the size of the representation in how easy it is to use, and how easy it is to generate. Static Single Assignment (SSA). k = 2 (2) if k > 5 then (3) k = k + 1

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Static Single Assignment

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  1. Static Single Assignment CS 540

  2. Efficient Representations for Reachability Efficiency is measured in terms of • the size of the representation • in how easy it is to use, and • how easy it is to generate. Static Single Assignment (SSA) CS540 Spring 2010

  3. k = 2 (2) if k > 5 then (3) k = k + 1 (4) m = k * 2 (5) else m = k / 2 (7) endif (8) k = k + m k = 2 (2) if k(1) > 5 then (3) k = k(1)+ 1 (4) m = k(3)* 2 (5) else m = k(1) / 2 (7) endif (8) k = k(1,3)+ m(4,6) Consider the following The uses in each statement have been marked with the statement number of all definitions that reach. CS540 Spring 2010

  4. Static Single Assignment Idea: • Each definition will be uniquely numbered. • There will be a single reaching definition for each point. Algorithms for static single assignment are space efficient and take control flow into account. CS540 Spring 2010

  5. k = 2 (2) if k > 5 then (3) k = k + 1 (4) m = k * 2 (5) else m = k / 2 (7) endif (8) k = k + m k1 = 2 (2) if k1 > 5 then (3) k2 = k1 + 1 (4) m1 = k2 * 2 (5) else m2 = k1 / 2 (7) endif (8) k3 = k??+ m?? SSA numbering Problem: Because of multiple reaching definitions, we can’t give each use a unique number without analysis. CS540 Spring 2010

  6. k = 2 (2) if k > 5 then (3) k = k + 1 (4) m = k * 2 (5) else m = k / 2 (7) Endif (8) k = k + m k1 = 2 (2) if k1 > 5 then (3) k2 = k1 + 1 (4) m1 = k2 * 2 (5) else m2 = k1 / 2 (7) Endif k3 = f (k1,k2) m3 = f (m1,m2) (8) k4 = k3+ m3 F Functions F functions - merge definitions, factoring in control flow CS540 Spring 2010

  7. SSA for Structured Code Associate with each variable x, a current counter xc. Assignment statement: x := y op z becomes xxc++ := yyc op zzc CS540 Spring 2010

  8. SSA for Structured Code Loops: Repeat S until E for all variables M with definition k in loop body if M has a definition j above the loop then generate MMc++ := f (Mk,Mj); at loop start s = 1 s1 = 1 repeat repeat … … s3 = f(s1,s2) s = s + 1 s2 = s3 + 1 until s > 5 until s2 > 5 CS540 Spring 2010

  9. Computing SSA for Structured Code • Conditionals: • if E then S1 else S2 for all variables M with definition in either S1 or S2 Case 1: M has definition j in S1 and definition k in S2 Generate MMc++ := f(Mk, Mj); after the conditional if … then if … then a := b aj := bn else else a := c ak := cm al =f(aj,ak) CS540 Spring 2010

  10. Computing SSA for Structured Code • Conditionals: • if E then S1 else S2 for all variables M with definition in either S1 or S2 Case 2: M has definition k in S1 or in S2, definition j above the conditional Generate MMc++ := f(Mk, Mj); after the conditional a := c aj := cm if … then if … then a := b ak := bn else … else … al =f(aj,ak) CS540 Spring 2010

  11. Computing SSA for Structured Code • Conditionals: • if E then S1 for all variables M with definition k in S1 and definition j that reaches the conditional, generate MMc++ := f(Mk, Mj); after the conditional a := c aj := cm if … then if … then a := b ak := bn al =f(aj,ak) CS540 Spring 2010

  12. 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 CS540 Spring 2010

  13. 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 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 Number existing defns CS540 Spring 2010

  14. 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 i1 = j = k = l = 1 repeat i3 = f() 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 Add f definitions where needed CS540 Spring 2010

  15. 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 i1 = j = k = l = 1 repeat i3 = f(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 Fill in the use numbers CS540 Spring 2010

  16. 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 = j1 = k = l = 1 repeat j2 = f(j1,j4) if (p) then begin j3 = i if Q then l = 2 else l = 3 k = k + 1 end else k = k + 2 j4 = f(j2,j3) print (i,j4,k,l) repeat if R then l = l + 4 until S i = i + 6 until T CS540 Spring 2010

  17. 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 CS540 Spring 2010

  18. 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 = l1 = 1 repeat l2 = f(l1,l9) if (p) then begin j = i if Q then l3 = 2 else l4 = 3 l5 = f(l3,l4) k = k + 1 end else k = k + 2 l6 = f(l2,l5) print (i,j,k,l6) repeat l7 = f(l6,l9) if R then l8 = l7 + 4 l9 = f(l7,l8) until S i = i + 6 until T CS540 Spring 2010

  19. 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 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 CS540 Spring 2010

  20. Using SSA for Constant Propagation • For statements xi := C, for some constant C, replace all xi with C and remove the statement • For xi := f(c,c,...,c), for some constant c, replace statement with xi := c • Can extend to evaluate conditional branches • Iterate Locates AND “Performs” the replacement

  21. Example: SSA a1 := 3 d1 := 2 a := 3 d := 2 d3 = f(d1,d2) a3 = f(a1,a2) f1 := a3 + d3 g1 := 5 a2 := g1 – d3 f1 <= g1 f := a + d g := 5 a := g – d f < = g F T F T f2 := g1 + 1 g1 < a2 f := g + 1 g < a T T F F f3 := f(f1,f2) d2 := 2 d := 2

  22. Example: SSA a1 := 3 d1 := 2 a1 := 3 d1 := 2 d3 = f(d1,d2) a3 = f(a1,a2) f1 := a3 + d3 g1 := 5 a2 := g1 – d3 f1 <= g1 d3 = f(2,2) a3 = f(3,a2) f1 := a3 + d3 g1 := 5 a2 := 5 – d3 f1 <= 5 F F T T f2 := g1 + 1 g1 < a2 f2 := 5 + 1 5 < a2 T T F F f3 := f(f1,f2) d2 := 2 f3 := f(f1,f2) d2 := 2

  23. Example: SSA a1 := 3 d1 := 2 a1 := 3 d1 := 2 d3 = f(2,2) a3 = f(3,a2) f1 := a3 + d3 g1 := 5 a2 := 5 – d3 f1 <= 5 d3 = 2 a3 = f(3,a2) f1 := a3 + 2 g1 := 5 a2 := 5 – 2 f1 <= 5 F F T T f2 := 5 + 1 5 < a2 f2 := 6 5 < a2 T T F F f3 := f(f1,f2) d2 := 2 f3 := f(f1,6) d2 := 2

  24. Example: SSA a1 := 3 d1 := 2 a1 := 3 d1 := 2 d3 = 2 a3 = f(3,a2) f1 := a3 + 2 g1 := 5 a2 := 5 – 2 f1 <= 5 d3 = 2 a3 = f(3,3) f1 := a3 + 2 g1 := 5 a2 := 3 f1 <= 5 F F T T f2 := 6 5 < a2 f2 := 6 5 < 3 T T F F f3 := f(f1,6) d2 := 2 f3 := f(f1,6) d2 := 2

  25. Example: SSA a1 := 3 d1 := 2 a1 := 3 d1 := 2 d3 = 2 a3 = f(3,3) f1 := a3 + 2 g1 := 5 a2 := 3 f1 <= 5 d3 = 2 a3 = 3 f1 := 3 + 2 g1 := 5 a2 := 3 f1 <= 5 F F T T f2 := 6 5 < 3 f2 := 6 T F f3 := f(f1,6) d2 := 2 f3 := f(f1,6) d2 := 2

  26. Example: SSA a1 := 3 d1 := 2 a1 := 3 d1 := 2 d3 = 2 a3 = 3 f1 := 3 + 2 g1 := 5 a2 := 3 f1 <= 5 d3 = 2 a3 = 3 f1 := 3 + 2 g1 := 5 a2 := 3 true F T f2 := 6 f2 := 6 f3 := f(f2) d2 := 2 f3 := f(6) d2 := 2

  27. Example: SSA a1 := 3 d1 := 2 a1 := 3 d1 := 2 d3 = 2 a3 = 3 f1 := 5 g1 := 5 a2 := 3 true d3 = 2 a3 = 3 f1 := 3 + 2 g1 := 5 a2 := 3 f2 := 6 f2 := 6 f3 := f(6) d2 := 2 d2 := 2

  28. ... X17  x10 ... X17  x11 X17Ø(x10,x11) ...  x17 ...  x17 SSA Deconstruction At some point, we need executable code • Can’t implement Ø operations • Need to fix up the flow of values Basic idea • Insert copies Ø-function pred’s • Simple algorithm • Works in most cases • Adds lots of copies • Most of them coalesce away CS540 Spring 2010 *

  29. i = j = k0 = l = 1 k1 = k0 repeat if (p) then begin j = i if Q then l = 2 else l = 3 k2 = k2 + 1 k4 = k2 end else k3 = k1 + 2 k4 = k3 print (i,j,k4,l) repeat if R then l = l + 4 until S i = i + 6 k1 = k4 until T k0 = 1 k1 = k0 1 2 k1 = f(k0,k4) 3 4 5 7 k3 =k1 + 2 k4 = k3 6 k2 =k1 + 1 k4 = k2 8 k4 = f(k2,k3) 9 10 11 Exit 12 CS540 Spring 2010 k1 =k4

  30. Final Exam • 75%-80% (ish) on material since the midterm • Syntax directed translation (a little of this on midterm) • Symbol table & types • Intermediate code • Runtime Environments • Code Generation • Code Optimization • Remaining – HL concepts from the first part of the semester CS 540 Spring 2010 GMU

  31. Final Exam: Syntax directed translation & Types • SDT: • Some on midterm already • Got lots of practice (program #3, #4) • Understanding/Creating • Symbol Tables & Types • Types Terminology • Scope • Table implementation CS 540 Spring 2010 GMU

  32. Final Exam: Intermediate Code & RT Environments • Intermediate Code • Expressions, Control constructs • Should be able to write/understand basic code • Don’t memorize spim – will put info on exam • RT Environments • Control flow • Data flow • Variable addressing • Parameter passing CS 540 Spring 2010 GMU

  33. Final Exam: Code Generation & Code Optimization • Code Generation • Instr. selection/Instruction scheduling: only what they are trying to accomplish • Register allocation: understand how to use liveness to allocation, graph coloring to assign • Review the example on the slides – could get a question like that • Code Optimization • Gave lots of examples of optimizations that are useful • Dataflow analysis basics – I won’t make you use the equations (would take too long) but you should have a general idea what is being done. CS 540 Spring 2010 GMU

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