1 / 31

Optimizing Compilers CISC 673 Spring 2011 Static Single Assignment III

Optimizing Compilers CISC 673 Spring 2011 Static Single Assignment III. John Cavazos University of Delaware. SSA Construction Algorithm. 1. Insert  -functions a.) calculate dominance frontiers b.) find global names for each name, build list of blocks that define it

joannao
Télécharger la présentation

Optimizing Compilers CISC 673 Spring 2011 Static Single Assignment III

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Optimizing CompilersCISC 673Spring 2011Static Single Assignment III John Cavazos University of Delaware

  2. SSA Construction Algorithm 1. Insert  -functions a.) calculate dominance frontiers b.) find global names for each name, build list of blocks that define it c.) insert  -functions

  3. Insert  -functions • global name n worklist ← Block(n) // blocks in which n is assigned  block b ∈ worklist  block d in b’s dominance frontier insert a  -function for n in d add d to worklist

  4. B0 B1 x (...) B2 B3 B4 B5 x ... B6 x (...) • DF(4) is {6}, so def in 4 forces -function in 6 B7 x (...) • def in 6 forces -function in DF(6) = {7} • def in 7 forces -function in DF(7) = {1} Dominance Frontiers & -Function Insertion • A definition at n forces a -function at m iff • n  DOM(m) but n DOM(p) for some p  preds(m) • DF(n ) is fringe just beyond region n dominates Dominance Frontiers • def in 1 forces -function in DF(1) = Ø (halt ) For each assignment, we insert the  -functions

  5. B0 i > 100 i  ... B1 a (a,a) b  (b,b) c  (c,c) d (d,d) i (i,i) a  ... c  ... B2 B3 b  ... c  ... d  ... a  ... d  ... d  (d,d) c  (c,c) b  ... B4 B5 B6 d  ... c  ... B7 a (a,a) b  (b,b) c (c,c) d (d,d) y  a+b z  c+d i  i+1 i > 100 Excluding local names avoids ’s for y & z • With all the -functions • Lots of new ops • Renaming is next Assume a, b, c, & d defined before B0

  6. SSA Construction Algorithm (Less high-level sketch) 2. Rename variables in a pre-order walk over dominator tree (uses counter and a stack per global name) Staring with the root block, b a.) generate unique names for each  -function and push them on the appropriate stacks

  7. SSA Construction Algorithm (Less high-level sketch) • Rename variables (cont’d) b.) rewrite each operation in the block i. Rewrite uses of global names with the current version (from the stack) ii. Rewrite definition by creating & pushing new name c.) fill in  -function parameters of successor blocks d.) recurse on b’s children in the dominator tree e.)<on exit from block b> pop names generated in b from stacks

  8. SSA Construction Algorithm Adding all the details ... Rename(b) for each  -function in b, x  (…) rename x as NewName(x) for each operation “x  y op z” in b rewrite y as top(stack[y]) rewrite z as top(stack[z]) rewrite x as NewName(x) for each successor of b in the CFG rewrite appropriate  parameters for each successor s of b in dom. tree Rename(s) for each operation “x  y op z” in b pop(stack[x]) for each global name i counter[i]  0 call Rename(n0) NewName(n) i  counter[n] counter[n]  counter[n] + 1 push ni onto stack[n] return ni

  9. B0 i > 100 i  ... B1 a (a,a) b  (b,b) c (c,c) d (d,d) i (i,i) a  ... c  ... Assume a, b, c, & d defined before B0 B2 B3 b  ... c  ... d  ... a  ... d  ... d  (d,d) c  (c,c) b  ... B4 B5 B6 d  ... c  ... B7 a (a,a) b  (b,b) c (c,c) d (d,d) y  a+b z  c+d i  i+1 i has not been defined i > 100 Before processing B0 a b c d i Counters Stacks 1 1 1 1 0 a0 b0 c0 d0

  10. B1 a (a0,a) b  (b0,b) c (c0,c) d (d0,d) i (i0,i) a  ... c  ... B2 B3 b  ... c  ... d  ... a  ... d  ... B4 B5 d  ... c  ... i > 100 B0 i > 100 i0 ... End of B0 d  (d,d) c  (c,c) b  ... a b c d i B6 Counters Stacks 1 1 1 1 1 B7 a (a,a) b  (b,b) c (c,c) d (d,d) y  a+b z  c+d i  i+1 a0 b0 c0 d0 i0

  11. B1 a1 (a0,a) b1  (b0,b) c1 (c0,c) d1 (d0,d) i1 (i0,i) a2 ... c2  ... B2 B3 b  ... c  ... d  ... a  ... d  ... B4 B5 d  ... c  ... i > 100 B0 i > 100 i0 ... End of B1 d  (d,d) c  (c,c) b  ... a b c d i B6 Counters Stacks 3 2 3 2 2 B7 a (a,a) b  (b,b) c (c,c) d (d,d) y  a+b z  c+d i  i+1 a0 b0 c0 d0 i0 a1 b1 c1 d1 i1 c2 a2

  12. B1 a1 (a0,a) b1 (b0,b) c1 (c0,c) d1 (d0,d) i1 (i0,i) a2 ... c2  ... B2 B3 b2 ... c3  ... d2  ... a  ... d  ... B4 B5 d  ... c  ... i > 100 B0 i > 100 i0 ... End of B2 d  (d,d) c  (c,c) b  ... a b c d i B6 Counters Stacks 3 3 4 3 2 B7 a (a2,a) b  (b2,b) c (c3,c) d (d2,d) y  a+b z  c+d i  i+1 a0 b0 c0 d0 i0 a1 b1 c1 d1 i1 d2 b2 c2 a2 c3 23

  13. B1 a1 (a0,a) b1 (b0,b) c1 (c0,c) d1 (d0,d) i1 (i0,i) a2 ... c2  ... B2 B3 b2 ... c3  ... d2  ... a  ... d  ... B4 B5 d  ... c  ... i > 100 B0 i > 100 i0 ... Before starting B3 d  (d,d) c  (c,c) b  ... a b c d i B6 Counters Stacks 3 3 4 3 2 B7 i ≤ 100 a (a2,a) b  (b2,b) c (c3,c) d (d2,d) y  a+b z  c+d i  i+1 a0 b0 c0 d0 i0 a1 b1 c1 d1 i1 c2 a2 24

  14. B1 a1 (a0,a) b1 (b0,b) c1 (c0,c) d1 (d0,d) i1 (i0,i) a2 ... c2  ... B2 B3 b2 ... c3  ... d2  ... a3 ... d3  ... B4 B5 d  ... c  ... i > 100 B0 i > 100 i0 ... End of B3 d  (d,d) c  (c,c) b  ... a b c d i B6 Counters Stacks 4 3 4 4 2 B7 a (a2,a) b  (b2,b) c (c3,c) d (d2,d) y  a+b z  c+d i  i+1 a0 b0 c0 d0 i0 a1 b1 c1 d1 i1 c2 d3 a2 a3 25

  15. B1 a1 (a0,a) b1 (b0,b) c1 (c0,c) d1 (d0,d) i1 (i0,i) a2 ... c2  ... B2 B3 b2 ... c3  ... d2  ... a3 ... d3  ... B4 B5 d4  ... c  ... i > 100 B0 i > 100 i0 ... End of B4 d  (d4,d) c  (c2,c) b  ... a b c d i B6 Counters Stacks 4 3 4 5 2 B7 a (a2,a) b  (b2,b) c (c3,c) d (d2,d) y  a+b z  c+d i  i+1 a0 b0 c0 d0 i0 a1 b1 c1 d1 i1 c2 d3 a2 a3 d4 26

  16. B1 a1 (a0,a) b1 (b0,b) c1 (c0,c) d1 (d0,d) i1 (i0,i) a2 ... c2  ... B2 B3 b2 ... c3  ... d2  ... a3 ... d3  ... B4 B5 d4  ... c4  ... i > 100 B0 i > 100 i0 ... End of B5 d  (d4,d3) c  (c2,c4) b  ... a b c d i B6 Counters Stacks 4 3 5 5 2 B7 a (a2,a) b  (b2,b) c (c3,c) d (d2,d) y  a+b z  c+d i  i+1 a0 b0 c0 d0 i0 a1 b1 c1 d1 i1 c2 d3 a2 a3 c4 27

  17. B1 a1 (a0,a) b1 (b0,b) c1 (c0,c) d1 (d0,d) i1 (i0,i) a2 ... c2  ... B2 B3 b2 ... c3  ... d2  ... a3 ... d3  ... B4 B5 d4  ... c4  ... i > 100 B0 i > 100 i0 ... End of B6 d5  (d4,d3) c5  (c2,c4) b3  ... a b c d i B6 Counters Stacks 4 4 6 6 2 B7 a (a2,a3) b  (b2,b3) c (c3,c5) d (d2,d5) y  a+b z  c+d i  i+1 a0 b0 c0 d0 i0 a1 b1 c1 d1 i1 c2 d3 a2 b3 a3 c5 d5 28

  18. B1 a1 (a0,a) b1 (b0,b) c1 (c0,c) d1 (d0,d) i1 (i0,i) a2 ... c2  ... B2 B3 b2 ... c3  ... d2  ... a3 ... d3  ... B4 B5 d4  ... c4  ... i > 100 B0 i > 100 i0 ... Before B7 d5  (d4,d3) c5  (c2,c4) b3  ... a b c d i B6 Counters Stacks 4 4 6 6 2 B7 a (a2,a3) b  (b2,b3) c (c3,c5) d (d2,d5) y  a+b z  c+d i  i+1 a0 b0 c0 d0 i0 a1 b1 c1 d1 i1 c2 a2 29

  19. B1 a1 (a0,a4) b1 (b0,b4) c1 (c0,c6) d1 (d0,d6) i1 (i0,i2) a2 ... c2  ... B2 B3 b2 ... c3  ... d2  ... a3 ... d3  ... B4 B5 d4  ... c4  ... i > 100 B0 i > 100 i0 ... End of B7 d5  (d4,d3) c5  (c2,c4) b3  ... a b c d i B6 Counters Stacks 5 5 7 7 3 B7 a4 (a2,a3) b4 (b2,b3) c6 (c3,c5) d6 (d2,d5) y  a4+b4 z  c6+d6 i2  i1+1 a0 b0 c0 d0 i0 a1 b1 c1 d1 i1 b4 c2 d6 a2 i2 a4 c6 30

  20. B1 a1 (a0,a4) b1 (b0,b4) c1 (c0,c6) d1 (d0,d6) i1 (i0,i2) a2 ... c2  ... B2 B3 b2 ... c3  ... d2  ... a3 ... d3  ... B4 B5 d4  ... c4  ... i > 100 B0 i > 100 i0 ... • After renaming • Semi-pruned SSA form • We’re done … d5  (d4,d3) c5  (c2,c4) b3  ... B6 B7 a4 (a2,a3) b4 (b2,b3) c6 (c3,c5) d6 (d2,d5) y  a4+b4 z  c6+d6 i2  i1+1 Semi-pruned  only names live in 2 or more blocks are “global names”. 31

  21. ... X17  x10 ... X17  x11 X17(x10,x11) ...  x17 ...  x17 SSA Deconstruction At some point, we need executable code • Few machines implement  operations • Need to fix up the flow of values Basic idea • Insert copies -function pred’s • Simple algorithm • Works in most cases • Adds lots of copies • Many of them coalesce away

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

  23. Constant Prop Example

  24. Constant Prop Example ⊥⊥⊥ Set Boundary Conditions XYZ 1. ⊥⊥⊥ 2. 3. ⊥⊥⊥ ⊥⊥⊥ 4. ⊥⊥⊥

  25. Constant Prop Example ⊥⊥⊥ XYZ out1= 1⊥⊥ out2= 023 out3= 12⊥ out4= ⊤23 1. 1⊥⊥ ⊥⊥⊥ 2. 3. ⊥⊥⊥ ⊥⊥⊥ We are propagating information through each node. 12⊥ 023 4. ⊥⊥⊥ ⊤23

  26. Sparse Constant Prop Example • Consider what happens when a variable gets updated during constant prop using worklist • Put all successors of CFG node into worklist • But if “x” is not used in immediate successors? • Lot of wasted time propagating data. • Update of “x” only matters at last node

  27. Sparse Constant Prop Example • Instead of propagating data along CFG, what if we just propagate along use-def edges? • When x is updated • propagate data directly to last node • bypasses all intermediate nodes!

  28. Problems with U-D chains • Can be expensive to represent • Each use can have multiple defs • Makes it difficult to keep u/d information accurate • Multiple defs make optimization harder • Use SSA!

  29. SSA vs U-D Chains • We have 16 u-d chains • In SSA form: place a phi-node in middle

  30. Problems with u-d chains • What happens if we statically know direction of branch? • Do no need to propagate information along that path • Easy to do with CFGs • U-D chains • Hard to tell which definitions to ignore

  31. U-D with SSA • SSA form shortens u-d chains • Chains terminate at merge points, rather than crossing them • Can simply ignore information merged from un-taken branches • Much easier to account for irrelevant information

More Related