SSA-Form Register Allocation

Transcription

1 SSA-Form Register Allocation Foundations Sebastian Hack Compiler Construction Course Winter Term 2009/2010 saarland university computer science

2 2 Overview 1 Graph Theory Perfect Graphs Chordal Graphs 2 SSA Form Dominance φ-functions 3 Interference Graphs Non-SSA Interference Graphs Perfect Elimination Orders Chordal Graphs 4 Interference Graphs of SSA-form Programs Dominance and Liveness Dominance and Interference Spilling Implementing φ-functions 5 Intuition

3 3 Overview 1 Graph Theory Perfect Graphs Chordal Graphs 2 SSA Form Dominance φ-functions 3 Interference Graphs Non-SSA Interference Graphs Perfect Elimination Orders Chordal Graphs 4 Interference Graphs of SSA-form Programs Dominance and Liveness Dominance and Interference Spilling Implementing φ-functions 5 Intuition

4 4 Complete Graphs and Cycles Complete Graph K 5 Cycle C 5

5 5 Induced Subgraphs Graph with a C 4 subgraph Graph with a C 4 induced subgraph

6 5 Induced Subgraphs Graph with a C 4 subgraph Graph with a C 4 induced subgraph Note Induced complete graphs are called cliques

7 6 Clique number and Chromatic number Definition ω(g) Size of the largest clique in G χ(g) Number of colors in a minimum coloring of G

8 6 Clique number and Chromatic number Definition ω(g) Size of the largest clique in G χ(g) Number of colors in a minimum coloring of G Corollary ω(g) χ(g) holds for each graph G

9 6 Clique number and Chromatic number Definition ω(g) Size of the largest clique in G χ(g) Number of colors in a minimum coloring of G Corollary ω(g) χ(g) holds for each graph G ω(g) χ(g)

10 7 Perfect Graphs Definition G is perfect χ(h) = ω(h) for each induced subgraph H of G

11 7 Perfect Graphs Definition G is perfect χ(h) = ω(h) for each induced subgraph H of G perfect?

12 7 Perfect Graphs Definition G is perfect χ(h) = ω(h) for each induced subgraph H of G perfect?

13 8 Chordal Graphs Definition G is chordal G contains no induced cycles longer than 3

14 8 Chordal Graphs Definition G is chordal G contains no induced cycles longer than 3 chordal?

15 8 Chordal Graphs Definition G is chordal G contains no induced cycles longer than 3 chordal? Theorem Chordal graphs are perfect

16 8 Chordal Graphs Definition G is chordal G contains no induced cycles longer than 3 chordal? Theorem Chordal graphs are perfect Theorem Chordal graphs can be colored optimally in O( V ω(g))

17 9 Overview 1 Graph Theory Perfect Graphs Chordal Graphs 2 SSA Form Dominance φ-functions 3 Interference Graphs Non-SSA Interference Graphs Perfect Elimination Orders Chordal Graphs 4 Interference Graphs of SSA-form Programs Dominance and Liveness Dominance and Interference Spilling Implementing φ-functions 5 Intuition

18 10 Dominance Definition Every use of a variable is dominated by its definition start v v

19 10 Dominance Definition Every use of a variable is dominated by its definition start v You cannot reach the use without passing by the definition Else, you could use uninitialized variables Dominance induces a tree on the control flow graph v Sometimes called strict SSA

20 11 What do φ-functions mean? z 1 φ(x 1, y 1 ) z 2 φ(x 2, y 2 ) z 3 φ(x 3, y 3 ) z 1 x 1 z 2 x 2 z 3 x 3 z 1 y 1 z 2 y 2 z 3 y 3 Frequent misconception Put a sequence of copies in the predecessors

21 11 What do φ-functions mean? z 1 φ(x 1, y 1 ) z 2 φ(x 2, y 2 ) z 3 φ(x 3, y 3 ) z 1 x 1 z 2 x 2 z 3 x 3 z 1 y 1 z 2 y 2 z 3 y 3 Frequent misconception Put a sequence of copies in the predecessors

22 12 What do φ-functions mean? Lost Copies B B z 1 y 1 z 2 y 2 z 3 y 3 A z 1 φ(, y 1 ) z 2 φ(, y 2 ) z 3 φ(, y 3 ) C A C Cannot simply push copies in predecessor Copies are also executed if we jump from B to C Need to remove critical edges (edge from B to A)

23 12 What do φ-functions mean? Lost Copies B B z 1 y 1 z 2 y 2 z 3 y 3 A z 1 φ(, y 1 ) z 2 φ(, y 2 ) z 3 φ(, y 3 ) C A C Cannot simply push copies in predecessor Copies are also executed if we jump from B to C Need to remove critical edges (edge from B to A)

24 13 What do φ-functions mean? φ-swap z 1 z 2 z 2 z 1 z 1 φ(, z 2 ) z 2 φ(, z 1 ) z 1 overwritten before used

25 13 What do φ-functions mean? φ-swap z 1 z 2 z 2 z 1 z 1 φ(, z 2 ) z 2 φ(, z 1 ) z 1 overwritten before used

26 14 What do φ-functions mean? (z 1, z 2, z 3 ) (x 1, x 2, x 3 ) (z 1, z 2, z 3 ) (y 1, y 2, y 3 ) z 1 φ(x 1, y 1 ) z 2 φ(x 2, y 2 ) z 3 φ(x 3, y 3 ) The Reality φ-functions correspond to parallel copies on the incoming edges

27 15 φ-functions and uses z 1 φ(x 1, y 1 ) z 2 φ(x 2, y 2 ) z 3 φ(x 3, y 3 ) Does not fulfill dominance property φs do not use their operands in the φ-block Uses happen in the predecessors

28 15 φ-functions and uses φ r (x 1, x 2, x 3 ) φ r (y 1, y 2, y 3 ) z 1 φ(x 1, y 1 ) z 2 φ(x 2, y 2 ) z 3 φ(x 3, y 3 ) (z 1, z 2, z 3 ) φ w Does not fulfill dominance property φs do not use their operands in the φ-block Uses happen in the predecessors Split φ-functions in two parts: Split critical edges Read part (φ r ) in the predecessors Write part (φ w ) in the block Correct modelling of liveness

29 16 Overview 1 Graph Theory Perfect Graphs Chordal Graphs 2 SSA Form Dominance φ-functions 3 Interference Graphs Non-SSA Interference Graphs Perfect Elimination Orders Chordal Graphs 4 Interference Graphs of SSA-form Programs Dominance and Liveness Dominance and Interference Spilling Implementing φ-functions 5 Intuition

30 17 Non-SSA Interference Graphs An inconvenient property Program Interference Graph a 1 c b d 1 e a + 1 d b a + a c a + 1 e b + 1 c e d a The number of live variables at each instruction (register pressure) is 2 However, we need 3 registers for a correct register allocation In theory, this gap can be arbitrarily large (Mycielski Graphs)

31 18 Graph-Coloring Register Allocation [Chaitin 80, Briggs 92, Appel & George 96, Park & Moon 04] program changed Spill Build IG Coalesce Color coloring heuristic failed Every undirected graph can occur as an interference graph = Finding a k-coloring is NP-complete Color using heuristic = Iteration necessary Might introduce spills although IG is k-colorable Rebuilding the IG each iteration is costly

32 18 Graph-Coloring Register Allocation [Chaitin 80, Briggs 92, Appel & George 96, Park & Moon 04] program changed Spill Build IG Coalesce Color coloring heuristic failed Spill-code insertion is crucial for the program s performance Hence, it should be very sensitive to the structure of the program Place load and stores carefully Avoid spilling in loops! Here, it is merely a fail-safe for coloring

33 19 Coloring Subsequently remove the nodes from the graph d e elimination order a b c

34 19 Coloring Subsequently remove the nodes from the graph a d b e c elimination order d,

35 19 Coloring Subsequently remove the nodes from the graph a d b e c elimination order d, e,

36 19 Coloring Subsequently remove the nodes from the graph a d b e c elimination order d, e, c,

37 19 Coloring Subsequently remove the nodes from the graph a d b e c elimination order d, e, c, a,

38 19 Coloring Subsequently remove the nodes from the graph a d b e c elimination order d, e, c, a, b

39 19 Coloring Subsequently remove the nodes from the graph Re-insert the nodes in reverse order Assign each node the next possible color a d b e c elimination order d, e, c, a, b

40 19 Coloring Subsequently remove the nodes from the graph Re-insert the nodes in reverse order Assign each node the next possible color a d b e c elimination order d, e, c, a,

41 19 Coloring Subsequently remove the nodes from the graph Re-insert the nodes in reverse order Assign each node the next possible color a d b e c elimination order d, e, c,

42 19 Coloring Subsequently remove the nodes from the graph Re-insert the nodes in reverse order Assign each node the next possible color a d b e c elimination order d, e,

43 19 Coloring Subsequently remove the nodes from the graph Re-insert the nodes in reverse order Assign each node the next possible color a d b e c elimination order d,

44 19 Coloring Subsequently remove the nodes from the graph Re-insert the nodes in reverse order Assign each node the next possible color d e elimination order a b c

45 19 Coloring Subsequently remove the nodes from the graph Re-insert the nodes in reverse order Assign each node the next possible color d e elimination order a b c But... this graph is 3-colorable. We obviously picked a wrong order.

46 19 Coloring Subsequently remove the nodes from the graph Re-insert the nodes in reverse order Assign each node the next possible color d e elimination order a b c But... this graph is 3-colorable. We obviously picked a wrong order.

47 20 Coloring PEOs Perfect Elimination Order (PEO) All not yet eliminated neighbors of a node are mutually connected d e elimination order a b c

48 20 Coloring PEOs Perfect Elimination Order (PEO) All not yet eliminated neighbors of a node are mutually connected a d b e c elimination order a,

49 20 Coloring PEOs Perfect Elimination Order (PEO) All not yet eliminated neighbors of a node are mutually connected a d b e c elimination order a, c,

50 20 Coloring PEOs Perfect Elimination Order (PEO) All not yet eliminated neighbors of a node are mutually connected a d b e c elimination order a, c, d,

51 20 Coloring PEOs Perfect Elimination Order (PEO) All not yet eliminated neighbors of a node are mutually connected a d b e c elimination order a, c, d, e,

52 20 Coloring PEOs Perfect Elimination Order (PEO) All not yet eliminated neighbors of a node are mutually connected a d b e c elimination order a, c, d, e, b

53 20 Coloring PEOs Perfect Elimination Order (PEO) All not yet eliminated neighbors of a node are mutually connected a d b e c elimination order a, c, d, e,

54 20 Coloring PEOs Perfect Elimination Order (PEO) All not yet eliminated neighbors of a node are mutually connected a d b e c elimination order a, c, d,

55 20 Coloring PEOs Perfect Elimination Order (PEO) All not yet eliminated neighbors of a node are mutually connected a d b e c elimination order a, c,

56 20 Coloring PEOs Perfect Elimination Order (PEO) All not yet eliminated neighbors of a node are mutually connected a d b e c elimination order a,

57 20 Coloring PEOs Perfect Elimination Order (PEO) All not yet eliminated neighbors of a node are mutually connected d e elimination order a b c

58 20 Coloring PEOs Perfect Elimination Order (PEO) All not yet eliminated neighbors of a node are mutually connected d e elimination order a b c From Graph Theory [Berge 60, Fulkerson/Gross 65, Gavril 72] A PEO allows for an optimal coloring in k V The number of colors is bound by the size of the largest clique

59 21 Coloring PEOs Graphs with holes larger than 3 have no PEO, e.g. G has a PEO G is chordal

60 21 Coloring PEOs Graphs with holes larger than 3 have no PEO, e.g. G has a PEO G is chordal Core Theorem of SSA Register Allocation The dominance relation in SSA programs induces a PEO in the IG Thus, SSA IGs are chordal

61 22 Overview 1 Graph Theory Perfect Graphs Chordal Graphs 2 SSA Form Dominance φ-functions 3 Interference Graphs Non-SSA Interference Graphs Perfect Elimination Orders Chordal Graphs 4 Interference Graphs of SSA-form Programs Dominance and Liveness Dominance and Interference Spilling Implementing φ-functions 5 Intuition

62 23 Liveness and Dominance Each instruction l where a variable v is live, is dominated by v start v l : v

63 23 Liveness and Dominance Each instruction l where a variable v is live, is dominated by v start v l : v Why? Assume l is not dominated by v Then there s a path from start to some usage of v not containing the definition of v This cannot be since each value must have been defined before it is used

64 23 Liveness and Dominance Each instruction l where a variable v is live, is dominated by v start v l : v Why? Assume l is not dominated by v Then there s a path from start to some usage of v not containing the definition of v This cannot be since each value must have been defined before it is used

65 24 Interference and Dominance Assume v, w interfere, i.e. they are live at some instruction l Then, v l and w l Since dominance is a tree, either v w or w v v {, } w

66 24 Interference and Dominance Assume v, w interfere, i.e. they are live at some instruction l Then, v l and w l Since dominance is a tree, either v w or w v v {, } w Consequences Each edge in the IG is directed by dominance The interference graph is an excerpt of the dominance relation

67 25 Interference and Dominance Assume v w Then, v is live at w v dominance subtree of v w

68 25 Interference and Dominance Assume v w Why? Then, v is live at w v w dominance subtree of v v both live w If v and w interfere then there is a place where both are live w dominates all places where w is live Hence, v is live inside w s dominance tree Thus, v is live at w

69 26 Interference and Dominance Consider three nodes u, v, w in the IG: u??? v w

70 26 Interference and Dominance Consider three nodes u, v, w in the IG: u??? v u is live at w v is live at w w

71 26 Interference and Dominance Consider three nodes u, v, w in the IG: u or v u is live at w v is live at w w Thus, they interfere

72 26 Interference and Dominance Consider three nodes u, v, w in the IG: u or v u is live at w v is live at w w Thus, they interfere Conclusion All variables that... interfere with w dominate w... are mutually connected in the IG

73 27 Dominance and PEOs Before a value v is added to a PEO, add all values whose definitions are dominated by v A post order walk of the dominance tree defines a PEO A pre order walk of the dominance tree yields a coloring sequence IGs of SSA-form programs can be colored optimally in O(ω(G) V ) Without constructing the interference graph itself

74 28 Spilling Theorem For each clique in the IG there is a program point where all nodes in the clique are live.

75 28 Spilling Theorem For each clique in the IG there is a program point where all nodes in the clique are live. a b c d Dominance induces a total order inside the clique There is a smallest value d All others are live at the definition of d

76 29 Spilling Consequences The chromatic number of the IG is exactly determined by the number of live variables at the labels Lowering the number of values live at each label to k makes the IG k-colorable We know in advance where values must be spilled = All labels where the pressure is larger than k Spilling can be done before coloring and coloring will always succeed afterwards

77 29 Spilling Consequences The chromatic number of the IG is exactly determined by the number of live variables at the labels Lowering the number of values live at each label to k makes the IG k-colorable We know in advance where values must be spilled = All labels where the pressure is larger than k Spilling can be done before coloring and coloring will always succeed afterwards Conclusion No iteration as in Chaitin/Briggs-allocators No interference graph necessary

78 30 Getting out of SSA We now have a k-coloring of the SSA interference graph Can we turn that program into a non-ssa program and maintain the coloring?

79 30 Getting out of SSA We now have a k-coloring of the SSA interference graph Can we turn that program into a non-ssa program and maintain the coloring? Central question What to do about φ-functions?

80 31 Φ-Functions Consider following example z 1 φ(x 1, y 1 ) z 2 φ(x 2, y 2 ) z 3 φ(x 3, y 3 )

81 31 Φ-Functions Consider following example (z 1, z 2, z 3 ) (x 1, x 2, x 3 ) (z 1, z 2, z 3 ) (y 1, y 2, y 3 ) z 1 φ(x 1, y 1 ) z 2 φ(x 2, y 2 ) z 3 φ(x 3, y 3 ) Φ-functions are parallel copies on control flow edges

82 31 Φ-Functions Consider following example (z 1, z 2, z 3 ) (x 1, x 2, x 3 ) (z 1, z 2, z 3 ) (y 1, y 2, y 3 ) z 1 φ(x 1, y 1 ) z 2 φ(x 2, y 2 ) z 3 φ(x 3, y 3 ) Φ-functions are parallel copies on control flow edges Considering assigned registers...

83 31 Φ-Functions Consider following example z 1 φ(x 1, y 1 ) z 2 φ(x 2, y 2 ) z 3 φ(x 3, y 3 ) Φ-functions are parallel copies on control flow edges Considering assigned registers Φs represent register permutations

84 32 Permutations A permutation can be implemented with copies if one auxiliary register is available Permutations can be implemented by a series of transpositions (i.e. swaps) = A transposition can be implemented by three xors without a third register

85 33 Intuition: Why do SSA IGs do not have cycles? Why are SSA IGs chordal? Program a b c d a + b e c + 1 Live Ranges a b c d e Interference Graph a b e c d How can we create a 4-cycle {a, c, d, e}?

86 33 Intuition: Why do SSA IGs do not have cycles? Why are SSA IGs chordal? Program a b c d a + b e c + 1 a Live Ranges a b c d e a Interference Graph a b e c d How can we create a 4-cycle {a, c, d, e}? Redefine a = SSA violated!

87 34 Intuition: φ-functions break cycles in the IG Program and live ranges a d e a + d b c a + e b c Interference Graph d a e b c

88 34 Intuition: φ-functions break cycles in the IG Program and live ranges a d e 1 a + d b c a + e 2 b c Interference Graph d a e 1 e 3 e 3 φ(e 1, e 2 ) b c e 2

89 35 Intuition: Why destroying SSA before RA is bad Parallel copies (a, b, c, d ) (a, b, c, d) Sequential copies d d c c b b a a

90 35 Intuition: Why destroying SSA before RA is bad Parallel copies (a, b, c, d ) (a, b, c, d) a b c d Sequential copies a b c d d d c c b b a a a b c d a b c d

91 35 Intuition: Why destroying SSA before RA is bad Parallel copies (a, b, c, d ) (a, b, c, d) a b c d Sequential copies a b c d d d c c b b a a a b c d a b c d

92 36 Summary IGs of SSA-form programs are chordal The dominance relation induces a PEO No further spills after pressure is lowered Register assignment optimal in linear time Do not need to construct interference graph Allocator without iteration Spill Color Φ-Impl. Coalesce