SSA Construction & Destruction - ethz.ch

Advanced Compiler Design – Homework 1

SSA Construction & Destruction

Michael [email protected]

(slides originally stolen from Luca Della Toffola)

SSA Pipeline

1. Build the control-flow graph of the program

2. Convert the program into SSA form

3. Apply optimizations

4. Convert the program back to non-SSA form

5. Generate code from the control-flow graph

2

Homework 1 Tasks

1. Build the control-flow graph of the program

2. Convert the program into SSA form

3. Apply optimizations

4. Convert the program back to non-SSA form

5. Generate code from the control-flow graph

(Due date: Mar 24, 2016)

3

SSA Construction

1. Compute Dominance Frontier

• Using Dominator Tree

2. Insert ɸ-operations where necessary

• In join nodes with different incoming assignments

• Use dominance frontier

3. Add versions to variables (renaming)

4

SSA Destruction

1. Replace ɸ-operations with copy statements in direct predecessor blocks

5

DOMINANCE RELATIONS REVIEWHomework 1

6

Dominator

A (CFG) node y dominates n if y lies in every path from the entry node to n.

DOM(n) = the set of all dominators of n

Note:• Every node dominates itself• The entry node dominates all nodes

Notation from Cooper et al.

7

n DOM(n)

A {entry, A}

B {entry, B}

C {entry, A, C}

D {entry, D}

E {entry, B, E}

F {entry, A, C, F}

G {entry, D, G}

H {entry, H}

A

D

B

EC

GF

H

entry

8

SDOM(n) = the set of strict dominators of n

Note:• SDOM(n) = DOM(n) \ {n}

Strict Dominator

A node y strictly dominates n

if y dominates n and n ≠ y.

9

A

D

B

EC

GF

H

n SDOM(n)

A {entry}

B {entry}

C {entry, A}

D {entry}

E {entry, B}

F {entry, A, C}

G {entry, D}

H {entry}

entry

10

IDOM(n) = the immediate dominator of n

Note:• Every node (except the entry) has exactly one

immediate dominator

Immediate Dominator

Node y immediately dominates n if y is the closest strict dominator of n on any path from entry to n.

11

A

D

B

EC

GF

H

n IDOM(n)

A {entry}

B {entry}

C {A}

D {entry}

E {B}

F {C}

G {D}

H {entry}

entry

12

Dominator Tree

• Contains all nodes of the CFG with edges that reflect the dominance relation

• Each node n is a child of its IDOM(n)

• Bottom-up traversal results in DOM(n)

13

A

D

B

EC

GF

H

CFG

A DB

EC G

F

H

Dominator Tree

14

DOMINATOR TREE CONSTRUCTIONHomework 1 – SSA Construction

15

Iterative Dominator Algorithm

Idea: Solve the following data-flow problem:

• For n n0, initialize DOM(n) with all nodes• Update DOM(n) according to equations• Walk over the CFG, repeat until no changes.

entry

16

direct!

A

C

B

D

E

DOM(n0) = {n0}

for n in nodes \ {n0}:DOM(n) = nodes

Dominator Algorithm: Example

17

A

C

B

D

E

DOM(n0) = {n0}



{A}

17

A

C

B

D

E

DOM(n0) = {n0}



{A}

{A,B,C,D,E}

{A,B,C,D,E}

{A,B,C,D,E}

{A,B,C,D,E}

17

A

C

B

D

E

DOM(n0) = {n0}


while changed:for n in nodes:pre = intersect(DOM(preds(n)))new = union(pre, n)DOM(n) = new


{A}

{A,B,C,D,E}

{A,B,C,D,E}

{A,B,C,D,E}

{A,B,C,D,E}

17

{A,C}

A

C

B

D

E

DOM(n0) = {n0}




{A}

{A,B,C,D,E}

{A,B,C,D,E}

{A,B,C,D,E}

17

{A,C,D}

{A,C}

A

C

B

D

E

DOM(n0) = {n0}




{A}

{A,B,C,D,E}

{A,B,C,D,E}

17

{A,C,D}

{A,B}

{A,C}

A

C

B

D

E

DOM(n0) = {n0}




{A}

{A,B,C,D,E}17

{A,C,D}

{A,E}

{A,B}

{A,C}

A

C

B

D

E

DOM(n0) = {n0}




{A}

17

Computing Dominators

• Shown algorithm

– Simple but not very efficient

– How to get IDOM(n) from DOM(n)?

• Efficient algorithm described in A Simple, Fast Dominance Algorithm by Cooper et al.

– Same idea, still very simple

– Not required for getting full marks

– Also covers shown algorithm

34

DOMINANCE FRONTIER CONSTRUCTION

Homework 1 – SSA Construction

35

Dominance Frontier

DF(n) = all nodes y such that n dominates a direct predecessor of y but does not strictly dominate y.

Intuition:The Dominance Frontier of n is where n’s dominance “ends”.

CFG!

36

Dominance Frontier: Intuition

H

A

B

F

C

D

J

I

G

E

37

Dominance Frontier: Example

A

C E

D

B

n dominates a direct pred. of y and n not strictly dominates y

38


A

C E

D

B


A

B DC

E

Dominator Tree

38


A

C E

D

B

DF(A) = {}


A

B DC

E

Dominator Tree

38


A

C E

D

B

DF(A) = {}

DF(B) = {D}


A

B DC

E

Dominator Tree

38


A

C E

D

B

DF(A) = {}

DF(B) = {D}

DF(C) = {D}


A

B DC

E

Dominator Tree

38

DF(C) = {C, D}


A

C E

D

B

DF(A) = {}

DF(B) = {D}


A

B DC

E

Dominator Tree

38

DF(C) = {C, D}


A

C E

D

B

DF(A) = {}

DF(B) = {D} DF(E) = {C}


A

B DC

E

Dominator Tree

38

DF(C) = {C, D}


A

C E

D

B

DF(A) = {}

DF(B) = {D}

DF(D) = {}

DF(E) = {C}


A

B DC

E

Dominator Tree

38

Computing the DF

• Idea:

– Given node n, do not compute DF(n) directly

– Instead, find nodes y for which n is in DF(y)

• Simple algorithm based on two observations

55

Computing the DF: Observation 1

Direct predecessors of a node n have n in theirDF unless they dominate n.

A

C

D

B

1

3

2

I

II

56

Computing the DF: Observation 2

Dominators of direct predecessors of a node nhave n in their DF unless they dominate n.

A

E

F

C

B D

… …

1

4

3

2

…

57

for n in nodes:for p in preds(n):r = pwhile r != IDOM(n):

DF(r) += nr = IDOM(r)

A

C E

D

B

A

D C

E

B

Dominator TreeCFG

n = DDF(B) = {D}DF(C) = {D}

n = CDF(E) = {C}DF(C) = {C,D}

DF(A) = {}DF(B) = {D}DF(C) = {C,D}DF(E) = {C}DF(D) = {}

Dominance Frontier Algorithm

58

direct!

ɸ-FUNCTION INSERTION & VARIABLE VERSIONING

Homework 1 – SSA Construction

59

ɸ-Function Insertion

• ɸ-functions need to be inserted in all join nodes with different incoming assignments

• ɸ-operations are assignments too!

Assignments require a ɸ-function to be inserted in the DF nodes of the block containing the assignment

BB4

x = ɸ(x,x)write(x);

BB1

x = 0

BB2

x = 1BB3

write(x)

60

DF and ɸ-Functions: Example

H

A

B

F

C

D

J

I

G

E

x = 1

x = 4

write(x)

write(x)

write(x)

61

DF and ɸ-Functions: Example

H

A

B

F

C

D

J

I

G

E

x = 1

x = 4

x = ɸ(x,x) x = ɸ(x,x)write(x)

x = ɸ(x,x,x)write(x)

write(x)

61

Variable Versioning

• Need to add versions to all variables

– S.t. each version is assigned only once (statically)

• Make sure that always the “newest” version of a variable is used

– Traverse the CFG, keep track of current (newest) and highest variable versions

– Replace all uses of unversioned variables

• Also in ɸ-functions!

66

Variable Versioning: Example

BB6

x = ɸ(x ,x )write(x );

BB1

x = 0

BB2

x = 1BB3

write(x )

BB5

x = x + 1BB4

x = ɸ(x ,x )

67


BB6


BB1

x = 0

BB2

x = 1BB3

write(x )

BB5

x = x + 1BB4

x = ɸ(x ,x )

1

67


BB6


BB1

x = 0

BB2

x = 1BB3

write(x )

BB5

x = x + 1BB4

x = ɸ(x ,x )

1

2

67


BB6


BB1

x = 0

BB2

x = 1BB3

write(x )

BB5

x = x + 1BB4

x = ɸ(x ,x )

1

2

2

67


BB6


BB1

x = 0

BB2

x = 1BB3

write(x )

BB5

x = x + 1BB4

x = ɸ(x ,x )

1

12

2

67


BB6


BB1

x = 0

BB2

x = 1BB3

write(x )

BB5

x = x + 1BB4

x = ɸ(x ,x )

1

1

1

2

2

67


BB6


BB1

x = 0

BB2

x = 1BB3

write(x )

BB5

x = x + 1BB4

x = ɸ(x ,x )

1

1

1

2

2

3

67


BB6


BB1

x = 0

BB2

x = 1BB3

write(x )

BB5

x = x + 1BB4

x = ɸ(x ,x )

1

1

1

2

2

3

3

67


BB6


BB1

x = 0

BB2

x = 1BB3

write(x )

BB5

x = x + 1BB4

x = ɸ(x ,x )

1

1

1

2

2

3 3

3

67


BB6


BB1

x = 0

BB2

x = 1BB3

write(x )

BB5

x = x + 1BB4

x = ɸ(x ,x )

1

1

1

2

2

3 34

3

67


BB6


BB1

x = 0

BB2

x = 1BB3

write(x )

BB5

x = x + 1BB4

x = ɸ(x ,x )

1

1

1

2

2

3 344

3

67


BB6


BB1

x = 0

BB2

x = 1BB3

write(x )

BB5

x = x + 1BB4

x = ɸ(x ,x )

1

1

1

2

2

3 344

35

67


BB6


BB1

x = 0

BB2

x = 1BB3

write(x )

BB5

x = x + 1BB4

x = ɸ(x ,x )

1

1

1

2

2

3 344

35

5

67

SSA DESTRUCTIONHomework 1

80

SSA Destruction

• Replace ɸ-operations with copy statements in direct predecessors

81

Removing ɸ-Functions: Example 1

BB1

x1 = 0

BB2

x2 = 1BB3

write(x1)

BB4

x3 = ɸ(x2,x1)write(x3)

82


BB1

x1 = 0

BB2

x2 = 1x3 = x2

BB3

write(x1)x3 = x1

BB4


82


BB1

x1 = 0

BB2

x2 = 1

BB3


85


BB1

x1 = 0x3 = x1

BB2

x2 = 1x3 = x2

BB3


85


BB1

x1 = 0x3 = x1

BB2

x2 = 1x3 = x2

BB3


• Process may insert redundantassignments:

x3 = x1...

x3 = x2

• Solution: Insert empty blocks where necessary

85


BB1

x1 = 0

BB2

x2 = 1

BB3



x3 = x1...

x3 = x2


BB4

89


BB1

x1 = 0

BB2

x2 = 1x3 = x2

BB3



x3 = x1...

x3 = x2


BB4

x3 = x1

89


BB1

x1 = 0

BB2

x2 = 1x3 = x2

BB3



x3 = x1...

x3 = x2

• Solution: Insert empty blocks where necessary– Already done in CFG construction

phase in HW1 fragment

BB4

x3 = x1

89

PRACTICAL TIPS(FRAMEWORK)

Homework 1

93

Framework: Dominator Tree

• Computation of IDOM(n) and DF(n) in cd.transform.Dominator class

• Dominator tree edges directly in BasicBlock:

dominatorTreeParent: BB

dominatorTreeChildren: List<BB>

dominanceFrontier: Set<BB>

cd.ir.BasicBlock

IDOM(n)

94

Framework: Variable Versions

• VariableSymbol has new field: version

• New constructor to create a new version:VariableSymbol(VariableSymbol v0sym, int version)

95

Framework: ɸ-Operations

• ɸ-operations represented by class cd.ir.Phi

write(x)


v0sym: VariableSymbol

lhs: VariableSymbol

rhs: List<Expr>

cd.ir.Phi

96

Framework: ɸ-Operations

• BasicBlock contains all attached ɸ-ops

• Map key is original variable (version 0)

• Add the new symbols to the method locals

phis: Map<VS,Phi>

cd.ir.BasicBlock

97

Testing & Debugging

• No comparison of SSA form between your and reference solution available

– Others still work, e.g., .exec vs .exec.ref

• Manually inspect .dom.dot, .ssa.dot, and

.dessa.dot files– Again, using Graphviz et al.

• Need to write test programs nevertheless!

– Or even a few little unit tests?...

98

?Questions

99

SSA Construction & Destruction - ethz.ch

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