Optimizing Compilers CISC 673 Spring 2009 More Data flow analysis

UUNIVERSITYNIVERSITY OFOF D DELAWARE ELAWARE • • C COMPUTER & OMPUTER & IINFORMATION NFORMATION SSCIENCES CIENCES DDEPARTMENTEPARTMENT

Optimizing CompilersCISC 673

Spring 2009More Data flow analysis

John CavazosUniversity of Delaware

UUNIVERSITYNIVERSITY OFOF D DELAWARE ELAWARE • • C COMPUTER & OMPUTER & IINFORMATION NFORMATION SSCIENCES CIENCES DDEPARTMENTEPARTMENT 2

Overview

Last Time: Avail Expressions Using Sets Using Bit Vectors Additional Data Flow Problem

Constant Propagation Generalization of Data Flow


Computing Available Expressions

AVAIL(b) = ppred(b) (DEEXPR(p) (AVAIL(p) EXPRKILL(p) ))

AVAIL(n0) = Ø

where preds(b) is the set of b’s predecessors in the CFG

Remember these!


Avail Expressions Using Bit Vectors

1 = avail; 0 = not Boundary condition

Forward Flow on Entry Backward Flow on Exit

Initial conditions Merge operator



1=avail0=not

(A+B)(X+Y)

10 = (A+B) avail (X+Y) not

11 = both avail

…

Note: Empty entry and exit blocks added to CFGTo save space not pictured here!



1=avail0=not

(A+B)(X+Y)

What is Merge Operator?Remember original equation

…



1=avail0=not

(A+B)(X+Y)

Merge: and

What is Boundary Condition?Remember original algorithm

(⋀)

…



1=avail0=not

(A+B)(X+Y)

Merge: and

Boundary Condition

Now must initializeOUT of blocks.

Initial Condition?(⋀)

00

…


Initial Conditions?

What is needed so loops work? Needs to be identify for merge function

X ⋀ ? = X


Initial Conditions?

What is needed so loops work? Needs to be identify for merge function

X ⋀ 1 = X



1=avail0=not

(A+B)(X+Y)

Merge: and(⋀)

00

1111

11

out=gen ⋁ (in ⋀ kill)─1.

2. 3.

4.

11

…

in[b]=⋀p pred(b)out[p]

gen and kill are constants!─



1=avail0=not

(A+B)(X+Y)

Merge: and(⋀)

00

1111

11


2. 3.

4.

out1= ??11

…




1=avail0=not

(A+B)(X+Y)

Merge: and(⋀)

00

1111

11


2. 3.

4.

out1= 11out2= ??

11

…




1=avail0=not

(A+B)(X+Y)

Merge: and(⋀)

00

1111

11


2. 3.

4.

out1= 11out2= 00 ⋁ (?? ⋀ ??)

11

…




1=avail0=not

(A+B)(X+Y)

Merge: and(⋀)

00

1111

11


2. 3.

4.

out1= 11out2= 00 ⋁ (11 ⋀ ??)

11

…




1=avail0=not

(A+B)(X+Y)

Merge: and(⋀)

00

11

11


2. 3.

4.

out1= 11out2= 00 ⋁ (11 ⋀ 01)

11

= 01out3= ??01

…




1=avail0=not

(A+B)(X+Y)

Merge: and(⋀)

00

11

11


2. 3.

4.

out1= 11out2= 00 ⋁ (11 ⋀ 01)

11

= 01out3= 01 ⋁ (?? ⋀ ??)01

…




1=avail0=not

(A+B)(X+Y)

Merge: and(⋀)

00

11

11


2. 3.

4.

out1= 11out2= 00 ⋁ (11 ⋀ 01)

11

= 01out3= 01 ⋁ (11 ⋀ 11)01

= 11

…




1=avail0=not

(A+B)(X+Y)

Merge: and(⋀)

00

11

11


2. 3.

4.

out1= 11out2= 00 ⋁ (11 ⋀ 01)

11

= 01out3= 01 ⋁ (11 ⋀ 11)01

= 11out4= ??

…




1=avail0=not

(A+B)(X+Y)

Merge: and(⋀)

00

11


2. 3.

4.

out1= 11out2= 00 ⋁ (11 ⋀ 01)

11

= 01out3= 01 ⋁ (11 ⋀ 11)01

= 11out4= 00 ⋁ (01 ⋀ 11)

= 01 01

…



Constant Propagation

Along every path to point p,variable v has same “known” value

Specialize computation at p based on v’s value


Constant Propagation

Lattice (V, ≤) Top ⊤

Not a Constant Bottom ⊥

Unknown

… -2 -1 0 1 2 ...

The Lattice of CP

Hasse diagram


Some Lattice Theory

Generalized Meet Operator ⊔ Used at “merge” points Avail Expression:

Least Upper Bound (LUB)

X ⊔ Y = least element Z s.t.

X ≤ Z and Y ≤ Z


Concrete Example: Available Expressions

{ }

{a+b} {c+d} {e+f} {a+17} {b+18}

{a+b,c+d} {a+b,a+17} {c+d,e+f} {c+d,b+18} {e+f,b+18} {a+b,e+f} {a+b,b+18} {c+d,a+17} {e+f,a+17} {a+17,b+18}

{a+b,c+d,e+f} {a+b,c+d,b+18} {a+b,c+d,a+17} {a+b,e+f,a+17} {a+b,e+f,b+18} {a+b,a+17,b+18} {c+d,e+f,a+17} {c+d,e+f,b+18} {c+d,a+17,b+18} {e+f,a+17,b+18},

{a+b,c+d,e+f,a+17} {a+b,c+d,e+f,b+18} {a+b,c+d,a+17,b+18} {a+b,e+f,a+17,b+18}

{c+d,e+f,a+17,b+18}

{a+b,c+d,e+f,a+17,b+18},

meet


More Lattice Theory: Const Prop

⊥ ≤ anything ∀x ⊥ ⊔ x = x anything ≤ ⊤ ∀x x ⊔ ⊤= ⊤ ∀x x ⊔ x = x

… -2 -1 0 1 2 ...

The Lattice of CP


Relate to Constant Prop?!

i and j are integer values i ⊔ i = i i ⊔ j = ⊤ if i ≠ j


What is X in Block 3?

1. 2.

3.


What is X in Block 3?

1. 2.

3.

x is ⊤


Top ⊤ versus Bottom ⊥

⊤ means definitely NOT a constant Inputs are ⊤

⊥ means Undefined Could be constant but we do not

know!


Constant Propagation Problem

Variable v has constant value c at point p, iff ∀ paths from entry to p, the most recently assigned value for v is c


Constant Prop Example

Meet Operator = ⊔ Identity= ? X ⊔ ? = X



Meet Operator = ⊔ Identify= ⊥ X ⊔ ⊥ = X Initial Condition?



Meet Operator = ⊔ Identify= ⊥ X ⊔ ⊥ = X Initial Condition Boundary Condition?

⊥⊥⊥

⊥⊥⊥⊥⊥⊥

⊥⊥⊥

XYZ



Meet Operator = ⊔ Identify= ⊥ X ⊔ ⊥ = X Initial Condition Boundary Condition

⊥⊥⊥

⊥⊥⊥⊥⊥⊥

⊥⊥⊥

⊥⊥⊥

1.

2. 3.

4.

XYZ



⊥⊥⊥

⊥⊥⊥⊥⊥⊥

⊥⊥⊥

1.

2. 3.

4.

out1= ???⊥⊥⊥XYZ



⊥⊥⊥⊥⊥⊥

⊥⊥⊥

1.

2. 3.

4.

out1= 1⊥⊥out2= ???

⊥⊥⊥

1⊥⊥

XYZ



⊥⊥⊥

⊥⊥⊥

1.

2. 3.

4.

out1= 1⊥⊥out2= 023out3= ???

⊥⊥⊥

1⊥⊥

023

XYZ



⊥⊥⊥

1.

2. 3.

4.

out1= 1⊥⊥out2= 023out3= 12⊥out4= ???

⊥⊥⊥

1⊥⊥

023

XYZ

12⊥



1.

2. 3.

4.

out1= 1⊥⊥out2= 023out3= 12⊥

out4= ⊤23

⊥⊥⊥

1⊥⊥

023

XYZ

12⊥

⊤23

Does this make sense?


Next Time

Static-Single Assignment (SSA Form)

Read Efficiently computing static single

assignment form and the control dependence graph, Cytron et al.

http://portal.acm.org/citation.cfm?id=115320

Documents

Optimizing Compilers CISC 673 Spring 2009 More Data flow analysis