Classical Optimization Techniques in Dataflow Analysis at UMich (EECS 483), Study Guides, Projects, Research of Electrical and Electronics Engineering

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Dataflow Analysis/Opti III Classical Optimization

EECS 483 – Lecture 20 University of Michigan Monday, November 22, 2004

• 1 -

Announcements

Y

Project 3 » Due tonight (Monday) » Can turn it anytime before 9am Tues » Grading will happen next week (Thurs/Fri)

y

Signup sheet available next Monday

Y

No class on Wednes, 11/

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Caveat

Y

Traditional compiler class » Fancy implementations of optimizations, efficient algorithms » Bla bla bla » Spend entire class on 1 optimization

Y

For this class – Go over concepts of eachoptimization » What it is » When can it be applied (set of conditions that must be satisfied)

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Constant Folding Y

Simplify operation based on values of src operands

» Constant propagation creates opportunities for this

Y

All constant operands

» Evaluate the op, replace with a move

y r1 = 3 * 4

Æ

r1 = 12 y r1 = 3 / 0

Æ

??? Don’t evaluate excepting ops !, what about FP?

» Evaluate conditional branch, replace with BRU or noop

y if (1 < 2) goto BB

Æ

BRU BB

y if (1 > 2) goto BB

Æ

convert to a noop

Y

Algebraic identities

» r1 = r2 + 0, r2 – 0, r2 | 0, r2 ^ 0, r2 << 0, r2 >> 0

Æ

r1 = r

» r1 = 0 * r2, 0 / r2, 0 & r

Æ

r1 = 0

» r1 = r2 * 1, r2 / 1

Æ

r1 = r

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Dead Code Elimination

Y

Remove any operation who’sresult is never consumed

Y

Rules

X can be deleted y no stores or branches » DU chain empty or dest not live Y

This misses some dead code!!

Especially in loops » Critical operation y store or branch operation » Any operation that does notdirectly or indirectly feed acritical operation is dead » Trace UD chains backwardsfrom critical operations » Any op not visited is dead r1 = 3 r2 = 10 r4 = r4 + 1 r7 = r1 * r r2 = 0 r3 = r3 + 1 r3 = r2 + r1 store (r1, r3)

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Class Problem

Optimize this applying 1. constant folding 2. strength reduction 3. dead code elimination r1 = 0 r4 = r1 | -1 r7 = r1 * 4 r6 = r r3 = 8 / r r3 = 8 * r6 r3 = r3 + r r2 = r2 + r1r6 = r7 * r6r1 = r1 + 1 store (r1, r3)

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Local Constant Propagation

Y

Consider 2 ops, X and Y ina BB, X is before Y

» 1. X is a move » 2. src1(X) is a literal » 3. Y consumes dest(X) » 4. There is no definition of

dest(X) between X and Y

y Defn is locally available!

» 5. Be careful if dest(X) is SP,

FP or some other specialregister – If so, no subroutinecalls between X and Y

1: r1 = 5 2: r2 = ‘_x’ 3: r3 = 7 4: r4 = r4 + r1 5: r1 = r1 + r2 6: r1 = r1 + 1 7: r3 = 12 8: r8 = r1 - r2 9: r9 = r3 + r5 10: r3 = r2 + 1 11: r10 = r3 – r

• 10 -

Global Constant Propagation

Y

Consider 2 ops, X and Y indifferent BBs

» 1. X is a move » 2. src1(X) is a literal » 3. Y consumes dest(X) » 4. X is in adef_IN(BB(Y)) » 5. dest(X) is not modified

between the top of BB(Y) and Y

y Rules 4/5 guarantee X is available

» 6. If dest(X) is SP/FP/..., no

subroutine call between X and Y

r1 = 5 r2 = ‘_x’ r1 = r1 + r r7 = r1 – r r8 = r1 * r r9 = r1 + r Note: checks for subroutine calls whenever SP/FP/etc. are involved is required for all optis. I will omit the check from here on!

• 12 -

Forward Copy Propagation Y

Forward propagation of theRHS of moves

X: r1 = r »

Y: r4 = r1 + 1

Æ

r4 = r2 + 1 Y

Benefits

Reduce chain of dependences » Possibly eliminate the move Y

Rules (ops X and Y)

X is a move » src1(X) is a register » Y consumes dest(X) » X.dest is an available def at Y » X.src1 is an available expr at Y r1 = r2 r3 = r r2 = 0 r6 = r3 + 1 r5 = r2 + r

• 13 -

Backward Copy Propagation Y

Backward prop. of the LHS of moves

X: r1 = r2 + r

Æ

r4 = r2 + r »

r5 = r1 + r

Æ

r5 = r4 + r »

Y: r4 = r

Æ

noop Y

Rules (ops X and Y in same BB)

dest(X) is a register » dest(X) not live out of BB(X) » Y is a move » dest(Y) is a register » Y consumes dest(X) » dest(Y) not consumed in (X…Y) » dest(Y) not defined in (X…Y) » There are no uses of dest(X) after the firstredefinition of dest(Y)

r1 = r8 + r9 r2 = r9 + r1 r4 = r2 r6 = r2 + 1 r9 = r1 r10 = r6 r5 = r6 + 1 r4 = 0 r8 = r2 + r

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Global CSE

Y

Rules (ops X and Y)

» X and Y have the same

opcode

» src(X) = src(Y), for all

srcs

» expr(X) is available at Y » if X is a load, then there

is no store that may writeto address(X) along anypath between X and Y

r1 = r2 * r6^ r3 = r4 / r r2 = r2 + 1 r1 = r3 * 7 r5 = r2 * r6^ r8 = r4 / r7^ r9 = r3 * 7 if op is a load, call it redundant load elimination rather than CSE

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Class Problem

Optimize this applying 1. constant propagation 2. constant folding 3. strength reduction 4. dead code elimination 5. forward copy propagation 6. backward copy propagation 7. CSE r1 = 9 r4 = 4 r5 = 0 r6 = 16 r2 = r3 * r4 r8 = r2 + r r9 = r r7 = load(r2)^ r5 = r9 * r4 r3 = load(r2)^ r10 = r3 / r6^ store (r8, r7) r11 = r r12 = load(r11) store(r12, r3)

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Operation Folding Y

Combine 2 dependent ops into 1complex op

Classic example is MPYADD » r1 = r2 * r »

r5 = r1 + r

Æ

r5 = r2 * r3 + r Y

First op often becomes dead

Y

Borders on machine dependent opti

Y

Rules (ops X and Y in same BB)

X is an arithmetic operation » dest(X) != any src(X) » Y is an arithmetic operation » Y consumes dest(X) » X and Y can be merged » src(X) not modified in (X…Y)

r1 = r2 & 4 r3 = r1 ^ -1 r2 = r3 < 6 r4 = r2 == 0 r5 = r6 << 1 r7 = r5 + r

• 19 -

Loop Optimizations

Y

The most important set of optimizations » Because programs spend so much time in loops

Y

Optimize given that you know a sequenceof code will be repeatedly executed

Y

Optis » Invariant code removal » Global variable migration » Induction variable strength reduction » Induction variable elimination