cmput680 - fall 2003

CMPUT 680 - Compiler Design and Optimization

1

CMPUT680 - Fall 2003

Topic 6: Optimal Code GenerationJosé Nelson Amaral

http://www.cs.ualberta.ca/~amaral/courses/680


2

Data Dependency Graph

(a) t1 := ld(x);(b) t2 := t1 + 4;(c) t3 := t1 * 8;(d) t4 := t1 - 4;(e) t5 := t1 / 2;(f) t6 := t2 * t3;(g) t7 := t4 - t5;(h) t8 := t6 * t7;(i) st(y,t8);

B3a


3



B3a

b c d e


4


B3a

b c d e

f



5


B3a

b c d e

f g



6


B3a

b c d e

f g

h



7


B3a

b c d e

f g

h

i



8

Code Generation

Problem: How to generate optimal code for a basic block specified by its DAG representation?

If the DAG is a tree, we can use Sethi-Ullman algorithm togenerate code that is optimal in terms of program length or number of registers used.

(Aho-Sethi-Ullman,pp. 557)


9

Example

t1 := a + bt2 := c + dt3 := e - t2t4 := t1 - t3 t4

+

-

-+

a b e

c d

t3

t2

t1

Generate code for amachine with tworegisters.


Assume that only t4is alive at the exitof the basic block.


10

Example

t1 := a + bt2 := c + dt3 := e - t2t4 := t1 - t3 t4

+

-

-+

a b e

c d

t3

t2

t1

LOAD a, R0ADD b, R0



11

Example

t1 := a + bt2 := c + dt3 := e - t2t4 := t1 - t3 t4

+

-

-+

a b e

c d

t3

t2

t1

LOAD a, R0ADD b, R0LOAD c, R1ADD d, R1


Assume that SUB only works with registers.

Can’t evaluate

t3 because thereare no available

registers!


12

Example

t1 := a + bt2 := c + dt3 := e - t2t4 := t1 - t3 t4

+

-

-+

a b e

c d

t3

t2

t1

LOAD a, R0ADD b, R0LOAD c, R1ADD d, R1STORE R0, t1LOAD e, R0SUB R1, R0



13

Example

t1 := a + bt2 := c + dt3 := e - t2t4 := t1 - t3 t4

+

-

-+

a b e

c d

t3

t2

t1

LOAD a, R0ADD b, R0LOAD c, R1ADD d, R1STORE R0, t1LOAD e, R0SUB R1, R0LOAD t1, R1SUB R0, R1STORE R1, t4



14

Example

t1 := a + bt2 := c + dt3 := e - t2t4 := t1 - t3 t4

+

-

-+

a b e

c d

t3

t2

t1

LOAD a, R0ADD b, R0LOAD c, R1ADD d, R1STORE R0, t1LOAD e, R0SUB R1, R0LOAD t1, R1SUB R0, R1STORE R1, t4

10 instructions

1 spill


Evaluation Order:t1, t2, t3, t4


15

Example(can we do better?)

t1 := a + bt2 := c + dt3 := e - t2t4 := t1 - t3 t4

+

-

-+

a b e

c d

t3

t2

t1

LOAD c, R0ADD d, R0



16


t1 := a + bt2 := c + dt3 := e - t2t4 := t1 - t3 t4

+

-

-+

a b e

c d

t3

t2

t1

LOAD c, R0ADD d, R0LOAD e, R1SUB R0,R1



17


t1 := a + bt2 := c + dt3 := e - t2t4 := t1 - t3 t4

+

-

-+

a b e

c d

t3

t2

t1

LOAD c, R0ADD d, R0LOAD e, R1SUB R0, R1LOAD a, R0ADD b, R0



18


t1 := a + bt2 := c + dt3 := e - t2t4 := t1 - t3 t4

+

-

-+

a b e

c d

t3

t2

t1

LOAD c, R0ADD d, R0LOAD e, R1SUB R0, R1LOAD a, R0ADD b, R0SUB R1, R0STORE R0, t4



19

Example(can we do better? Yes!!!)

t1 := a + bt2 := c + dt3 := e - t2t4 := t1 - t3 t4

+

-

-+

a b e

c d

t3

t2

t1

8 instructions

no spills!!!


LOAD c, R0ADD d, R0LOAD e, R1SUB R0, R1LOAD a, R0ADD b, R0SUB R1, R0STORE R0, t4

Evaluation Order:t2, t3, t1, t4


20

Example:Why the improvement?

t1 := a + bt2 := c + dt3 := e - t2t4 := t1 - t3 t4

+

-

-+

a b e

c d

t3

t2

t1

We evaluated t4 immediatelyafter t1 (its leftmost argument).



21

Heuristic Node Listing Algorithm for a DAG

(1) while unlisted interior nodes remain(2) select an unlisted node n, all of whose parents

have been listed;(3) list n;(4) while the leftmost child m of n has no unlisted parents,

and is not a leaf node do/* since n was just listed, m is not yet listed */

(5) list m;(6) n := m;

endwhile endwhile



22

Node Listing Example

+ -

a b

+ c

- +

ed

1

2 3

4

8

11 12

5

6 7

9 10(1) while unlisted interior nodes remain(2) select an unlisted node n, all of whose parents

have been listed;(3) list n;


23


+ -

a b

+ c

- +

ed

1

2 3

4

11 12

5

6 7

9 10

List:1



8

n


24


+ -

a b

+ c

- +

ed

1

2 3

4

11 12

5

6 7

9 10(4) while the leftmost child m of n has no unlisted parents,


(5) list m;(6) n := m;

8

m

n

List:1


25


+ -

a b

+ c

- +

ed

1

2 3

4

11 12

5

6 7



(5) list m;(6) n := m;

List:12

8

m

n


26


+ -

a b

+ c

- +

ed

1

2 3

4

11 12

5

6 7



(5) list m;(6) n := m;

List:12

8m

n


27


+ -

a b

+ c

- +

ed

1

2 3

4

11 12

5

6 7

9 10

List:123



8

n


28


+ -

a b

+ c

- +

ed

1

2 3

4

11 12

5

6 7



(5) list m;(6) n := m;

8

n

m

List:123


29


+ -

a b

+ c

- +

ed

1

2 3

4

11 12

5

6 7

9 10

List:1234

(4) while the leftmost child m of n has no unlisted parents,and is not a leaf node do/* since n was just listed, m is not yet listed */

(5) list m;(6) n := m;

8

n

m


30


+ -

a b

+ c

- +

ed

1

2 3

4

11 12

5

6 7



(5) list m;(6) n := m;

8

n

m

List:1234


31


+ -

a b

+ c

- +

ed

1

2 3

4

11 12

5

6 7

9 10

List:12345


(5) list m;(6) n := m;

8

n

m


32


+ -

a b

+ c

- +

ed

1

2 3

4

11 12

5

6 7



(5) list m;(6) n := m;

8

n

m

List:12345


33


+ -

a b

+ c

- +

ed

1

2 3

4

11 12

5

6 7

9 10

List:123456


(5) list m;(6) n := m;

8

n

m


34


+ -

a b

+ c

- +

ed

1

2 3

4

11 12

5

6 7

9 10

List:123456


(5) list m;(6) n := m;

8n

m


35


+ -

a b

+ c

- +

ed

1

2 3

4

11 12

5

6 7

9 10

List:1234568

8



n


36


+ -

a b

+ c

- +

ed

1

2 3

4

11 12

5

6 7

9 10

List:1234568

Therefore the optimal evaluation order (regardless of thenumber of registers available) for the internal nodes is 8654321.

8


37

Optimal Code Generationfor Trees

The order is optimal in the sense that it yields the shortest instruction sequence over allinstruction sequences that evaluate the tree.

If the DAG representing the data flow in a basicblock is a tree, then for some machine models,there is a simple algorithm (the SethiUllmanalgorithm) that gives the optimal order.


38

Sethi-Ullman Algorithm

Intuition:1. Label each node according to the number of registers that are required to generate code for the node.2. Generate code from top down always generating code first for the child that requires the most registers.


39

Sethi-Ullman Algorithm(Intuition)

Leftleaf

Rightleaves

Bottom-Up Labeling: visit a node after all its children are labeled.


40

Labeling Algorithm

end

begin else

else

thenif

thenif

1max (6)

that so

by ordered ofchildren thebe let (5)

/* nodeinterior an is * /

0 (4)

1 (3)

parent its of childleftmost theis (2)

leaf a is (1)

1

21

21

iclabel :nlabel

clabelclabelclabel

labeln, c, , cc

n

label(n) :

label(n) :

n

n

iki

k

k


41

Labeling Algorithm

)()( if1)(

)()( if)(),(max)(

:becomes

1max

relation following then thechildren), with twonode (a 1k If

211

2121

11

21

clabelclabelclabel

clabelclabelclabelclabelnlabel

iclabel :nlabel

clabelclabelclabel

iki

k


42

Example

a b t2

c d

t1 t3

e

t4


43

Example

a b t2

c d

t1 t3

e

t4Labeling leaves:leftmost is 1, others are 0

1 0

0

1

1


44

Example

a b t2

c d

t1 t3

e

t4

Labeling t2:

Thus 1 0

0

1

1

112)(

111)(

)()(

dlabel

clabel

dlabelclabel

1)2( tlabel


45

Example

a b t2

c d

t1 t3

e

t4

1 0

0

1

1

1

2)3( Thus

212)2(

111)(

)2()(

tlabel

tlabel

elabel

tlabelelabel


46

Example

a b t2

c d

t1 t3

e

t4

1 0

0

1

1

1

2

Likewise labeling t2


47

Example

a b t2

c d

t1 t3

e

t4

1 0

0

1

1

1

2)4( Thus

212)1(

211)2(

)1()2(

tlabel

tlabel

tlabel

tlabeltlabel

21


48

Example

a b t2

c d

t1 t3

e

t4

1 0

0

1

1

1

21

2

See Aho, Sethi, Ullman (pp. 564)for code generating algorithm.

cmput680 - fall 2003

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