Basic Block and Trace

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Chapter 8

Tree IR

(1) Semantic gap

(2) IR is not proper for optimization analysis

Machine Languages

Eg: - Some expressions have side effects

ESEQ, CALL Tree representation => no execution order is assumed.

- Semantic Gap CJUMP vs. Jump on Condition 2 targets 1 target + “fall through”

Semantic Gap Continued

- ESEQ within expression is inconvenient- evaluation order matters

- CALL node within expression causes side effect ! - CALL node within the argument – expression of other CALL nodes will cause problem if the args of result are passed in the same (one) register.

- Rewrite Tree into an equivalent tree(Canonical Form)

SEQ SEQ

S2 S3 S4 S5

=> S1;S2;S3;S4;S5

Transformation

Step 1: A tree is rewritten into a list of “canonical trees” without SEQ or ESEQ nodes.

-> Tree. StmList linearize(Tree.Stm S);

Step 2: Grouping into a set of “basic blocks” which contains no internal jumps or labels

-> BasicBlocks

Step 3: Basic Blocks are ordered into a set of “traces” in which every CJUMP is immediately followed by

false label.-> Trace Schedule(BasicBlock b)

8.1 Canonical Trees

Def : canonical trees as having following properties:1. No SEQ or ESEQ2. The parent of each CALL is either EXP(..) or MOVE(TEMP t, ….)

=> Separate SEQs and EXPressions

Transformations on ESEQ-move ESEQ to higher level.

ESEQS1

e2op ESEQ

op l1 l2e

BINOPS

CJUMPS

e1 e2 l2l1

Case 1:

Case 2:

Case 3:

op ESEQ

op ㅣ1

ㅣ 2e1

ESEQMOVE

TEMP e2

TEMP e1

S CJUMP

e2 ㅣ1

TEMP e1

Case 4: When S does not affect e1 in case 3 and (s and e1 have no side effect( e.g., I/O))

S BINOP

op ㅣ1

ㅣ 2e1

S CJUMP

e1 e2 ㅣ1

if s,e1 commute if s,e1 commute

How can we tell if two Exp/Stm commute?

MOVE(MEM(x),y) MEM(z)x = z (aliased) not commutex z commute

CONST(n) can commute with any Statement!!

NOP (= ExpStm(CONST(0)) ) can commute with any Exp!!

=> Be Conservative !!

We don’tknow yet!

General Rewriting Rules

1. Identify the subexpressions.2. Pull the ESEQs out of the stm or exp.

Ex: [e1,e2,ESEQ(S,e3)]

-> (s1;[e1,e2,e3]) if s1, e1,e2 commute -> (SEQ(MOVE(t1,e1),s); [TEMP(t1),e2,e3])

if s1, e2 commute-> (SEQ (MOVE(t1, e1),SEQ(MOVE(t2,e2),s));

[TEMP(t1),TEMP(t2),e3]) o/w

Reorder(ExpList exps) => (stms; ExpList)

MOVING CALLS TO TOPLEVEL

CALL returns its result in the same register TEMP(Ri) BINOP(+,CALL(…),CALL(…))

SolutionCALL(fun,args) -> ESEQ(MOVE(TEMP t,CALL()),TEMP t)

Then eliminate ESEQ.=> need extra TEMP(t) (registers)

do_stm(MOVE(TEMP tnew, CALL(f, args))) do_stm (EXP(CALL(f, args)))

- will not reorder on CALL node ( so infinite recursion can be avoided) - will reorder on f and args as the children of MOVE

overwrite TEMP(RV)

A LINEAR LIST OF STATEMENTS

S0 S0’ (right linear)

SEQ(a, SEQ(b, c) )

=> [a,b,c]

linearlize(stm s0) :StmList

8.2 TAMING CONDITIONAL BRANCHES

BASIC BLOCKa sequence of statements entered at the beginning

exited at the end- The 1st stmt is a LABEL- The last stmt is a JUMP or

a CJUMP- no other LABELs, JUMPs, CJUMPs..

F: …… t

Algorithm

Scan from beginning to end- when Label is found, begin new Block- when JUMP or CJUMP is found, a block is ended- If block ends without JUMP or CJUMP, insert JUMP LABEL, LABEL ;

Epilogue block of Function.Label it as DONEand put JUMP DONE at the end of body of the function.

-> Canon.BasicBlocks.

Example:

L3: if v >= n goto L15

if r < n goto L9

v v + 1

goto L3

L9: x M[r]

s s + x

if s <= m got L13

L13: r r + 1

goto L6

L15: rv m

Example: (end of blocks)

if r < n goto L9

v v + 1

goto L3

L9: x M[r]

s s + x

if s <= m got L13

L13: r r + 1

goto L6

L15: rv m

JUMP done

Done: (function Epilogue)

Example: (start of blocks)

if r < n goto L9

v v + 1

goto L3

L9: x M[r]

s s + x

if s <= m got L13

L13: r r + 1

goto L6

L15: rv m

JUMP done

Example: (add label and JUMP)

Lb1: m 0

Lb2: r v

if r < n goto L9

Lb3: v v + 1

goto L3

L9: x M[r]

s s + x

if s <= m got L13

Lb4: m s ; JUMP L13

L13: r r + 1

goto L6

L15: rv m

JUMP done

Trace:a sequence of stmts that could beconsecutively executed during the execution of the program.

We want a set of traces that exactly covers the program

: one block in one trace.

To reduce JUMPs, fewer traces are preferred !!

Traces

T->FF->T

remove JUMP

JUMP on False

Idea !!

Algorithm 8.2 (Canon.Trace Schedule)

Put all the blocks of the Program into a list Q.while Q is not empty

Start a new(empty) trace, call it T.Remove the head element b from Q.while b is not marked Mark b; T <- T;b. Examine the successors of b. if there is any unmarked successor C

b <- C. END the current trace T.

Finishing Up

- analysis and optimizations are efficient for basic blocks (not for stmts level)

- Some local arrangement(1)CJUMP + false Label => OK(2)CJUMP + true Label => reverse condition CJUMP(>=,e1,e2, Lt,Lf) ; Lt: => CJUMP(<,e1,e2,Lf,Lt); Lt: …(1)CJUMP lt,lf + no lt lf

CJUMP lt, lf’; LABEL lf’; JUMP lf JUMP on true lt; JUMP lf chance to optimized !!

Finding optimal trace is not easy !!

Instruction Selection

Chapter 9

What we are going to do.

Tree IR machine Instruction

(Jouette Architecture or SPARC or MIPS or Pentium or T )

=> LOAD r1 M[ fp + c]

Jouette Architecture

Name Effect Trees

ri rj + rk ADD

*ri rj * rk MUL

-ri rj - rk SUB

/ri rj / rk DIV

CONSTri rj + c

SUBI ri rj - c -

LOAD M[rj + c] ri

MEM MEM MEM MEM

Jouette Architecture

MEM MEM MEM MEM

MOVE MOVE MOVE MOVE

Name Effect Trees

STORE M[rj + c] ri

MOVEM M[rj]

• Register r0 always contains zero• Instructions produces a result in a register => EXP• instructions produce side effects on Mem => Stm

Tiling the IR tree ex: a[i]:= x i:register a,x:frame var

2 LOAD r1<-M[fp+a]4 ADDI r2<- r0 + 4

5 MUL r2 <- ri*r2

6 ADD r1 <- r1+r28 LOAD r2<-M[fp+x]9 STORE M[r1+0] <- r2

* CONST x

CONST a

CONST 4TEMP i1

Another Solution ex: a[i]:= x i:register a,x:frame var

2 LOAD r1<-M[fp+a]4 ADDI r2<- r0 + 4

5 MUL r2 <- ri*r2

6 ADD r1 <- r1+r28 ADDI r2<- fp+x

9 MOVEM M[r1] <- M[r2 ]

* CONST x

CONST a

CONST 4TEMP i1

Or Another Tiles with a different set of tile-pattern

3 LOAD r1<-M[r1+0]4 ADDI r2<- r0 + 4

5 MUL r2 <- ri*r2

6 ADD r1 <- r1+r2

8 ADD r2<- fp+ r2

10 STORE M[r1+0] <- r2

1 ADDI r1<- r0 + a

2 ADD r1 <- fp +r1

7 ADDI r2<- r0 + x

9 LOAD r2<-M[r2+0]

* CONST x

CONST a

CONST 4TEMP i12

OPTIMAL and OPTIMUM TILINGS

Optimum Tiling : one whose tiles sum to the lowest possible value.

cost of tile : instr. exe. time, # of bytes, ......

Optimal Tiling : one where no two adjacent tiles can be combined into a single tile of lower cost.

then why we keep ?

are enough.30 25

Algorithms for Instruction Selection

1. Optimal vs Optimum simple maybe hard

2. CISC vs RISC (Complex Instr. Set Computer) tile size large small

optimal >= optimum optimal ~= optimum

instruction cost varies almost same! on addressing mode

Maximal Munch – optimal tiling algorithm

1. starting at root, find the largest tile that fits.

2. repeat step 1 for several subtrees which are generated(remain)!!

3. Generate instructions for each tile (which are in reverse order)

=> traverse tree of tiles in post-orderWhen several tiles can be matched, select the largest tile(which covers the most nodes).

If same tiles are matched, choose an arbitrary one.

Implementation

See Program 9.3 for example(p181)

case statements for each root type!!

There is at least one tile for each type of root node!!

B30+2 A

Dynamic Programming – finding optimum tiling finding optimum solutions based on

optimum solutions of each subproblem!!

1. Assign cost to every node in the

2. Find several matches.

3. Compute the cost for each match.

4. Choose the best one.

5. Let the cost be the value of node.

10+20+40

30+2+40+5=?

ADDIADDI

CONSTCONST2CONST1+

TileCost LeavesCost Total

1 1+1 3

LOAD ri<-M[rj] LOAD ri<-M[rj+c] LOAD ri<-M[rj+c]

cost 1+2 1+1 1+1

Example

MEM node

Tree Grammars

Example : Schizo-Jouette machine

ADD di <- dj +dk

MUL di <- dj *dk

SUB di <- dj - dk

DIV di <- dj /dk

ADDI di <- dj +C

SUBI di <- dj -C

CONST CONST

d CONSTd

MOVEA dj<- ai

MOVED aj<- di

A generalization of DP for machines with complex instruction set and several classes of registers and addressing modes.

ai : address register

dj : data register

LOAD di<-M[aj+C]

STORE M[aj+C]<- di

MOVEM M[aj] <- M[ai ]

dMEM dMEM

MEM MEM

MOVE MOVE MOVE MOVE

d d d d

Use Context-free grammar to describe the tiles;ex: nonterminal s : statement

d : data a : address

d -> MEM(+(a,CONST))d-> MEM(+(CONST,a))d-> MEM(CONST)d-> MEM(a)

d -> a

a -> d

=> ambiguous grammar!!

-> parse based on the minimum cost!!

s MOVE(MEM(+(a,CONST)), d)STORE

s MOVE(M(a),M(a))MOVEM

Efficiency of Tiling Algorithms

Order of Execution Cost for “Maximal Munch & Dynamic Programming”

T : # of different tiles.K : # of non-leaf node of tile (in average)K’: largest # of node that need to be examined to choose the right tile ~= the size of largest tileT’: average # of tile-patterns which matches at each tree node

Ex: for RISC machineT = 50, K = 2, K’= 4, T’ = 5 ,

N : # of input nodes in a tree.

complexity = N/K * ( K’ + T’) ofmaximal Munch # of node

(#of patterns)to be examined

to find matched pattern

to findminimum

complexity ofDynamic Programming

= N * (K’ + T’)

“linear to N”

9.2 RISC vs CISC

1. 32registers.

2. only one class of integer/pointer

registers.

3. arithmetic operations only between

registers.

4. “three-address” instruction form r1<-r2

5. load and store instructions with only the

M[reg+const] addressing mode.

6. every instruction exactly 32 bits long.

7. One result or effect per instruction.

CISC(Complex Instruction Set Computers)

Complex Addressing Mode

1. few registers (16 or 8 or 6).2. registers divided into different classes.3. arithmetic operations can access registers or memory through “addressing mode”.4. “two-address” instruction of the form r1<-r1 & r2.5. several different addressing modes.6. variable length instruction format.7. instruction with side effects. eg: auto-increment/decrement.

Solutions for CISC1. Few registers.

- do it in register allocation phase.2. Classes of registers.

- specify the operands and result explicitly. - ex: left opr of arith op (e.g. mul) must be eax

- t1 t2 x t3 ==>- move eax, t2 eax t2- mul t3 eax eax x t3; edx

garbage - mov t1 eax t1 eax3. Two addressing instructions

- add extra move instruction -> resgister allocation

t1 <- t2+t3 move t1,t2 t1<- t2 add t1,t3 t1<- t1+t3

4. Arithmetic operations can address memory.- actually handled by “register spill” phase.

- load memory operand into register and

store back into memory -> may trash registers!!

-ex: add [ebp – 8,] ecx is equivalent to - mov eax, [ebp –8] - add eax, ecx - mov [ebp – 8], eax

5. several addressing modes- takes time to execute (no faster than multiInstr seq) “trash” fewer registers short instruction sequence select appropriate patterns for addressing mode.6. Variable Length Instructions

- let assembler do generate binary code.7. Instruction with Side effect

eg: r2 <- M[r1]; r1<- r1 + 4; - difficult to model!!

(a) ignore the auto increment-> forget it! (b) try to match special idioms

(c) try to invent new algorithms.

• assembly language instruction without register assignment.

package assem;

public abstract class Instr {

public String assem; // instr template

public abstract temp.TempList use(); // retrun src list

public abstract temp.TempList def(); // return dst list

public abstract Targets jumps(); // return jump

public String format(temp.tempMap m); // txt of assem instr

public Targets(temp.LabelList labes);

Abstract Assembly Language Instructions

// dst, src and jump can be null.

public OPER(String assem, TempList dst, TempList src, temp.LabelList jump);

public OPER(String assem, TempList dst, TempList src);

public MOVE(String assem, Temp dst, Temp src)

public LABEL(String assem, temp.Label label);

Example

• assem.Instr is independent of th etarget machine assembly.

ex: MEM( +( fp, CONST(8)) ==>

new OPER(“LOAD ‘d0 <- M[‘s0 + 8]”,

new TempList(new Temp(),null),

new TempList(frame.FP(), null));

call format(…) on the above Instr. we get

LOAD r1 <- M[r27+8]

assume reg. allocator assign r1 to the new Temp and r27 is the frame pointer register.

Another Example• *(+(Temp(t87), CONST(3)), MEM(temp(t92))• assem dst src• ADDI ‘d0 <- ‘s0 + 3 t908 t87• LOAD ‘d0 <- M[‘s0+0] t909 t92• MUL ‘d0 <- ‘s0*’s1 t910 t908,t909• after register allocation, the instr look like:

– ADDI r1 <- r12 + 3 t908/r1 t87/r12– LOAD r2 <- M[r13+0] t909/r2 t92/r13– MUL r1 <- r1*r2 t910/r1

• Two-address instructions– t1 t1 + t2 ==>– assem dst src– add ‘d0 ‘s1 t1 t1,t2

Basic Block and Trace

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