basic block and trace
DESCRIPTION
Basic Block and Trace. Chapter 8. (1) Semantic gap. Tree IR. Machine Languages. (2) IR is not proper for optimization analysis. Eg: - Some expressions have side effects ESEQ, CALL Tree representation => no execution order is assumed. - Semantic Gap - PowerPoint PPT PresentationTRANSCRIPT
1
Basic Block and Trace
Chapter 8
2
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”
3
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
SEQ SEQS1
S2 S3 S4 S5
=> S1;S2;S3;S4;S5
4
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)
5
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
6
Transformations on ESEQ-move ESEQ to higher level.
Eg.ESEQ
ESEQS1
S2e
ESEQ
SEQ
S1 S2
e
ESEQ
S e1
BINOPe2op ESEQ
S e1
MEM
ESEQ
S e1
JUMP
ESEQ
S e1
CJUMPop l1 l2e
ESEQBINOPS
op
e1 e2
ESEQ
S MEM
e1
ESEQJMPS
e1
SEQCJUMPS
op
e1 e2 l2l1
Case 1:
Case 2:
7
Case 3:
ESEQ
S
e1
BINOP
e2
op ESEQ
S1 e2
CJUMPop ㅣ
1
ㅣ 2e1
ESEQESEQMOVE
S
t
BINOP
op
TEMP e2
TEMP e1
t
SEQ
S CJUMP
op
e2 ㅣ1
ㅣ 2
ESEQMOVE
TEMP e1
t
t
TEMP
8
Case 4: When S does not affect e1 in case 3 and (s and e1 have not I/O)
ESEQ
S
e1
BINOP
e2
op
ESEQ
S BINOP
op
e1 e2
ESEQ
S1 e2
CJUMPop ㅣ
1
ㅣ 2e1
SEQ
S CJUMP
op
e1 e2 ㅣ1
ㅣ 2
if s,e1 commute if s,e1 commute
9
How can we tell if tow expressions commute?
MOVE(MEM(x),y) MEM(z)x = z (aliased) not commutex z commute
CONST(n) can commute with any Expression!!
=> Be Conservative !!
?? We don’tknow yet!
10
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]) s1, e1,e2 commute
-> (SEQ (MOVE(t1, e1),SEQ(MOVE(t2,e2),s));
-> (SEQ(MOVE(t1,e1),s); [TEMP(t1),e2,e3]
Reorder(ExpListexps) => (stms; ExpList)
11
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 - will reorder on f and args as the children of MOVE
overwrite TEMP(RV)
12
A LINEAR LIST OF STATEMENTS
S0 [S0’ ; ]
SEQ
SEQ
a b
c
SEQSEQa
b c
SEQ(SEQ(SEQ…())) => a;b;c
linear(stm s)
13
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
a CJUMP- no other LABELs, JUMPs, CJUMPs..
FT
CJUMP
Cond
CondCJUMP
T:F: …… t
.
.
.JUMP
LABEL
14
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.
15
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
Exit
16
5
34
2
4
7
6
3
2
1
5
T F
7
6
1
JUMP
T->FF->T
remove JUMP
JUMP on False
Idea !!
T F
17
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 nay unmarked successor C
b <- C. END the current trace T.
18
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(3)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 !!
19
Instruction Selection
Chapter 9
20
What we are going to do.
Tree IR machine Instruction(Jouette Architecture or SPARC or MIPS or Pentium or T )
LOAD R1,ea(c);MEM
CONST
BINOP
+ ea
C
21
Machine Example - Jouette ArchitectureRegister R0 always contains zero
ADD ri <- rj+rk+
MUL ri <- rj*rk*
SUB ri <- rj-rk-
DIV ri <- rj/rk
BINOP/
ADDI ri <- rj+C+
CONST+
CONSTCONST
SUBI ri <- rj-C-
CONSTLOAD ri <- M[rj+C]
+CONST
CONST
MEM MEM
+CONST
MEM MEM
Instructions produces a result in a register => EXP
ri TEMP
22
STORE M[rj+C] <-ri
+CONST
CONST
MEM MEM
+CONST
MEM MEMMOVE MOVE MOVE MOVE
MOVEM M[rj] <-M[ri]
MEMMOVE
MEM
Execution of instructions produce side effects on Mem.=> Stm
23
Tiling the IR tree ex: a[i]:= x i:register a,x:frame var
2 LOAD r1<-M[fp+a]4 ADDI r2<- r0 + 45 MUL r2 <- ri*r2
6 ADD r1 <- r1+r28 LOAD r2<-M[fp+x]9 STORE M[r1+0] <- r2
* CONST x
MEM
FP
+CONST a
MEM
FP
+
MOVE
+
MEM
CONST 4TEMP i1
23 4
5
67
8
9
24
Another Solution ex: a[i]:= x i:register a,x:frame var
2 LOAD r1<-M[fp+a]4 ADDI r2<- r0 + 45 MUL r2 <- ri*r2
6 ADD r1 <- r1+r28 ADDI r2<- fp+x9 MOVEM M[r1] <- M[r2 ]
* CONST x
MEM
FP
+CONST a
MEM
FP
+
MOVE
+
MEM
CONST 4TEMP i1
23 4
5
67
8
9
25
Or Another Tiles with a different set of tile-pattern
3 LOAD r1<-M[r1+0]4 ADDI r2<- r0 + 45 MUL r2 <- ri*r2
6 ADD r1 <- r1+r2
8 ADD r2<- fp+ r2
10 STORE M[r1+0] <- r2
1 ADDI r1<- r0 + a2 ADD r1 <- fp +r1
7 ADDI r2<- r0 + x
9 LOAD r2<-M[r2+0]
* CONST x
MEM
FP
+CONST a
MEM
FP
+
MOVE
+
MEM
CONST 4TEMP i12
3
45
67
8
9
10
26
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
27
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
28
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.
29
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!!
30
MEM
B30+2 A40
10 20
Dynamic Programming – finding optimum tiling finding optimum solutions based on
optimum solutions of each subproblem!!
1. Assign cost to every node in the
tree.
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
+4=
30+2+40+5=?
31
+
MEM
ADDIADDI
+
+CONSTCONST2CONST1
+CONST
TileCost LeavesCost Total
1 1+1 3
1 1 2
1 1 2
CONST
MEM
+CONST
MEM
+
MEM
2
1 1
LOAD ri<-M[rj] LOAD ri<-M[rj+c] LOAD ri<-M[rj+c]
cost 1+2 1+1 1+1
Example
MEM node
32
Tree Grammars
Example : Schizo-Jouette machine
ADD di <- dj +dk
MUL di <- dj *dk
SUB di <- dj - dk
DIV di <- dj /dk
d
d+
d
d
d*
d
d
d-
d
d
d/
d
ADDI di <- dj +C
SUBI di <- dj -C
d
d+
CONST CONST
d+
d CONSTd
d
d-
CONST
MOVEA dj<- ai
MOVED aj<- di
da
ad
A generalization of DP for machines with complex instruction set and several classes of registers and addressing modes.
ai : address register
dj : data register
33
LOAD di<-M[aj+C]
STORE M[aj+C]<- di
MOVEM M[aj] <- M[ai ]
+CONST
CONST
dMEM dMEM
+CONST
dMEM
a a
a
dMEM
+CONST
CONST
MEM MEM
+CONST
MEM MEMMOVE MOVE MOVE MOVE
a a
ad d d d
MEMMOVE
a
MEM
a
34
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
MOVEA
MOVED
LOAD
=> ambiguous grammar!!
-> parse based on the minimum cost!!
s MOVE(MEM(+(a,CONST)), d)STORE
s MOVE(M(a),M(a))MOVEM
35
Efficiency of Tiling AlgorithmsOrder 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 ,
36
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
cost
complexity ofDynamic Programming
= N * (K’ + T’)
“linear to N”
37
9.2 RISC vs CISC
RISC 1. 32registers.2. only one class of integer/pointer registers.3. arithmetic operations only between registers.4. “three-address” instruction form r1<-r2 & r35. 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.
38
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.
39
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
40
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
41
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.
42
• 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
43
// 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);
44
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.
45
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