Advanced Compilers - Hyperblocks Control Height Reduction | EECS 583, Study notes of Electrical and Electronics Engineering

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EECS 583 – Lecture 5 Hyperblocks Control Height Reduction

University of Michigan January 26, 2004

• 1 -

Recap: Backedge Coalescing

BB2^ c <= 0 c > 0

b <= 13

c <= 25

c > 25 d++

b++^

c++

c > 25 c <= 25 c++

BB

BB

BB

BB

BB

b < 0^

b >= 0^ BB3 BB

b > 13

e < 34 e >= 34 a++

e++^ BB

BB2^ c <= 0 c > 0

b <= 13 BB

BB

BB

BB

BB

b < 0^

b >= 0^ BB^

e++

b > 13 b++

d++^ BB^

a++ e < 34

e >= 34

• 3 -

Recap: CDs Via Algorithm^ Entry

BB

Control deps (left is taken) BB1: none BB2: -1 BB3: 1 BB4: -2 BB5: -4 BB6: 2, 4 BB7: -1 BB8: -1, -3 BB9: none

b < 0^

b >= 0 BB2^ c <= 0 c > 0

b <= 13

c <= 25

c > 25 d++

e++ BB3 (^) c++

BB4 b > 13 BB^

BB

b++

BB

BB

a++

e < 34^

BB

Exit

• 4 -

Recap: CMPP Creation^ Entry

K^ = {{-1}, {1}, {-2}, {-4}, {2,4}, {-1,-3}} p’s =^

p1,^ p2,

p3,

p4,

p5,

p

BB

b < 0^

b >= 0 BB2^ c <= 0 c > 0

b <= 13

c <= 25

c > 25 d++

e++ BB3 (^) c++

p1 = cmpp.ON (b < 0) if T

Æ^ BB

p2 = cmpp.ON (b >= 0) if T

Æ^ BB

p3 = cmpp.ON (c > 0) if T

Æ^ BB

p4 = cmpp.ON (b > 13) if T

Æ^ BB

p5 = cmpp.ON (c <= 0) if T

Æ^ BB

p5 = cmpp.ON (b <= 13) if T

Æ^ BB

p6 = cmpp.ON (b < 0) if T

Æ^ BB

p6 = cmpp.ON (c <= 25) if T

Æ^ BB

BB4 b > 13 BB^

BB

b++

BB

BB

a++

e < 34^

BB

Exit

• 6 -

Running Example – Control Flow Substitution

Loop:^ p1 = p2 = p3 = p4 = p5 = p6 = 0^ b = load(a) if T^ p1 = cmpp.ON (b < 0) if T^ p2 = cmpp.ON (b >= 0) if T^ p6 = cmpp.ON (b < 0) if T^ p3 = cmpp.ON (c > 0) if p1^ p5 = cmpp.ON (c <= 0) if p1^ p4 = cmpp.ON (b > 13) if p3^ p5 = cmpp.ON (b <= 13) if p3^ b = b + 1 if p4^ c = c + 1 if p5^ d = d + 1 if p1^ p6 = cmpp.ON (c <= 25) if p2^ e = e + 1 if p2^ a = a + 1 if p6^ bge e, 34, Done if p6^ jump Loop if T Done:

BB2^ c <= 0 c > 0

b <= 13

c <= 25

c > 25 d++

c++ BB b < 0^

b >= 0^ BB^

e++

BB4 b > 13 BB^

BB

b++

BB

BB

a++

e < 34^

BB

e >= 34

• 7 -

Step 4: CMPP Compaction^ Y^

Convert ON CMPPs to UN^ »^ All singly defined predicates don’t need to be OR-type^ »^ OR of 1 condition

Æ^ Just compute it !!!

»^ Remove initialization (Unconditional don’t require init) Y Reduce number of CMPPs »^ Utilize 2

nd^ destination slot

»^ Combine any 2 CMPPs with:^ y

Same source operands y Same guarding predicate y Same or opposite compare conditions

• 9 -

Class Problem^ if (a > 0) {

r = t + s if (b > 0 || c > 0)^ u = v + 1 else if (d > 0)^ x = y + 1 else^ z = z + 1 } a. Draw the CFG b. Compute CD c. If-convert the code

• 10 -

Region Formation + If-conversion

BB2 BB4 BB

BB5’

BB

BB

64.^

BB4’^2028 BB6’

2.^

Y^ Control flow representation^ »

branches » predicated operations

Y^ If-conversion not all all ornothing deal^ »

Often bad to apply in blanketmode » Selectively apply

Y^ Regions^ »

Extend a superblock tocontain if-converted code » Convert off-trace transitionsto on-trace » A hyperblock is born » Superblock is a special caseHB where all guardingpredicates are True

• 12 -

Negative 1: Resource Usage Resource usage is additive for all BBs that are if-converted

Case 1: Each BB requires 3 resources Assume processor has 2 resources No IC: 13 + .63 + .43 + 13 = 9

9 / 2 = 4.5 = 5 cycles IC: 1(3 + 3 + 3+ 3) = 12

12 / 2 = 6 cycles

BB

60 BB

BB^

BB

BB2 if p BB^

Case 2: Each BB requires 3 resources Assume processor has 6 resources No IC: 13 + .63 + .43 + 13 = 9

9 / 6 = 1.5 = 2 cycles IC: 1(3+3+3+3) = 12

12 / 6 = 2 cycles

BB3 if p2^ BB

100

• 13 -

Negative 2: Dependence Height Dependence height is max of for all BBs that are if-converted (dep height = schedule length with infinite resources)

Case 1: height(bb1) = 1, height(bb2) = 3 Height(bb3) =

^9 , height(bb4) = 2 No IC: 11 + .63 + .49 + 12 = 8.4 IC: 11 + 1MAX(3,9) + 1*3 = 13

BB

60 BB

BB^

Case 2: height(bb1) = 1, height(bb2) = 3 Height(bb3) =

^3 , height(bb4) = 2 No IC: 11 + .63 + .43 + 12 = 6 IC: 11 + 1MAX(3,3) + 1*2 = 6 BB

BB2 if p BB^

BB3 if p2^ BB

100

• 15 -

When To If-convert^ Y^

Resources^ »^ Small resource usage ideal forless important paths Y Dependence height^ »^ Matched heights are ideal^ »^ Close to same heights is ok Y Remember everything is relative for resources and dependenceheight! Y Hazards^ »^ Avoid hazards unless on mostimportant path Y Estimate of benefit^ »^ Branches/Mispredicts removed^ »^ Fudge factor

BB

BB1 BB

BB

BB1 BB2 if p1 BB3 if p2 BB

• 16 -

The Hyperblock^ Y^

Hyperblock

• Collection of

basic blocks in which controlflow may only enter at the firstBB.^ All internal control flowis eliminated via if-conversion^ »^ “Likely control flow paths

»^ Acyclic (outer backedge ok) »^ Multiple intersecting traceswith no side entrances »^ Side exits still exist Y Hyperblock formation »^ 1. Block selection »^ 2. Tail duplication »^ 3. If-conversion

BB

BB4 BB

BB

BB

BB

• 18 -

Block Selection^ Y^

Create a trace

Æ“main path”

»^ Use a heuristic function to select other blocks that are“compatible” with the main path »^ Consider each BB by itself for simplicity^ y

Compute priority for other BB’s y Normalize against main path.

Y^ BSVi = (K x (weight_bbi / size_bbi) x (size_main_path /weight_main_path) x bb_chari)^ »

weight = execution frequency » size = number of operations » bb_char = characteristic value of each BB^ y^

Max value = 1, Hazardous instructions reduce this to 0.5, 0.25, ...

»^ K = constant to represent processor issue rate Y Include BB when BSVi > Threshold

• 19 -

Example - Step 1 - Block Selection

main path = 1,2,4,

num_ops = 5 + 8 + 3 + 2 = 18 weight = 80 Calculate the BSVs for BB3, BB5 assuming no hazards, K = 4 BSV3 = 4 x (20 / 2) x (18 / 80) = 9 BSV5 = 4 x (10 / 5) x (18 / 80) = 1.8 If Threshold = 2.0, select BB3 along with main path 10 BB2 - 8

80 BB4 - 3^ BB6 - 2

BB5 - 5

BB1 - 5 80

BB3 – 2