register allocation graph

8/9/2019 Register Allocation Graph

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Register allocation problems(via graph colouring)

by: Navdeep singh


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Lecture Outline

• Memory Hierarchy Management

• Register Allocation

– Register interference graph

– Graph coloring heuristics

– Spilling


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Current Trends to manage memory hierarchy

• .ache and register si/es are gro#ing slo#ly

• "rocessor speed improves faster than memory speed anddis$ speed– 0he cost of a cache miss is gro#ing– 0he #idening gap is bridged #ith more caches

• (t is very important to:

– Manage registers properly– Manage caches properly

• .ompilers are good at managing registers


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The Register Allocation ProblemThe Register Allocation Problem

• Recall that intermediate code uses as many temporariesas necessary– 0his complicates final translation to assembly– 1ut simplifies code generation and optimi/ation

– 0ypical intermediate code uses too many temporaries

• 0he register allocation problem:– Re#rite the intermediate code intermediate code to use e!er temporaries than

there are machine registers

– Method"Method" assign more temporaries to a register• 1ut #ithout changing the program behavior


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%asic Register Allocation &dea

• 0he value in a dead temporary is not neededfor the rest of the computation– A dead temporary can be reused

• %asic rule%asic rule:

– 0emporaries t and t % can share the same

register if at any point in the program atmost one of t or t % is live 7


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Algorithm to minimi'e the number oregisters" Part &

• .ompute live variables for each point:

a :2 b 3 cd :2 4a

e :2 d 3 f

f :2 % 8 eb :2 d 3 e

e :2 e 4

b :2 f 3 c

9b

9c'e

9b

9c'f 9c'f

9b'c'e'f

9c'd'e'f

9b'c'f

9c'd'f9a'c'f

a

f

e

d

c

b

{a,b} not

included in

any subset


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The Register &ntererence raphThe Register &ntererence raph

• 0#o temporaries that are live simultaneouslycannot be allocated in the same register same register

• ;e construct an undirected graph– A node for each temporary

– An edge bet#een t and t% if they are livesimultaneously at some point in the program

• 0his is the register interference graph &R(G+– 0#o temporaries can be allocated to the same

register if there is no edge connecting them


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Register &ntererence raph #$ample

• !g!' b and c cannot be in the same register

• >!g!' b and d can be in the same register

a :2 b 3 cd :2 4a

e :2 d 3 f

f :2 % 8 eb :2 d 3 e

e :2 e 4

b :2 f 3 c


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Register &ntererence raphRegister &ntererence raph Properties

! (t e=tracts e=actly the information neededto characteri/e legal register assignments

%! (t gives a global &i!e!' over the entire flo#graph+ picture of the register re?uirements

,! After R(G construction the registerallocation algorithm is architectureindependent


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raph Coloring *einitionsraph Coloring *einitions

• A coloring of a graph is an assignment ofcolors to nodes' such that nodes connected by

an edge have different colors

• A graph is $4colorable if it has a coloring #ith$ colors


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Register Allocation Through raph Coloring

• (n our problem' colors + registers colors + registers – ;e need to assign colors &registers+ to graph nodes

&temporaries+

• @et $ 2 number of machine registers

• (f the R(G is $4colorable then there is aregister assignment that uses no more than $registers

Register Interference Graph


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raph Coloring #$ample

• .onsider the e=ample R(G

a

f

e

d

c

b

• 0here is no coloring #ith less than - colors

• 0here are -4colorings of this graph

r-

r

r%

r,

r%

r,


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raph Coloring #$ample

• nder this coloring the code becomes:

r% :2 r, 3 r-r, :2 4r%

r% :2 r, 3 r

r :2 % 8 r%r, :2 r, 3 r%r% :2 r% 4

r, :2 r 3 r-


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Computing raph ColoringsComputing raph Colorings

• 0he remaining problem is to compute acoloring for the interference graph

• 1ut:

! 0his problem is very hard &N"4hard+! No efficientalgorithms are $no#n!

%! A coloring might not e=ist for a given number orregisters

• 0he solution to &+ is to use heuristics• ;eBll consider later the other problem


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raph Coloring Heuristic

• *bservation:– "ic$ a node t #ith fe#er than $ neighbors in R(G– >liminate t and its edges from R(G– (f the resulting graph has a $4coloring then so does

the original graph

• ;hy:

– @et c'C'cn be the colors assigned to the neighbors oft in the reduced graph– Since n D $ #e can pic$ some color for t that is

different from those of its neighbors


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raph Coloring Heuristic

• 0he follo#ing #or$s #ell in practice:– "ic$ a node t #ith fe#er than $ neighbors

– "ut t on a stac$ and remove it from the R(G

– Repeat until the graph has one node

• 0hen start assigning colors to nodes on thestac$ &starting #ith the last node added+– At each step pic$ a color different from those

assigned to already colored neighbors


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raph Coloring #$ample (,)raph Coloring #$ample (,)

• Remove a and then d

a

f

e

d

c

b

• Start #ith the R(G and #ith $ 2 -:

Stac$: 9


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raph Coloring #$ample (-)

• No# all nodes have fe#er than - neighborsand can be removed: c' b' e' f

f

e c

b

Stac$: 9d' a


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raph Coloring #$ample (-)

• Start assigning colors to: f' e' b' c' d' a

b

a

e c r-

fr

r%

r,

r%

r,

d


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.hat i the Heuristic /ails0.hat i the Heuristic /ails0

• ;hat if during simplification #e get to a state #here allnodes have $ or more neighbors E

• >=ample: try to find a ,4coloring of the R(G:

a

f

e

d

c

b


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.hat i the Heuristic /ails0

• Remove a and get stuc$ &as sho#n belo#+

f

e

d

c

b

• "ic$ a node as a candidate for spilling– A spilled temporary 1lives2 in memory

• Assume that f is pic$ed as a candidate


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• Remove f and continue the simplification– Simplification no# succeeds: b' d' e' c

e

d

c

b


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• *n the assignment phase #e get to the point#hen #e have to assign a color to f

• ;e hope that among the - neighbors of f #euse less than , colors ⇒ optimistic coloring

f

e

d

c

b r,

rr%

r,

E

ll3 illi


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3pilling3pilling

• Since optimistic coloring failed #e must spill temporaryf

• ;e must allocate a memory location as the home of f– 0ypically this is in the current stac$ frame– .all this address fa

• 1efore each operation that uses f' insert f :2 load fa

• After each operation that defines f' insert store f' fa


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3pilling #$ample

• 0his is the ne# code after spilling f

a :2 b 3 cd :2 4af :2 load fae :2 d 3 f

f :2 % 8 estore f' fa

b :2 d 3 e

e :2 e 4

f :2 load fab :2 f 3 c


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Recomputing Liveness &normationRecomputing Liveness &normation

• 0he ne# liveness information after spilling:

a :2 b 3 cd :2 4af :2 load fae :2 d 3 f

f :2 % 8 estore f' fa

b :2 d 3 e

e :2 e 4

f :2 load fab :2 f 3 c

9b

9c'e

9b

9c'f9c'f

9b'c'e'f

9c'd'e'f

9b'c'f

9c'd'f9a'c'f

9c'd'f

9c'f

9c'f


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Recomputing Liveness &normation

• 0he ne# liveness information is almost as before

• f is live only– 1et#een a f :2 load fa and the ne=t instruction

– 1et#een a store f' fa and the preceding instructiom!

• Spilling reduces the live range of f

• And thus reduces its interferences

• ;hich result in fe#er neighbors in R(G for f


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Recompute R& Ater 3pilling

• 0he only changes are in removing some of the edges

of the spilled node

• (n our case f still interferes only #ith c and d

• And the resulting R(G is ,4colorable

a

f

e

d

c

b


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register allocation graph

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