Interprocedural Analysis and Optimization Seminar on Optimizations for Modern Architectures “...

Interprocedural Interprocedural Analysis and Analysis and OptimizationOptimization

Seminar on Optimizations for Modern Architectures

“Optimizing Compilers for Modern Architectures”,

Allen and Kennedy, Chapter 11 - through section 11.2.4

Presented by Gabi Kliot

©Gabi Kliot, Technion, 2006 2

Roadmap Examples of Interprocedural problems Classification of Interprocedural problems Solve two Interprocedural problems

MOD Analysis Alias Analysis

More by Li-Tal: Additional Interprocedural problems

Constant propagation, Kill Analysis, Symbolic analysis, Array section analysis, Call graph construction

Interprocedural Optimization Managing whole program compilation

Introduction

Interprocedural Analysis Gathering information about the whole

program instead of a single procedure Interprocedural Optimization

Program transformation that involves more than one procedure in the program (uses interprocedural analysis)

Today’s topics

Examples of Interprocedural problems Classification of Interprocedural

problems Solve two Interprocedural problems

Some Interprocedural Problems

Modification and Reference Side-effectCOMMON X,Y...DO I = 1, N

S0: CALL PS1: X(I) = X(I) + Y(I)

Can vectorize if P 1. neither modifies nor uses X2. does not modify Y

MOD(s)MOD(s): set of variables that may be modified as a side effect of call at s

REF(s)REF(s): set of variables that may be referenced as a side effect of call at s

DO I = 1, N

S0: CALL PS1: X(I) = X(I) + Y(I)

Can vectorize S1 if ( 0) ( 0) ( 0)x REF S x MOD S y MOD S

Modification and Reference Side Effect

Alias AnalysisSUBROUTINE S(A,X,N)

COMMON Y /* Y is global variable */DO I = 1, N

S0: X = X + Y*A(I)ENDDO

It would be efficient to keep X and Y in different registers and store in X outside the loop

What happens when there is a call, CALL S(A,Y,N)? Then Y is aliased to X on entry to S Can’t delay update to X in the loop any more

ALIAS(ALIAS(pp,x),x): set of variables that may refer to the same location as formal parameter x on entry to p

Call Graph Construction

Call Graph G=(N,E) N: one vertex for each procedure E: one edge for each possible call

Edge (p,q) is in E if procedure p calls procedure q Looks easy Construction difficult in presence of procedure

variables Fortan: procedure parameters C: pointers to function C++: virtual functions

Call Graph Construction

SUBROUTINE S(X,P)S0: CALL P(X)

RETURNEND

P is a procedure parameter to S What values can P have on entry to S? CALL(s)CALL(s): set of all procedures that may be

invoked at s Resembles closely to the alias analysis problem Call graph construction will be presented later

Live and Use AnalysisDO I = 1, N

T = X(I)*CA(I) = T + B(I)C(I) = T + D(I)

This loop can be parallelized by making T a local variable in the loop

PARALLEL DO I = 1, NLOCAL tt = X(I)*CA(I) = t + B(I)C(I) = t + D(I)IF(I.EQ.N) T = t

ENDDO Copy of local version of T to the global

version of T is required to ensure correctness

What if T was not live outside the loop? x is Live at point sx is Live at point s : there is a control-

flow path from s to a use of x that contains no definition of x prior to usage

Live and Use Analysis Solve Live analysis using Use analysis USE(s) USE(s) : set of variables having an upward

exposed use in procedure p called at site s upward exposed useupward exposed use : use of a variable

that is not proceeded by a definition of the variable on some path to the use from the point of invocation

x is Live at a call site sx is Live at a call site s iff x in USE(s) or there is a path through p that doesn’t assign a new

value to x and x is live in some successor(s)

Kill AnalysisDO I = 1, N

S0: CALL INIT(T,I)T = T + B(I)A(I) = A(I) + T

To parallelize the loop: INIT must not create dependence with respect to

loop It must be possible to recognize that variable T

can be made private to each iteration It means that T is assigned before being used on

every path through INIT

Kill Analysis

DO I = 1, N

S0: CALL INIT(T,I)T = T + B(I)A(I) = A(I) + T

T has to be assignedbefore being used on every path through INIT

SUBROUTINE INIT(T,I)

REAL TINTEGER IT = X(I)

For this INIT we can seethat T can be made private to each iteration

KILL(s)KILL(s) : set of variables assigned on every path through procedure p called at s and through procedures invoked in p

T in the previous example can be privatized under the following condition:

Also we can express LIVE(s) as following:T (KILL(S0)USE(S0))

LIVE(s) USE(s) (KILL(s) LIVE(b))

bsucc(s)

Kill Analysis

Constant PropagationSUBROUTINE S(A,B,N,IS,I1)

REAL A(*), B(*)DO I = 0, N-1

S0: A(IS*I+I1) = A(IS*I+I1) + B(I+1)ENDDO

END If IS==0 the loop around S0 always accesses A(I1)

(output dependence) If IS!=0 the loop can be vectorized CONST(p)CONST(p) : set of variables with known

constant values on every invocation of procedure p

Knowledge of CONST(p) is useful for interprocedural constant propagation

Today’s topics

Interprocedural Problem Classification

May and Must problems MOD, REF and USE are ‘May’ problems KILL is a ‘Must’ problem NOT May is Must, Not Must is May The solution of the inverse problem is usually a

universe complement Flow sensitive and flow insensitive problems

Flow sensitiveFlow sensitive: control flow info important Flow insensitiveFlow insensitive: control flow info unimportant

(in compilation time we assume all code is reachable)

MOD(R1) MOD(A) MOD(B)

MOD(R2) MOD(A) MOD(B)

REF(R1) REF(A) REF(B)

REF(R2) REF(A) REF(B)

KILL(R1) KILL(A) KILL(B)

KILL(R2)KILL(A) KILL(B)

USE(R1)USE(A) (KILL(A)USE(B))

USE(R2)USE(A)USE(B)

Flow Sensitive vs Flow Insensitive

Classification (continued)

A problem is flow insensitiveA problem is flow insensitive iff solution of both sequential and alternately composed regions is determined by taking union of subregions

Approximation of flow-sensitive problems with flow-insensitive analysis:

Possible approximation: Flow-insensitive problems are easier

( ) ( )APUSE p REF p( ) ( ) ( ) ( )USE p APUSE p APUSE p USE p

Classification (continued)

Side-effect vs Propagation problems Side-effectSide-effect – what may or must happen as a side

effect of a this procedure call (backward effects) PropagationPropagation – what conditions may or must hold

upon entry to the current procedure (forward context)

problem type side effect propagation

flow-insensitive

MOD, REF ALIAS, CALLSCONST

flow-sensitive KILL, USE

Three Graphs Call Graph G=(N,E)

N: one vertex for each procedure E: one edge for each possible call

Edge (p,q) is in E if procedure p calls procedure q

Binding graph GB=(NB,EB) One vertex for each formal parameter of each

procedure Directed edge from formal parameter, f1 of p to formal

parameter, f2 of q if there exists a call site s=(p,q) in p such that f1 is bound to f2

|NB| <= |E|, |EB| <= |E|, while G=(N,E) is a call graph

Three Graphs – cont.

Control-Flow graph G=(N,E) N: the set of all instructions (or basic blocks) E: the set of control flow edges

Edge (p,q) is in E if instruction p may be followed by instruction q

Today’s topics Examples of Interprocedural problems Classification of Interprocedural

Assumptions No procedure nesting and no scope nesting

only global and local variables All parameters passed by reference No pointers or reference variables

Aliases are introduced only at call sites Size of the parameter list bounded by a constant

We will formulate and solve the MOD(s) problem

Flow InsensitiveSide-effect Analysis

Solving MOD DMOD(s)DMOD(s) : set of variables which are directly

modified as side-effect of call at s

GMOD(p)GMOD(p) : set of global variables and formal parameters of p that are modified, either directly or indirectly as a result of invocation of p

MOD(s) DMOD(s) ALIAS( p,x)

xDMOD(s)

DMOD(s) {v | s p,v s w,w GMOD(p)}

Example: DMOD and GMOD

S0: CALL P(A,B,C)

…SUBROUTINE P(X,Y,Z)

INTEGER X,Y,ZX = X*ZY = Y*Z

GMOD(P)={X,Y} DMOD(S0)={A,B}

Solving GMOD – method A

GMOD(p) contains two types of variables Variables explicitly modified in body of p

This constitutes the set IMOD(p)IMOD(p) Variables modified as a side-effect of some

procedure q invoked in p Includes formal parameters Includes global variables, which are viewed as

parameters to a called procedure Does not include local variables of q

GMOD(p) IMOD( p) {z | z s

s(p ,q) w,w GMOD(q)}

The previous iterative method may take a long time to converge No limit on the number of times analysis must

traverse a loop in the problem graph Problem with recursive calls

Recursive callsSUBROUTINE P(F0,F1,F2,…,Fn)INTEGER X,F0,F1,F2,..,Fn…IF(Fk==<some>) RETURN

S0: F0 = <some expr>…S1: CALL P(F1,F2,…,Fn,X)…END

P0 F0 F0+F1+.. Fk

P1 F0 F0 +.. Fk-1

P2 F0 F0 +.. Fk-2

Pk-2 F0 F0+F1+F2

Pk-1 F0 F0+F1

Pk F0 F0

Decompose GMOD(p) differently to get an efficient solution

Key: Treat side-effects to global Treat side-effects to global variables and reference formal variables and reference formal parameters separatelyparameters separately

IMOD(p)IMOD(p)+ + : IMOD(p) + all actual parameters modified in the methods invoked inside p

Solving GMOD – method B

( ) ( ) _

( ) ( ) ( )s p q

GMOD p IMOD p Global Variables

GMOD p IMOD p GMOD q LOCAL

IMOD(p)IMOD(p)+ + : Formally defined

RMOD(p)RMOD(p) : set of formal parameters in p that may be modified in p, either directly or by assignment to a reference formal parameter of q as a side effect of a call of q in p

IMOD( p) IMOD( p) {z | z s

s( p,q) w,wRMOD(q)}

Solving for IMOD+ by RMOD

Some Order

IMOD(p) : globals + formals + locals explicitly modified in a body of p

RMOD(p) : only formals modified directly in p or as parameters to q (p calls q)

IMOD(p)+ : everything directly in p + parameters to q

GMOD(p) :IMOD( p) IMOD( p) {z | z s

s( p,q) w,wRMOD(q)}

( ) ( ) _GMOD p IMOD p Global Variables

Some Order - example

CALL p(X, Y, Z)COMMON G1, LOCAL t, nX = 3G1 = 4t = 5 CALL q(Y, n)

SUBROUTINE q(A, B)COMMON G2, LOCAL mA = 16B = 17G2 = 18m = 19

IMOD RMOD IMOD+ GMOD

qA,B,G2,

A,B,G2,m

p X, G1, t X, YX, G1, t,

Y, nX, G1, t, Y, n, G2

IMOD(p) : globals + formals + locals explicitly modified in a body of p

RMOD(p) : only formals modified directly in p or as parameters to q (p calls q)

IMOD(p)+ =

GMOD(p) = ( ) _IMOD p Global Variables ( , )

( ) { | , ( )}s

IMOD p z z w w RMOD q

Solving for RMOD - 1

RMOD(p) construction needs binding graph GB=(NB,EB) One vertex for each formal parameter of each

procedure Directed edge from formal parameter, f1 of p to

formal parameter, f2 of q if there exists a call site s=(p,q) in p such that f1 is bound to f2

|NB| = |E|, |EB| = |E|, while G=(N,E) is a call graph

Use a marking algorithm to solve RMOD

Solving for RMOD - 2 Construct the binding graph GB=(NB,EB) with

forward and backward edges Calculate IMOD(p) for each procedure Mark all nodes that are in their IMOD and add

them to the worklist Run DFS/BFS from marked nodes on backward

edges From marked v visit every u, s.t. (u,v) є EB (from formal v back

to its actual u; u is formal of the inclosing method) Mark every reached, yet unmarked node u Add u to the list

Add every marked node to its RMOD(p)

SUBROUTINE p(A,B)INTEGER P,Q,II = 2CALL q(A,B,I)CALL q(B,A,I)

SUBROUTINE q(X,Y,Z)INTEGER X,Y,ZX = Z + 1Y = Z + 2

IMOD(p)={I} IMOD(q)={X,Y}

(0) (0) (0)

(0)(0)

IMOD(p)={I} IMOD(q)={X,Y} Worklist={X,Y}

(1) (1) (0)

SUBROUTINE p(A,B)INTEGER P,Q,II = 2 CALL q(A,B,I) CALL q(B,A,I)

RMOD(p)={A,B} RMOD(q)={X,Y} Complexity:

( ) ( ) ( )B BO N E O N E O N E

(1) (1) (0)

SUBROUTINE p(A,B)INTEGER P,Q,II = 2 CALL q(A,B,I) CALL q(B,A,I)

Solving for IMOD+

After gathering RMOD(p) for all procedures, update RMOD(p) to IMOD+(p) using this equation

For each call site in p and each actual parameter Z to q , add Z to IMOD+(p) if Z->W and W є RMOD(q)

This can be done in O(NV+E) time Conversion from RMOD(p) to IMOD+(p): O(N+E) Initialization of IMOD+(p): O(NV)

N – vertices in call graph (number of procedures) V – all global variables

IMOD( p) IMOD( p) {z | z s

s( p,q) w,wRMOD(q)}

Solving for GMOD - 1 After gathering IMOD+(p) for all procedures,

calculate GMOD(p) according to the following equation:

Global x is in GMOD(p) if there is a nonempty path in the control flow graph from p to q, where x є IMOD+(q)

This can be solved using a DFS algorithm based on Tarjan’s SCR algorithm on the Call Graph

GMOD(p) IMOD( p)

s( p,q) GMOD(q) LOCAL

Solving for GMOD – 2

If we wanted to find global part of GMOD(p) only for procedure p – just run DFS from p on the call graph and collect all global parts of GMODs from the children

( ( )) ( )G GMOD p GMOD q LOCAL

But we want to calculate global part of GMOD(p) for EVERY procedure p

G(GMOD(p)) is the global part of GMOD(p)

All procedures in the same strongly connected component of the Call Graph have the same set of global modified variables

Recall we are solving flow-insensitive problem Init: GMOD(p) = IMOD+(p) Run DFS and update G(GMOD(p)) at every node

from its children G(GMOD(q))s When the head p of SCC is reached back, update

all nodes in this component with G(GMOD(p)) O((N+E)V) complexity

Solving for GMOD - 3 - FindGMOD

dfn=low= GGMOD = {P}

dfn=low= GGMOD = {Q}

dfn=low= GGMOD = {R}

dfn=low= GGMOD = {S}

findGMOD runs on call graph method i modifies global I eventually low will be the

same for all nodes in SCC

Stack = {}p

FindGMOD – the call graph

dfn=1low= 1 GGMOD = {P}

dfn=2low= 2 GGMOD = {Q}

dfn= 3low= 3 GGMOD = {R}

dfn=4low= 4 GGMOD = {S}

findGMOD run on call graph Method i modifies global I

Stack = {s,r,q,p}pp

dfn= 3low= 1 GGMOD = {R}

dfn=2low= 1 GGMOD = {Q, R}

FindGMOD – forward steps

dfn=2low= 1 GGMOD = {Q, R, S}q

dfn=1low= 1 GGMOD = {P}

dfn=4low= 4 GGMOD = {S}

Stack = {r,q,p}pp

r dfn= 3low= 1 GGMOD = {R}

dfn=2low= 1 GGMOD = {Q, R}

dfn=1low= 1 GGMOD = {P, Q, R, S}

FindGMOD – updating from sons

sdfn=4low= 4 GGMOD = {S}

Stack = {r,q,p}pp

r dfn= 3low= 1 GGMOD = {R}

dfn=2low= 1 GGMOD = {Q, R, S}

dfn=1low= 1 GGMOD = {P, Q, R, S}

FindGMOD – updating SC components

dfn=2low= 1 GGMOD = {Q, R, S, P}

dfn= 3low= 1 GGMOD = {R, P, Q, S}

Solving for DMOD

After gathering GMOD(p) for all procedures, update DMOD(p) by using this equation

The entire computation of DMOD(p) takes O((N+E)V) steps

Still have to update DMOD to MOD with help of Alias analysis

DMOD(s) {v | s p,v s w,w GMOD(p)}

Today’s topics

Recall definition of MOD(s)

Task is to Compute ALIAS(p,x) Update DMOD to MOD using ALIAS(p,x)

( ) ( ) ( , )x DMOD s

MOD s DMOD s ALIAS p x

Alias Analysis

GMOD(Q)={Z}

MOD(S1)={X,Y}

DMOD(S1)={X}

Update DMOD to MOD - example

SUBROUTINE SINTEGER A

S0: CALL S(A,A)END

SUBROUTINE P(X,Y)INTEGER X,Y

S1: CALL Q(X)END

SUBROUTINE Q(Z)INTEGER ZZ = 0

ALIAS requires forward analysis MOD requires backward analysis

procedure findMod(N,E,n,IMOD+,LOCAL)for each call site s do begin

MOD[s]:=DMOD[s]; for each x in DMOD[s] do

for each v in ALIAS[p,x] doMOD[s]:=MOD[s] {v};

endend findMod O(EV2) algorithm; need to do better

Naïve Update of DMOD to MOD

S0: CALL P(G,B)

SUBROUTINE P(X,Y)COMMON G

S1: G = 1END

G DMOD p

X ALIAS p G

S2: CALL P(G,B)

S3: X = 1END

X DMOD p

G ALIAS p X

Aliases examples

S4: CALL P(A,A)

S5: X = 1END

X DMOD p

Y ALIAS p X

Global is aliased to formal

Formal is aliased to global

Formal is aliased to formal

Actually these two are the same

Key Observations Two global variables can never be aliases of each

other. Global variables can only be aliased to a formal parameter

No reference variables in Fortran Only passing parameters by reference

ALIAS(p, G) contains less than entries if G is global ALIAS(p, x) for formal parameter x can contain any

global variable or other formal parameters (O(V))

Update of DMOD to MOD - better analysis

Break down the update analysis into two parts If x is global we need to add less than elements

but need to do it O(V) times If x is a formal parameter we need to add O(V)

elements but need to do it only O( ) times This gives O( V)=O(V) update time per call,

implying O(VE) for all calls

Update of DMOD to MOD - better analysis

Computing Aliases

Compute A(f): the set of global variables a formal parameter f can be aliased to Use binding graph [with cycles reduced to single

nodes]

DFS from the place where ALIAS is introduced (call p(G)) on the binding graph edges in topological order and update ALIASes to formals

O((N+E)V) A(G): the set of formal variables a global G can be

aliased to. It is the same set, just interpreted reversely

g f0 f1 ... fn f

Computing Aliases

Expand A(f) to Alias(p,f) for formal parameters Find alias introduction sites of formal to formal

e.g. CALL P(X,X) Propagate aliases

DFS is your best friend - again…

1 2 3 4 5 6, , ... ,f f f f f f f

Summary

Some Interprocedural problems and their application in parallelization and vectorization

Classified Interprocedural problems Solved the modification side-effect problem

and the alias propagation problem Flow insensitive side-effect problems are

solvable in the O((N+E)V) time

BACKUP

DFS numbering dfsnum(v) is the number of vertices visited before v in the DFS

If there is a back or cross edge out of the subtree of v, it is a node visited before v and therefore with a smaller dfsnum

low value low(v) is the smallest dfsnum of a vertex reachable by a back or cross edge from the subtree of v

Tarjan SCC algorithm

visit(p)

add p to L

low(p) = dfsnum(p) = N , N++

for each edge p->q

if q is not already in T // tree edge

add p->q to T

visit(q)

low(p) = min(low(p), low(q))

else // back or cross edge

low(p) = min(low(p), dfsnum(q))

if low(p)=dfsnum(p) // component

repeat

remove last element v from L

until v=p

Tarjan SCC algorithm

Solving for GMOD

Initialize GMOD(p) to IMOD+(p) on discovery

Update GMOD(p) computation while backing up

Solving for GMOD

Initialize GMOD(p) to IMOD+(p) on discovery

Update GMOD(p) computation while backing up

For each node u in a SCC update GMOD(u) in a cycle

O((N+E)V) Algorithm

Interprocedural Analysis and Optimization Seminar on Optimizations for Modern Architectures “...

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