Upper Bounds on Memory Usage for Garbage-Collected Languages

(1)

Upper Bounds on Memory Usage for

Garbage-Collected Languages

Elvira Albert

1 _{, Samir Genaim}

1 _{, Miguel G´}

_{omez-Zamalloa}

1 Puri Arenas

1 _{, Germ´an Puebla}

2 _{and Damiano Zanardini}

2 (1)Complutense University of Madrid (Spain)

(2) Technical University of Madrid (Spain)

(2)

Cost Analysis

is the automatic study of

program efficiency

(or the

resource consumption

).

◮

Its aim is to statically

estimate the cost of a program execution in

(3)

Introduction

Cost Analysis

is the automatic study of

program efficiency

(or the

resource consumption

).

◮

Its aim is to statically

estimate the cost of a program execution in

terms of the size of its input args.

The cost can be defined w.r.t. different

cost models

:

◮

number of instructions executed

◮

memory allocated

(4)

Cost Analysis

is the automatic study of

program efficiency

(or the

resource consumption

).

◮

Its aim is to statically

estimate the cost of a program execution in

terms of the size of its input args.

The cost can be defined w.r.t. different

cost models

:

◮

number of instructions executed

◮

memory allocated

◮

number calls to certain methods: billable events on a mobile

Finding the exact cost of programs is

undecidable, but it is possible to

infer useful information (

bounds

):

(5)

Introduction

Cost Analysis

is the automatic study of

program efficiency

(or the

resource consumption

).

◮

Its aim is to statically

estimate the cost of a program execution in

terms of the size of its input args.

The cost can be defined w.r.t. different

cost models

:

◮

number of instructions executed

◮

memory allocated

◮

number calls to certain methods: billable events on a mobile

Finding the exact cost of programs is

undecidable, but it is possible to

infer useful information (

bounds

):

◮

Upper bounds

: a program runs within the resources available.

◮

Lower bounds

: useful for scheduling distributed execution.

(6)

0: new Cons

Cons.copy()

3: dup

7: astore

4: invokespecial

. . .

27: aload

28: areturn

Size

SOLVER

CFG

_Relations

Cost Relations

BYTECODE PROGRAM

RBR

COST MODEL

THE COSTA SYSTEM

UPPER BOUND

a

**C(a) = 8*2 O(2)**

(7)

COSTA: Cost Analyzer for Java Bytecode

0: new Cons

Cons.copy()

3: dup

7: astore

4: invokespecial

. . .

27: aload

28: areturn

Size

SOLVER

CFG

_Relations

Cost Relations

RBR

COST MODEL

THE COSTA SYSTEM

UPPER BOUND

a

**C(a) = 8*2 O(2)**

a

For mobile code, we do not have access to source code.

(8)

0: new Cons

Cons.copy()

3: dup

7: astore

4: invokespecial

. . .

27: aload

28: areturn

Size

SOLVER

CFG

_Relations

Cost Relations

RBR

COST MODEL

THE COSTA SYSTEM

UPPER BOUND

a

**C(a) = 8*2 O(2)**

a

For mobile code, we do not have access to source code.

(9)

HEAP:

Total Allocation Analysis

Symbolic

cost model

for heap consumption,

size

(

Tree

),

size

(

Integer

)..

We get upper bounds of the

total

allocated memory.

(10)

Symbolic

cost model

for heap consumption,

size

(

Tree

),

size

(

Integer

)..

We get upper bounds of the

total

allocated memory.

In presence of

garbage collection (GC)

, it is a too pessimistic

estimation of the actual memory consumption.

PEAK:

Live Heap Space Analysis

Aims at approximating the maximum or

peak

heap usage.

Scope-based GC model:

1

Reclaims unreachable memory when methods return. This assumption

can be refined up to an ideal GC.

2

_{Collects unreachable objects which have been created during the}

execution of the corresponding method call and not before.

(11)

From Java to Intermediate representation

class

Test

{

staticTree m(intn){

if( n>0 )return

newTree(m(n-1),m(n-1),f(n));

elsereturn null;

}

static intf(intn){

inta=0,i=n; for(; n>1 ; n=n/2 )

a += g(n).intValue(); for(; i>1; i=i/2)

a *= h(i).intValue(); returna;

}

staticInteger g(int n){

Integer x=newInteger(n); returnnewInteger(x.intValue()+1);

}

(12)

staticTree m(intn){

if( n>0 )return

elsereturn null;

}

static intf(intn){

inta=0,i=n; for(; n>1 ; n=n/2 )

}

staticLong h(int n){

returnnewLong(n-1);

n

>

0 ,

s

0 := new

Tree

1 ;

s

1 :=

n

−

1 ,

m

(

h

s

1 i

,

h

s

1 i

)

,

s

2 :=

n

−

1 ,

m

(

h

s

2 i

,

h

s

2 i

)

,

f

(

h

n

i

,

h

s

3 i

)

,

init

(

h

s

0 ,

s

1 ,

s

2 ,

s

3 i

,

hi

)

,

r

=

s

0 .

m

(

h

n

i

,

h

r

i

)::=

n

≤

0 ,

r

:= null

.

a

:=

0 ,

i

:=

n

,

f

c

(

h

n

,

a

i

,

h

n

,

a

i

)

,

f

d

(

h

i

,

a

i

,

h

i

,

a

i

)

,

r

:=

a

.

f

c

(

h

n

,

a

i

,

h

n

,

a

i

)::=

n

>

1 ,

g

(

h

n

i

,

h

s

0 i

)

,

intValue

1 (

h

s

0 i

,

h

s

0 i

)

a

:=

a

+

s

0 ,

n

:=

n

/

2 ,

f

c

(

h

n

,

a

i

,

h

n

,

a

i

)

.

(13)

From Java to Intermediate representation

class

Test

{

staticTree m(intn){ if( n>0 )return

elsereturn null; }

static intf(intn){

inta=0,i=n; for(; n>1 ; n=n/2 )

}

m

(

h

n

i

,

h

r

i

)::=

n

>

0 ,

s

0 := new

Tree

1 ;

s

1 :=

n

−

1,

m

(

h

s

1 i

,

h

s

1 i

)

,

s

2 :=

n

−

1,

m

(

h

s

2 i

,

h

s

2 i

)

,

f

(

h

n

i

,

h

s

3 i

)

,

init

(

h

s

0 ,

s

1 ,

s

2 ,

s

3 i

,

hi

)

,

r

=

s

0 .

m

(

h

n

i

,

h

r

i

)::=

n

≤

0 ,

r

:= null

.

f

(

h

n

i

,

h

r

i

)::=

a

:=

0 ,

i

:=

n

,

f

c

(

h

n

,

a

i

,

h

n

,

a

i

)

,

f

d

(

h

i

,

a

i

,

h

i

,

a

i

)

,

r

:=

a

.

f

c

(

h

n

,

a

i

,

h

n

,

a

i

)::=

n

>

1 ,

g

(

h

n

i

,

h

s

0 i

)

,

intValue

1 (

h

s

0 i

,

h

s

0 i

)

a

:=

a

+

s

0 ,

n

:=

n

/

2 ,

(14)

staticTree m(intn){

if( n>0 )return

elsereturn null;

}

static intf(intn){

inta=0,i=n; for(; n>1 ; n=n/2 )

}

returnnewLong(n-1);

n

>

0 ,

s

0 := new

Tree

1 ;

s

1 :=

n

−

1 ,

m

(

h

s

1 i

,

h

s

1 i

)

,

s

2 :=

n

−

1 ,

m

(

h

s

2 i

,

h

s

2 i

)

,

f

(

h

n

i

,

h

s

3 i

)

,

init

(

h

s

0 ,

s

1 ,

s

2 ,

s

3 i

,

hi

)

,

r

=

s

0 .

m

(

h

n

i

,

h

r

i

)::=

n

≤

0 ,

r

:= null

.

a

:=

0 ,

i

:=

n

,

f

c

(

h

n

,

a

i

,

h

n

,

a

i

)

,

f

d

(

h

i

,

a

i

,

h

i

,

a

i

)

,

r

:=

a

.

f

c

(

h

n

,

a

i

,

h

n

,

a

i

)::=

n

>

1 ,

g

(

h

n

i

,

h

s

0 i

)

,

intValue

1 (

h

s

0 i

,

h

s

0 i

)

a

:=

a

+

s

0 ,

n

:=

n

/

2 ,

f

c

(

h

n

,

a

i

,

h

n

,

a

i

)

.

(15)

From Java to Intermediate representation

class

Test

{

staticTree m(intn){

if( n>0 )return

elsereturn null;

}

static intf(intn){

inta=0,i=n; for(; n>1 ; n=n/2 )

}

m

(

h

n

i

,

h

r

i

)::=

n

>

0 ,

s

0 := new

Tree

1 ;

s

1 :=

n

−

1,

m

(

h

s

1 i

,

h

s

1 i

)

,

s

2 :=

n

−

1,

m

(

h

s

2 i

,

h

s

2 i

)

,

f

(

h

n

i

,

h

s

3 i

)

,

init

(

h

s

0 ,

s

1 ,

s

2 ,

s

3 i

,

hi

)

,

r

=

s

0 .

m

(

h

n

i

,

h

r

i

)::=

n

≤

0 ,

r

:= null

.

f

(

h

n

i

,

h

r

i

)::=

a

:=

0 ,

i

:=

n

,

f

c

(

h

n

,

a

i

,

h

n

,

a

i

)

,

f

d

(

h

i

,

a

i

,

h

i

,

a

i

)

,

r

:=

a

.

f

c

(

h

n

,

a

i

,

h

n

,

a

i

)::=

n

>

1 ,

g

(

h

n

i

,

h

s

0 i

)

,

intValue

1 (

h

s

0 i

,

h

s

0 i

)

a

:=

a

+

s

0 ,

n

:=

n

/

2 ,

(16)

staticTree m(intn){

if( n>0 )return

elsereturn null; }

static intf(intn){

inta=0,i=n; for(; n>1 ; n=n/2 )

}

returnnewLong(n-1);

n

>

0 ,

s

0 := new

Tree

1 ;

s

1 :=

n

−

1 ,

m

(

h

s

1 i

,

h

s

1 i

)

,

s

2 :=

n

−

1 ,

m

(

h

s

2 i

,

h

s

2 i

)

,

f

(

h

n

i

,

h

s

3 i

)

,

init

(

h

s

0 ,

s

1 ,

s

2 ,

s

3 i

,

hi

)

,

r

=

s

0 .

m

(

h

n

i

,

h

r

i

)::=

n

≤

0 ,

r

:= null

.

a

:=

0 ,

i

:=

n

,

f

c

(

h

n

,

a

i

,

h

n

,

a

i

)

,

f

d

(

h

i

,

a

i

,

h

i

,

a

i

)

,

r

:=

a

.

f

c

(

h

n

,

a

i

,

h

n

,

a

i

)::=

n

>

1 ,

g

(

h

n

i

,

h

s

0 i

)

,

intValue

1 (

h

s

0 i

,

h

s

0 i

)

a

:=

a

+

s

0 ,

n

:=

n

/

2 ,

f

c

(

h

n

,

a

i

,

h

n

,

a

i

)

.

(17)

From Java to Intermediate representation

class

Test

{

staticTree m(intn){

if( n>0 )return

elsereturn null;

}

static intf(intn){ inta=0,i=n; for(; n>1 ; n=n/2 )

}

m

(

h

n

i

,

h

r

i

)::=

n

>

0 ,

s

0 := new

Tree

1 ;

s

1 :=

n

−

1 ,

m

(

h

s

1 i

,

h

s

1 i

)

,

s

2 :=

n

−

1 ,

m

(

h

s

2 i

,

h

s

2 i

)

,

f

(

h

n

i

,

h

s

3 i

)

,

init

(

h

s

0 ,

s

1 ,

s

2 ,

s

3 i

,

hi

)

,

r

=

s

0 .

m

(

h

n

i

,

h

r

i

)::=

n

≤

0 ,

r

:= null

.

f

(

h

n

i

,

h

r

i

)::=

a

:=

0,

i

:=

n,

f

c

(

h

n,

a

i

,

h

n,

a

i

)

,

f

d

(

h

i,

a

i

,

h

i,

a

i

)

,

r

:=

a.

f

c(

h

n,

a

i

,

h

n,

a

i

)::=

n

>

1 ,

g

(

h

n

i

,

h

s

0 i

)

,

intValue

1 (

h

s

0 i

,

h

s

0 i

)

a

:=

a

+

s

0 ,

(18)

staticTree m(intn){

if( n>0 )return

elsereturn null;

}

static intf(intn){

inta=0,i=n;

for(; n>1 ; n=n/2 ) a += g(n).intValue();

for(; i>1; i=i/2) a *= h(i).intValue(); returna;

}

returnnewLong(n-1);

n

>

0 ,

s

0 := new

Tree

1 ;

s

1 :=

n

−

1 ,

m

(

h

s

1 i

,

h

s

1 i

)

,

s

2 :=

n

−

1 ,

m

(

h

s

2 i

,

h

s

2 i

)

,

f

(

h

n

i

,

h

s

3 i

)

,

init

(

h

s

0 ,

s

1 ,

s

2 ,

s

3 i

,

hi

)

,

r

=

s

0 .

m

(

h

n

i

,

h

r

i

)::=

n

≤

0 ,

r

:= null

.

a

:=

0 ,

i

:=

n

,

f

c

(

h

n,

a

i

,

h

n,

a

i

)

,

f

d

(

h

i

,

a

i

,

h

i

,

a

i

)

,

r

:=

a

.

f

c(

h

n,

a

i

,

h

n,

a

i

)::=

n

>

1 ,

g

(

h

n

i

,

h

s

0 i

)

,

intValue

1 (

h

s

0 i

,

h

s

0 i

)

a

:=

a

+

s

0 ,

n

:=

n/2,

f

c

(

h

n,

a

i

,

h

n,

a

i

)

.

(19)

From Intermediate representation to cost equations

m

(

h

n

i

,

h

r

i

)::=

n

>

0 ,

s

0 := new

Tree

1 ;

s

1 :=

n

−

1 ,

m

(

h

s

1 i

,

h

s

1 i

)

,

s

2 :=

n

−

1 ,

m

(

h

s

2 i

,

h

s

2 i

)

,

f

(

h

n

i

,

h

s

3 i

)

,

init

(

h

s

0 ,

s

1 ,

s

2 ,

s

3 i

,

hi

)

,

r

=

s

0 .

m

(

h

n

i

,

h

r

i

)::=

n

≤

0 ,

r

:= null

.

m

(

n

)=

size

(

Tree

1 )+

{n

>

0 }

m

(

n

−

1 )+

m

(

n

−

1 )+

f

(

n

)+

init

(

1 ,

n

−

1 ,

n

−

1 ,

s

3 )

(20)

m

(

h

n

i

,

h

r

i

)

::=

n

>

0 ,

s

0 := new

Tree

1 ;

s

1 :=

n

−

1 ,

m

(

h

s

1 i

,

h

s

1 i

)

,

s

2 :=

n

−

1 ,

m

(

h

s

2 i

,

h

s

2 i

)

,

f

(

h

n

i

,

h

s

3 i

)

,

init

(

h

s

0 ,

s

1 ,

s

2 ,

s

3 i

,

hi

)

,

r

=

s

0 .

m

(

h

n

i

,

h

r

i

)

::=

n

≤

0 ,

r

:= null

.

m

(

n

)

=

size

(

Tree

1 )+

{n

>

0 }

m

(

n

−

1 )+

m

(

n

−

1 )+

f

(

n

)+

init

(

1 ,

n

−

1 ,

n

−

1 ,

s

3 )

(21)

From Intermediate representation to cost equations

m

(

h

n

i

,

h

r

i

)::=

n

>

0 ,

s

0 := new

Tree

1 ;

s

1 :=

n

−

1 ,

m

(

h

s

1 i

,

h

s

1 i

)

,

s

2 :=

n

−

1 ,

m

(

h

s

2 i

,

h

s

2 i

)

,

f

(

h

n

i

,

h

s

3 i

)

,

init

(

h

s

0 ,

s

1 ,

s

2 ,

s

3 i

,

hi

)

,

r

=

s

0 .

m

(

h

n

i

,

h

r

i

)::=

n

≤

0 ,

r

:= null

.

m

(

n

)=

size

(

Tree

1 )+

{

n

>

0 }

m

(

n

−

1 )+

m

(

n

−

1 )+

f

(

n

)+

init

(

1 ,

n

−

1 ,

n

−

1 ,

s

3 )

(22)

m

(

h

n

i

,

h

r

i

)::=

n

>

0 ,

s

0 :=

new

Tree

1 ;

s

1 :=

n

−

1 ,

m

(

h

s

1 i

,

h

s

1 i

)

,

s

2 :=

n

−

1 ,

m

(

h

s

2 i

,

h

s

2 i

)

,

f

(

h

n

i

,

h

s

3 i

)

,

init

(

h

s

0 ,

s

1 ,

s

2 ,

s

3 i

,

hi

)

,

r

=

s

0 .

m

(

h

n

i

,

h

r

i

)::=

n

≤

0 ,

r

:= null

.

m

(

n

)=

size

(

Tree

1 )

+

{n

>

0 }

m

(

n

−

1 )+

m

(

n

−

1 )+

f

(

n

)+

init

(

1 ,

n

−

1 ,

n

−

1 ,

s

3 )

(23)

From Intermediate representation to cost equations

m

(

h

n

i

,

h

r

i

)::=

n

>

0 ,

s

0 := new

Tree

1 ;

s

1 :=

n

−

1 ,

m

(

h

s

1 i

,

h

s

1 i

)

,

s

2 :=

n

−

1 ,

m

(

h

s

2 i

,

h

s

2 i

)

,

f

(

h

n

i

,

h

s

3 i

)

,

init

(

h

s

0 ,

s

1 ,

s

2 ,

s

3 i

,

hi

)

,

r

=

s

0 .

m

(

h

n

i

,

h

r

i

)::=

n

≤

0 ,

r

:= null

.

m

(

n

)=

size

(

Tree

1 )+

{n

>

0 }

m

(

n

−

1 )

+

m

(

n

−

1 )

+

f

(

n

)

+

init

(

1,

n

−

1,

n

−

1,

s

3 )

(24)

m

(

h

n

i

,

h

r

i

)::=

n

>

0 ,

s

0 := new

Tree

1 ;

s

1 :=

n

−

1 ,

m

(

h

s

1 i

,

h

s

1 i

)

,

s

2 :=

n

−

1 ,

m

(

h

s

2 i

,

h

s

2 i

)

,

f

(

h

n

i

,

h

s

3 i

)

,

init

(

h

s

0 ,

s

1 ,

s

2 ,

s

3 i

,

hi

)

,

r

=

s

0 .

m

(

h

n

i

,

h

r

i

)::=

n

≤

0 ,

r

:= null

.

m

(

n

)=

size

(

Tree

1 )

+

{n

>

0 }

m

(

n

−

1 )+

m

(

n

−

1 )+

f

(

n

)+

init

(

1 ,

n

−

1 ,

n

−

1 ,

s

3 )

m

(

n

)=

0 {n≤

0 }

f

(

n

)=

f

c

(

n

,

0 ) +

f

d

(

n

,

a

′

)

{}

f

c

(

n

,

a

)=

g

(

n

)+

f

c

(

n

/

2 ,

a

′

)

{n

>

1 }

f

c

(

n

,

a

)=

0 {n≤

1 }

f

d

(

i

,

a

)=

h

(

i

)+

f

d

(

i

/

2 ,

a

′

)

{i

>

1 }

f

d

(

i

,

a

)=

0 {i≤

0 }

g

(

n

)=

size

(

Integer

2 )

+

size

(

Integer

3 )

{}

h

(

n

)=

size

(

Long

4 )

{}

(25)

From Cost Relations to Upper Bounds

solutions computed for f

total

(

f

) =

total

(

f

c

) +

total

(

f

d

)

for

(; n

>

1 ; n=n/2 ) a += g(n).intValue();

for

(; i

>

1 ; i=i/2 ) a *= h(i).intValue();

f

c

(

n

) =

log

(

n

)

∗

(

size

(

Integer

3 ) +

size

(

Integer

2 ))

(26)

solutions computed for f

total

(

f

) =

total

(

f

c

) +

total

(

f

d

)

for

(; n

>

1 ; n=n/2 ) a += g(n).intValue();

for

(; i

>

1 ; i=i/2 ) a *= h(i).intValue();

f

c

(

n

) =

log

(

n

)

∗

(

size

(

Integer

3 ) +

size

(

Integer

2 ))

f

d

(

i

) =

log

(

i

)

∗

size

(

Long

4 )

solutions computed for m

m

(

n

) = 2

n

_∗

₍

_total

₍

_f

₎

|

{z

}

exp times

+

size

(

Tree

1 ))

(27)

Why Live Heap Space Analysis is Different?

Basic idea in total allocation:

(28)

Basic idea in total allocation:

total

(

{

m

1 ;

m

2 }

) =

total

(

m

1 ) +

total

(

m

2 )

(29)

Why Live Heap Space Analysis is Different?

Basic idea in total allocation:

total

(

{

m

1 ;

m

2 }

) =

total

(

m

1 ) +

total

(

m

2 )

While total memory allocation is an

accumulative

resource, the live heap

space

increases and decreases

along an execution

Basic idea in live Heap Space Analysis:

(30)

Basic idea in total allocation:

total

(

{

m

1 ;

m

2 }

) =

total

(

m

1 ) +

total

(

m

2 )

While total memory allocation is an

accumulative

resource, the live heap

space

increases and decreases

along an execution

Basic idea in live Heap Space Analysis:

peak

(

{

m

1 ;

m

2 }

) =

max

(

peak

(

m

1 )

,

escaped

(

m

1 )

+

peak

(

m

2 ))

1

STEP 1:

Escaped memory analysis

(31)

Example

class

Test

{

staticTree m(intn){

if( n>0 )return newTree(m(n-1),m(n-1),f(n)); elsereturn null;

}

static intf(intn){

inta=0,i=n;

•for(;n>1;n=n/2)

•a += g(n).intValue();

•for(; i>1; i=i/2)

•a *= h(i).intValue();

•returna;

}

Integer x=newInteger(n); return newInteger(x.intValue()+1);

(32)

staticTree m(intn){

}

static intf(intn){

inta=0,i=n;

•for(;n>1;n=n/2)

•a +=g(n).intValue();

•for(; i>1; i=i/2)

•returna;

}

return newLong(n-1);

1

(33)

Example

class

Test

{

staticTree m(intn){

}

static intf(intn){

inta=0,i=n;

•for(;n>1;n=n/2)

•for(; i>1; i=i/2)

•returna;

}

1 2

(34)

staticTree m(intn){

}

static intf(intn){

inta=0,i=n;

•for(;n>1;n=n/2)

•for(; i>1; i=i/2)

•returna;

}

1 2

(35)

Example

class

Test

{

staticTree m(intn){

}

static intf(intn){

inta=0,i=n;

•for(;n>1;n=n/2)

•for(; i>1; i=i/2)

•returna;

}

1 2 3

(36)

staticTree m(intn){

}

static intf(intn){

inta=0,i=n;

•for(;n>1;n=n/2)

•for(; i>1; i=i/2)

•returna;

}

1 2 3

(37)

Example

class

Test

{

staticTree m(intn){

}

static intf(intn){

inta=0,i=n;

•for(;n>1;n=n/2)

•for(; i>1; i=i/2)

•returna;

}

1 2 3

_log(n

)

(38)

staticTree m(intn){

}

static intf(intn){

inta=0,i=n;

•for(;n>1;n=n/2)

•for(; i>1; i=i/2)

•returna;

}

1 2 3

_log(n

)

(39)

Example

class

Test

{

staticTree m(intn){

}

static intf(intn){

inta=0,i=n;

•for(;n>1;n=n/2)

•for(; i>1; i=i/2)

•returna;

}

1 2 3

_log(n

)

1

(40)

staticTree m(intn){

}

static intf(intn){

inta=0,i=n;

•for(;n>1;n=n/2)

•for(; i>1; i=i/2)

•a *=h(i).intValue();

•returna;

}

1 2 3

_log(n

)

1

(41)

Example

class

Test

{

staticTree m(intn){

}

static intf(intn){

inta=0,i=n;

•for(;n>1;n=n/2)

•for(; i>1; i=i/2)

•returna;

}

1 2 3

_log(n

)

1 2

(42)

staticTree m(intn){

}

static intf(intn){

inta=0,i=n;

•for(;n>1;n=n/2)

•for(; i>1; i=i/2)

•returna;

}

1 2 3

_log(n

)

1 2

(43)

Example

class

Test

{

staticTree m(intn){

}

static intf(intn){

inta=0,i=n;

•for(;n>1;n=n/2)

•for(; i>1; i=i/2)

•returna;

}

1 2 3

_log(n

)

1 2 3

(44)

staticTree m(intn){

}

static intf(intn){

inta=0,i=n;

•for(;n>1;n=n/2)

•for(; i>1; i=i/2)

•returna;

}

1 2 3

_log(n

)

1 2 3

(45)

Example

class

Test

{

staticTree m(intn){

}

static intf(intn){

inta=0,i=n;

•for(;n>1;n=n/2)

•for(; i>1; i=i/2)

•returna;

}

1 2 3

_log(n

)

1 2 3

_log(n

)

(46)

staticTree m(intn){

}

static intf(intn){

inta=0,i=n;

•for(;n>1;n=n/2)

•for(; i>1; i=i/2)

•returna;

}

1 2 3

_log(n

)

1 2 3

_log(n

)

(47)

Example

class

Test

{

staticTree m(intn){

}

static intf(intn){

inta=0,i=n;

•for(;n>1;n=n/2)

•for(; i>1; i=i/2)

•returna;

}

1 2 3

_log(n

)

1 2 3

_log(n

)

(48)

Infer upper bounds on the total memory consumption

Remove from the bounds the

collectable

objects

(49)

Step 1: Inference of Escaped Memory Bounds

Infer upper bounds on the total memory consumption

Remove from the bounds the

collectable

objects

The set of collectable objects can be approximated from the

information computed by

escape

analysis

collectable objects from

_m

(50)

• static

Tree m(

int

n)

{

if

( n

>

0 )

return new

Tree(m(n-1),m(n-1),f(n));

else

return null

;

}

• static int

f(

int

n)

{

int

a=0,i=n;

• for

(; n

>

1;n=n/2 )

a += g(n).intValue();

• for

(; i

>

1; i=i/2)

a *= h(i).intValue();

return

a;

}

• static

Integer g(int n)

{

Integer x=

new

Integer

2

_(n);

return new

Integer

3

_{(x.intValue()+1);}

}

• static

Long h(int n)

{

return new

Long

4

_(n-1);

}

// end of class Test

c ollectable

_{(g) =}

_{Integer

2

}

c ollectable

_{(h) =}

_∅

c ollectable

_(f

_{) =}

_{Integer

2

,

Integer

3

,

Long

(51)

Collectable Objects

class

Test

{

• static

Tree m(

int

n)

{

if

( n

>

0 )

return new

Tree(m(n-1),m(n-1),f(n));

else

return null

;

}

• static int

f(

int

n)

{

int

a=0,i=n;

• for

(; n

>

1;n=n/2 )

a += g(n).intValue();

• for

(; i

>

1; i=i/2)

a *= h(i).intValue();

return

a;

}

• static

Integer g(int n)

{

Integer x=

new

Integer

2

_(n)

_;

return

new

Integer

3

_{(x.intValue()+1)}

_;

}

• static

Long h(int n)

{

return new

Long

4

_(n-1);

}

// end of class Test

c ollectable

₍

_g

_{) =}

_{

_Integer

2

(52)

• static

Tree m(

int

n)

{

if

( n

>

0 )

return new

Tree(m(n-1),m(n-1),f(n));

else

return null

;

}

• static int

f(

int

n)

{

int

a=0,i=n;

• for

(; n

>

1;n=n/2 )

a += g(n).intValue();

• for

(; i

>

1; i=i/2)

a *= h(i).intValue();

return

a;

}

• static

Integer g(int n)

{

Integer x=

new

Integer

2

_(n);

return new

Integer

3

_{(x.intValue()+1);}

}

• static

Long h(int n)

{

return

new

Long

4

_(n-1)

_;

}

// end of class Test

c ollectable

_{(g) =}

_{Integer

2

}

c ollectable

₍

_h

_{) =}

_∅

c ollectable

_(f

_{) =}

_{Integer

2

,

Integer

3

,

Long

(53)

Collectable Objects

class

Test

{

• static

Tree m(

int

n)

{

if

( n

>

0 )

return new

Tree(m(n-1),m(n-1),f(n));

else

return null

;

}

• static int

f(

int

n)

{

int

a=0,i=n;

• for

(; n

>

1;n=n/2 )

a += g(n).intValue();

• for

(; i

>

1; i=i/2)

a *= h(i).intValue();

return

a;

}

• static

Integer g(int n)

{

Integer x=

new

Integer

2

_(n)

_;

return

new

Integer

3

_{(x.intValue()+1)}

_;

}

• static

Long h(int n)

{

return

new

Long

4

_(n-1)

_;

}

// end of class Test

(54)

• static

Tree m(

int

n)

{

if

( n

>

0 )

return

new

Tree(m(n-1),m(n-1),f(n));

else

return null

;

}

• static int

f(

int

n)

{

int

a=0,i=n;

• for

(; n

>

1;n=n/2 )

a += g(n).intValue();

• for

(; i

>

1; i=i/2)

a *= h(i).intValue();

return

a;

}

• static

Integer g(int n)

{

Integer x=

new

Integer

2

_(n)

_;

return

new

Integer

3

_{(x.intValue()+1)}

_;

}

• static

Long h(int n)

{

return

new

Long

4

_(n-1)

_;

}

// end of class Test

c ollectable

_{(g) =}

_{Integer

2

}

c ollectable

_{(h) =}

_∅

c ollectable

_(f

_{) =}

_{Integer

2

,

Integer

3

,

Long

(55)

Bounds on Escaped Memory

escaped memory of a procedure

p

given the set of collectable objects after return from p,

collectable

(

p

),

given the upper-bound for the total memory allocation

total

(

p

) =

exp

the escaped memory upper-bound:

(56)

given the set of collectable objects after return from p,

collectable

(

p

),

given the upper-bound for the total memory allocation

total

(

p

) =

exp

the escaped memory upper-bound:

escaped

(

p

) =

exp

[

∀

c

i

∈

collectable

(

p

)

.

size

(

c

i

)

7→

0].

g

(

n

) =

size

(

Integer

2 )+

size

(

Integer

3 )

h

(

n

) =

size

(

Long

4 )

f

(

n

) = log

₂

(

n

)

∗

(

size

(

Integer

2 )+

size

(

Integer

3 )+

size

(

Long

4 ))

m

(

n

) =(2

n

₎

_∗

₍

_size

₍

_Tree

1 )+

log

₂

(

n

)

∗

(

size

(

Integer

2 )+

size

(

Integer

3 )+

size

(

Long

4 )))

c ollectable

_{(g) =}

_{Integer

2

_}

c ollectable

_{(h) =}

_∅

c ollectable

_(f

_{) =}

_{Integer

2

,

Integer

3

,

Long

(57)

Bounds on Escaped Memory

escaped memory of a procedure

p

given the set of collectable objects after return from p,

collectable

(

p

),

given the upper-bound for the total memory allocation

total

(

p

) =

exp

the escaped memory upper-bound:

escaped

(

p

) =

exp

[

∀

c

i

∈

collectable

(

p

)

.

size

(

c

i

)

7→

0].

g

(

n

) =

size

(

Integer

2 )

+

size

(

Integer

3 )

h

(

n

) =

size

(

Long

4 )

f

(

n

) = log

₂

(

n

)

∗

(

size

(

Integer

2 )+

size

(

Integer

3 )+

size

(

Long

4 ))

m

(

n

) =(2

n

₎

_∗

₍

_size

₍

_Tree

1 )+

log

₂

(

n

)

∗

(

size

(

Integer

2 )+

size

(

Integer

3 )+

size

(

Long

4 )))

c ollectable

₍

_g

_{) =}

_{

_Integer

2

_}

(58)

given the set of collectable objects after return from p,

collectable

(

p

),

given the upper-bound for the total memory allocation

total

(

p

) =

exp

the escaped memory upper-bound:

escaped

(

p

) =

exp

[

∀

c

i

∈

collectable

(

p

)

.

size

(

c

i

)

7→

0].

ˇ

g

(

n

) =

size

(

Integer

3 )

h

(

n

) =

size

(

Long

4 )

f

(

n

) = log

₂

(

n

)

∗

(

size

(

Integer

2 )+

size

(

Integer

3 )+

size

(

Long

4 ))

m

(

n

) =(2

n

₎

_∗

₍

_size

₍

_Tree

1 )+

log

₂

(

n

)

∗

(

size

(

Integer

2 )+

size

(

Integer

3 )+

size

(

Long

4 )))

c ollectable

_{(g) =}

_{Integer

2

_}

c ollectable

₍

_h

_{) =}

_∅

c ollectable

_(f

_{) =}

_{Integer

2

,

Integer

3

,

Long

(59)

Bounds on Escaped Memory

escaped memory of a procedure

p

given the set of collectable objects after return from p,

collectable

(

p

),

given the upper-bound for the total memory allocation

total

(

p

) =

exp

the escaped memory upper-bound:

escaped

(

p

) =

exp

[

∀

c

i

∈

collectable

(

p

)

.

size

(

c

i

)

7→

0].

ˇ

g

(

n

) =

size

(

Integer

3 )

ˇ

h

(

n

) =

size

(

Long

4 )

f

(

n

) = log

₂

(

n

)

∗

(

size

(

Integer

2 )

+

size

(

Integer

3 )

+

size

(

Long

4 )

)

m

(

n

) =(2

n

₎

_∗

₍

_size

₍

_Tree

1 )+

log

₂

(

n

)

∗

(

size

(

Integer

2 )+

size

(

Integer

3 )+

size

(

Long

4 )))

(60)

given the set of collectable objects after return from p,

collectable

(

p

),

given the upper-bound for the total memory allocation

total

(

p

) =

exp

the escaped memory upper-bound:

escaped

(

p

) =

exp

[

∀

c

i

∈

collectable

(

p

)

.

size

(

c

i

)

7→

0].

ˇ

g

(

n

) =

size

(

Integer

3 )

ˇ

h

(

n

) =

size

(

Long

4 )

ˇ

f

(

n

) =0

m

(

n

) =(2

n

)

∗

(

size

(

Tree

1 )+

log

₂

(

n

)

∗

(

size

(

Integer

2 )

+

size

(

Integer

3 )

+

size

(

Long

4 )

))

c ollectable

_{(g) =}

_{Integer

2

_}

c ollectable

_{(h) =}

_∅

c ollectable

_(f

_{) =}

_{Integer

2

,

Integer

3

,

Long

(61)

Bounds on Escaped Memory

escaped memory of a procedure

p

given the set of collectable objects after return from p,

collectable

(

p

),

given the upper-bound for the total memory allocation

total

(

p

) =

exp

the escaped memory upper-bound:

escaped

(

p

) =

exp

[

∀

c

i

∈

collectable

(

p

)

.

size

(

c

i

)

7→

0].

ˇ

g

(

n

) =

size

(

Integer

3 )

ˇ

h

(

n

) =

size

(

Long

4 )

ˇ

f

(

n

) =0

ˇ

m

(

n

) =(2

n

)

∗

size

(

Tree

1 )

(62)

Build cost relations which capture the basic idea

(63)

Step 2: Inference of Live Memory Bounds

Build cost relations which capture the basic idea

peak

(

{

m

1 ;

m

2 }

) =

max

(

peak

(

m

1 )

,

escaped

(

m

1 )

+

peak

(

m

2 ))

Consider a rule

p

::=

g,

b1, . . . ,

bn

. Its

peak consumption

equation is

(64)

Build cost relations which capture the basic idea

peak

(

{

m

1 ;

m

2 }

) =

max

(

peak

(

m

1 )

,

escaped

(

m

1 )

+

peak

(

m

2 ))

Consider a rule

p

::=

g,

b1, . . . ,

bn

. Its

peak consumption

equation is

peak

(

p

) =

T

(

b

1 , . . . ,

b

n

)

, ϕ

r

where

T

is defined as follows:

peak cost relation

T

(

b

1 , . . . ,

b

n

)::=

◮

if

b

1 is a call, then

max

(

peak

(

b

1 )

,

escaped

(

b

1 )

+

T

(

b

2 , . . . ,

b

n

))

◮

if

b

(65)

Detailed Example

For the following code in m:

(66)

For the following code in m:

new Tree(

m(n-1)

,

m(n-1)

,

f(n)

);

We produce a peak consumption relation:

(67)

Detailed Example

For the following code in m:

new Tree(

m(n-1)

,

m(n-1)

,

f(n)

);

We produce a peak consumption relation:

peak

m

(

n

) =

size

(

Tree

1 )+

max

(

peak

m

(

n

−

1)

,

escaped

m

(

n

−

1)+