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63/76

DATA

FLO\il

TESTING

Data

flow

testing

focuses

on the

points

at

which

variables

receive

values

and the

points at

which

these

values are used

(or

referenced),

It

detects

improper

use

of

data values

(data

flow

anomalies)

due

to coding

errors.

Rapps

and Weyukers Motivation*:

"

it

is

our

belief

that,

iust

as one

would

not

feel

confident

about

a

program

without

executing

every

statement

in it

as

part

of

some test,

onc

should

not

feel

confident

about a

program

without having

seen

the

effect of[Jsingthe

value

produced

by

each

and

every

computation.

Early

data

flow

analyses often centered

on a

set

of

faults

that

are

known as

define/referen

ce

an

om

al

ies,

.

A

variable that is

defined but

never used

(referenced')

.

A

variable that

is

used

but

never

defined

.

A

variable

that

is defined

twice

before it is

used

r

Data flow

testing

1.

Define

/

Use Testing

2. Slice-Based

Testing

DEFINE/USE

TESTING

The

following

refers to

program

P that

has

program

graph

G

(P)

and

the

set of

program

variables V. In

a

program

graph

statement

fragments

are

nodes and

edges

represents

nodc

sequcncc .G

(P)

has

single entry

a:rd

single

exit

node. We also

disallow

ctlgcs

fiuttt

ttudc

to itself.

The

set of all

paths

in

P

is

PATHS

(P).

Dgfinition.of

usage

node and

defining

node

Definlng and

Usage

Nodes

"

Definin$

node

(e.g.

input x,

v

=

2,

etc.):

DEF(v,

n): Node

n

in G(P) is a

defining

node

of

var v

in V

iff value

of

v

is

defined

at

n.

'

Usage

node

(e.g.

output x,

?

=

2\v,

etc.):

USE(v,

n):

Node

n

in

G(P) is

a

usage

node

of

var

v in

V iff

value of

v

is

used at n.


64/76

Definition

of

du-oath:-

Definition-use

(du) path

(wrt.

variable

v)

A

path

in

PATHS(P)

such

that

for

sorne

v

in

V

There

exist

DEF(v,

m), USE{v,

n) nodes

s.t"

m and

n are

initial

and

final

nodes

of

the

path

respectively.

eTrnlfion-ctear

tcfc)

pa

wrt. variable

A

du-path

in PATHS(P)

where

the

initial

node

of

the

path

is

the

only

defining

node

of

v

{in

the

path).

Definition

of

P-use.,

C-use

A

usage

node

USE(v,n)

is

predicate

use

denotsd

as

P-case

,

if

statement

n

is

preclicate

statement

(example

if

a


65/76

1.

Dim

totallocks

,totalStocks

,

totalBarrels

As

Ilteger

5.

Dim lockSales,

stocksales,ba:relsSales

As

Real

6.

Dim

sales

,

commission

As

Real

7.

lockprice=4s.0

8

stockprice= 30.0

9. barrelpriee:25.0

10. totallocks:0

11. totalstocks:0

12. totalbarrels:0

13

.input(locks)

14.Whle

not

(locks:

-l)

I

5.Input

(stock,barrel)

l6.Totallocks

:total

locks

+locks

17.

Totalstocks

:total

stocks

+stocks

18.

Totalbarrels

:

totaibarrels

+barrels

19.

input

Qocks)

20.End

While

21.

output

(

"Locks

sold",

total

locks)

22.

output

("Stocks

sold",

total

stocks)

23.

output

("Barrels

sold", totalbarrels)

24.locksales:

lockprice

*totallocks

25.

stocksales:

siockprice

*totalstocks

26.barrelsales:

barrelprice

*totalbarrels

27 sales=

locksales

*

stocksales+barrelsales

28.

output(

"Tota1sales",

sales)

29.

if

(sales

>

1800.0)

30. then

31. commission:0.i0

*

1000

32.

commission

:commission

+0.

I

5

*800.0

33.commissi.on

-

commission

+0,20

*(

sales

>1000)

34. Else if

(sales

>1000)

/d-:-->.

qlf.r


66/76

35.Then

36.

commission:0.10

*

1000.0

37.

commission

:

commission

+0.

I

5

*(sales_l

000.

0)

38.

Else

39.

commission

=0.10

*

sales

40.

End

If

41.

End

If

42.

ouput

("commission

is

S',,

commission)

43.

End

commission.


67/76

DD-paths

and

Nodes

for

above

example

D

Nodes

A

7,8,9,1q,1

1,72,13,

B

14

c

l5,l6,l7,l

g,19,20

D

42225,2.+2s2627

28

E

29

F

GI

30.31.32.33

34

rii

II

I

35,36,3.1

38.39

t

J

40

K

41,42,42

(3

du-Pathr

for

Stock$j

First, let

us

look

at

a

simple

path:the

du-path

or

the

variable

stocks.Wehac

DEF(stocks,l5)

andUSE(stocks,l7),so

the

path

is

aclu-path

with

respect

to

stooks.

No

othcr

is

defining

nodes

are

used

for

stocks;

therefore,

this

path

also

definition

clear.


68/76

du-Paths

for Locks:

Two

defining

and

two usage

nodes

make the

locks

variable

more

interesting:

we

have

DEF(locks,13),

DEF(locks,19),USE(locks,14),and

USE(locks,16).

These

yreld

four

du-paths:

P1:

P1:

P1:

Pl:

Du-paths

pl

and

p2

refer

to

the

priming

Value

of

locks,

which

is

read

at

node

13:

locks has

a

predicate

use

in the

while

statement(node

14),

and if the

condition

is

true

(as

in

path

p2),

acomputation

use

at

statement

16.The

other

fwo

du-paths

start

near the

end

of the

while

loop

and occur

when

the

loop

repeats.

du-Paths

fo4

lgjallocksi

The du-paths

for totallocks

will

be

lead us

to fypical

test cases

for

computations.

With

two

defining

nodes(DEF(toallocks,

10) and

DEF(totallocks, 16))

and

three usage

nodes

(USE(totallocks,l6),USE(totailocks,2l),

USE(totallocks,24)),

Wc

might

expect

six

du-

paths.

Let

us

take

a

closer

look. Path

p5:

is

a

du-path

in

which

the

initial

value

of

totailocks(0)

has

computation

use.

du-Paths

for

Sales:-

Oniy one defining

node

is

used

for

sales; therefore,

all

the

du-paths

with

respect

to sales

must

be

definition-clear.

They

are

interesting

because they

illustrate

predicate and

computation

uses.

The

First 3

du-paths

are

easy:

PI0:

Pll:

Notice

thatplZ

is

a

definition-clearpath with 3 usage

nodes;

it

also contain

paths

p10

and

pl1.

If

we

were

testing

withp12,

we

know

P

12:


69/76


70/76

comrnission

Yes

Yes

commtsslon

36.42

commission

3rc2

No

mfilmlSSlOn

Yes

n/a

37,37

Yes

Yes

coillmNslon

37,42

Yes

Yes

commission

3gS-

comrtusslon

39,33

No

nla

commission

3W

Yes

Yes

du-path

Test

Coverage

Meticq

All-Du-Paths

All

C-Uses/Some

p-Uses

All

PusesiSome

C-Uses

lo,,-"-ur".

lou-"o**,

SLICE-BASED

TESTING


71/76

e9

DATAFLOW

TESTII\G

Data

flow

testing

focuses

on

the

points

at

which

variables

receive

values

and

the

points

at

which

these

values

are

used

(oi

referenced).

It

detects

improper

use

of

data

values

(data

flow

anomalies)

due

to coding

errors.

Rapps

and

Weyukers

Motivation*:

"

it

is

our

belief

that,

just

as

one

would

not

feel

confident

about

a

program without

executing

every

statement

in

it

as

part of

some

test, one

should

not

feel

confident

about

u

p.ogru-

without

having

seen

the

effect

oflJsingthe

vaiue

produced by

each

and

every

computation.

o

Dataflow

testing

1. Define

/

Use

Testing

2.

Slice-Based

Testing

Slice-Based

Testine

The

following

refers

to

program

P that

has

program

graph

G

(P)

and

the

set

of

program

variables

V.

In

a

program

graph

statement

fragments

are

nodes

and

edges

represents

node

sequence

.G

(p)

has single

entry

and

singie

exit

node.

We

also

disallow

edges

from

node

to itself.

The

set

of

all

paths

in

P

is PATHS

(P)

Definition:-Given

a

program

P and

a

set

V

of variables

in P,

a

slice

"t

rt.t.*."1n,wdtte;

S(V,n),

is the

set

of all

statements

in

P

prior to

to

the

values

of

variables

in

V

at

node

n.Listing

elements

of

a

on

the

variable

set

V

node

n

that

contribute

slice

S(V,

n)

will

be

cumbersome

because the elements

are

program

statement

fragments'

list

fragrnent

numbers

in

P(G).

It

is

much

simpler

to

USE

T\?ES

EFT

P-use

-

Used

in

a

predicate

stmt

I-def-Defined

by

input

A-def-Defi

ned

bY

assignment

C-use

-

Used

in comPutation

O-use

-Used

for oulPut

L-use

- used

for

location

(pointers)

I-use

Iteration

(Internal

counters,

loop

indices)

i

ontains

interesting dataflow

properties

,

and

these

are

not

present

in

the

triangle

problem(

or

in

next

date

function).Follow

these examples

while

looking

at

the

source

code

for

the

commission

problem that

we used

to

analyse

in

terms

of

define/use

paths'

(

Commission

Problem)

i

.

program

commission(INPUT.

OUTPUT)


72/76

2, Dim

lock,

stock,

barrels

as lnteger

3.

Dim

lockprice

,stockprice,

ba:relprice

As

Real

4. Dim

totallocks

,totalStocks

,

totalBarrels

As lnteger

5.

Dim

lockSales,

stocksales

,barrelsSales

As

Real

6.

Dim

sales

,

commission

As Real

7.

lockprice-45.0

8

stockprice=

30.0

9.

barrelprice=25.0

10.

totallocks:0

i

1.

totalstocks=0

12.

totalbarrels:O

13 .input(1ocks)

14.Whle

not

(locks:

-1)

1

5.Input

(stock,barrel)

l6.Totallocks

:total

locks

+locks

17.

Totalstocks

:total

stocks

*stocks

i

8. Totalbarrels

:

totalbarrels

*barrels

19.

input

(locks)

20.End

While

21.

output

(

"Locks

sold",

total

locks)

22.

output

("Stocks

sold", total

stocks)

23"

output ("Barrels

sold",

totalbarrels)

24.locksales:

lockprice

*totallocks

25.stocksales:

stoclqpdce

*totalstocks

26.banel

sal

es:

barrelpriee

*totalbarrel

s

27

sales:

locksales

*

stocksales*barrelsales

28.

output(

"Totalsales",

sales)

29.

if

(sales

>

I

800,0)

30.

then

31.

commission:0.10

*

1000

32.

commissicn

:commission

+0.

1 5

*800.0


73/76

6@

-:3.commission:

commission

+0'20

*(

sales

>1000)

3-1. Else

if

(sales

>1000)

36.

commission:

0.10

*

1000'0

37.

commission

=

commission

+0.15

*(sales-i000.0)

38.

EIse

39.

commission:0.10

*

sales

40.

End

If

41.

End

If

42.

output

("commission

is

$",

commission)

43.

End

commission.

Program

grap-h of

the

the

commission

problem


74/76


75/76

(

?.frq)

.

The next six

slices

demonstrate

our convention

regarding

values defined

b]-

assignment

stzuements

(Adefs).

Str:

S0ockPrice,

24)

-

l7l

S,r:

S(stockPrice,

25)

=

[81

Sr:

S(barrelPrice,

26)

-

l9l

S4:

S0ockSales,

24)

=

{7,

lO,

13,

14,

16,

19,

20,241

Srr:

S(stockSales,

25)

=

lB,

11,

13,

14, 15,

17, 19,20,25l.

S23:

S(barrelsales,

26)

=

t9,

12, 13,

74,15,

18, 19, 2A,

261

The

slices

on sales and

commission

"r.

th.

interesting ones.

Only

one

defining

node

exists

for sales,

the Adef at node

27,

The remaining

slices

on

sales

show

the

P-uses,

C-uses,

and

the

O-use

in

definition

clear

paths.

Sra:

S(sales,27)

- 17,8,9

10,

ll, 12,

13,

14, 15,

16,

17,

18,

19,20,21t,25,

25, Z7l

Srr:

s(sales,

28)

=

{7,

8,

g

10,

i.l,

12,

13,

14,

15,

16,

17, 18,

79,

20,24,

25,

26,

271'-

526:

S(sales,29)

=

ii7,8,9

10, 71,

12, 13,

14,75,76,77,18,

19,20,24,25,

26,271

Srr: S(sales,

33)

:

{7,

8,

9

10,

\1,

12,

13,

4,

15, 76,

17,

18, 19,20,

24,

25,

26,

?71

S2r:

S(sales,34):17,8,910,

11,'1.2,

13,

1,4,15, 16.

17, 18,

79,20,24,25,

26,27l|

Srr:

S(sales,

37)

:

'7,8,9

10,

1' .,

12, 13,74,

15,

16,

17,

18,

79,'20,

24,25,

26,271

S3s:

S(sales,

38)

=

|-7,8,9

1A,

\'),,' .2,

3,

14,

15,

1.6,

77,

18, 19,

20,

24,25,

26,27l.

Think about

slice

S2a

in

terms

of

its

"components,"

which

are

the

slices

on

the

C-use

variables.

\We

can

write

_Szl

i

Sro

(J

9,1

u

Sro

u

Sz, U S,

U

Srr.

Notice

how

the

fiormalism

corresponds

to

our

inrui$on:

if the

vaiue of

sales

is

wrong,

we

first

look at

how

it

is computed;

if this

is

OK,

we

check

how the

components

are

computed.

Everything comes

together

(literally)

with

the

slices

on

commissioq.

Six

A-def

nodes are used for

commission

(corresponding

to the

six

du-paths

we

identified

earlier).

Three computations

of

commission are

controlled by

P-uses

of

sales in

the IF, EISE

IF

logic.

This

yields three

'paths"

of

slices

that

compute

conurlission.

Su,:

S(commission,

31)

= {31}

Srr:

S(commission,

32)

=

131,32J


76/76

a

Srr:

S(commission,

411

=

.'7,

g,910,

11,

12,

13,

14,

j,

16,

17,7g,

79,

zo,

4,

25,

26,

27,2g,

30,

31,

32,

33,

34, 35"

i6,

37,

3gl

The

slice

information

irnproves

our

rrxighc.

Look

ar

the

iattice

in

Figure

10.4;

t

is

a

direcred

aryclic

sraph

in

which

slicJs

"..

,ofr;';fi;

JrJ';"o*senrs

he

proper

subset

retationsf,ip.

This

lanice

is

dr*awn,so

rhat

the

position

of

rhe

srice

nodes

roughly

corresponds

ith

their

position

in

rhe

source

.fa.,

The

definiJor-,

.t.",

parhs

,

'

and

correspond

to

the

edges

that

show

srices

sle,

sr,

and

Sro

are

ubsets

of

slice

soo'

Figure

10.5

shows

a

lattice

of

slices

for

rhe

entire

progranl.

ome

slices

(those

thar

are

idenrical

to

otheJ

fr"rr"-t"r.,

delered

for

clariry.

vhichever

coq-npqq4tioa-isjaken*alr

come

rogerher_in_the-.last_sliee,

Figure

10,4

lattice

of

slices

on commission.

Srr:

S(comrnission,

33) =

-

24,25,26,27,29,30,

Sro:

S(commission,

36)

-

Srr:

S(commission,

3;7)

-

24,

25,26,27,35,

371

Srr:

S(commission,

3g)

=

24,

25,

26,27,2g,

34,

Y:8:910,

11,

12,

13,

74,

.t5,

16,

17',

i,B,

tg,

Z0,

31, 32,33l,

136l

17,

8,9

10,

11,

lz,

13,

{7,

8,

g

r0,

l"t,

j.z,

j,

381

74,

1.5,

16,

17,

18,

"Ig,

2A;

14,75,

16,

17,

18,

7g,

2A,

st notes(unit2&3)

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