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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.
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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
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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
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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.
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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.
8/16/2019 St Notes(Unit2&3)
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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:
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8/16/2019 St Notes(Unit2&3)
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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
8/16/2019 St Notes(Unit2&3)
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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)
8/16/2019 St Notes(Unit2&3)
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
8/16/2019 St Notes(Unit2&3)
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
$",
commission)
43.
End
commission.
Program
grap-h of
the
the
commission
problem
8/16/2019 St Notes(Unit2&3)
74/76
8/16/2019 St Notes(Unit2&3)
75/76
(
?.frq)
.
The next six
slices
demonstrate
our convention
regarding
values defined
b]-
assignment
stzuements
(Adefs).
Str:
S0ockPrice,
24)
-
l7l
S,r:
S(stockPrice,
25)
=
[81
Sr:
S(barrelPrice,
26)
-
l9l
S4:
S0ockSales,
24)
=
{7,
lO,
13,
14,
16,
19,
20,241
Srr:
S(stockSales,
25)
=
lB,
11,
13,
14, 15,
17, 19,20,25l.
S23:
S(barrelsales,
26)
=
t9,
12, 13,
74,15,
18, 19, 2A,
261
The
slices
on sales and
commission
"r.
th.
interesting ones.
Only
one
defining
node
exists
for sales,
the Adef at node
27,
The remaining
slices
on
sales
show
the
P-uses,
C-uses,
and
the
O-use
in
definition
clear
paths.
Sra:
S(sales,27)
- 17,8,9
10,
ll, 12,
13,
14, 15,
16,
17,
18,
19,20,21t,25,
25, Z7l
Srr:
s(sales,
28)
=
{7,
8,
g
10,
i.l,
12,
13,
14,
15,
16,
17, 18,
79,
20,24,
25,
26,
271'-
526:
S(sales,29)
=
ii7,8,9
10, 71,
12, 13,
14,75,76,77,18,
19,20,24,25,
26,271
Srr: S(sales,
33)
:
{7,
8,
9
10,
\1,
12,
13,
4,
15, 76,
17,
18, 19,20,
24,
25,
26,
?71
S2r:
S(sales,34):17,8,910,
11,'1.2,
13,
1,4,15, 16.
17, 18,
79,20,24,25,
26,27l|
Srr:
S(sales,
37)
:
'7,8,9
10,
1' .,
12, 13,74,
15,
16,
17,
18,
79,'20,
24,25,
26,271
S3s:
S(sales,
38)
=
|-7,8,9
1A,
\'),,' .2,
3,
14,
15,
1.6,
77,
18, 19,
20,
24,25,
26,27l.
Think about
slice
S2a
in
terms
of
its
"components,"
which
are
the
slices
on
the
C-use
variables.
\We
can
write
_Szl
i
Sro
(J
9,1
u
Sro
u
Sz, U S,
U
Srr.
Notice
how
the
fiormalism
corresponds
to
our
inrui$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
8/16/2019 St Notes(Unit2&3)
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,