Softwaretesting 06 eng

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　　 [6] ホワイトボックステス
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Software Testing
[6] White Box Testing Techniques
あまん ひろひさ ひろひさ

阿萬 裕久 裕久（ AMAN

Hirohisa ）

(C) 2007-2022 Hirohisa AMAN

1


Classification of tests
 black box testing

Explained in Part 4, Exercise in
Session 5

The contents of the program are not seen
(black box), and operation tests are perfor
med based on specifications

 white box testing
Perform operation tests based on the intern
al structure (mainly flowchart)

 random testing

Create test cases randomly and test them
(C) 2007-2022 Hirohisa AMAN

2


white box test method
 How to design test cases by focusing on
the structure of the source program rat
her than the specification
 Test not in line with required specifications
 Dealing with structural complexity (combin
ation of conditions, etc.) in the source prog
ram
 Tests that supplement black box testing

(C) 2007-2022 Hirohisa AMAN

3


White box test method (1)
Statement coverage method
 Execute all statements at least once
 It may not be possible with only one test c
ase (execution path)
 With different test cases, and consider thei
r collection of test cases, It is possible to c
over them all
 The percentage of instructions that could b

e executed is called the instruction covera
ge rate.
 This is also known as C0
(C) 2007-2022 Hirohisa AMAN

4


Example to test
void foo(int x, int y){
int sum, n;
sum = 0;
for ( n = 0; n < x; n++ ){
sum += n;
}
if ( sum < y ){
printf("%dn", sum);
}
else{
printf("%dn", y);
}
}

[input]
Arguments x, y
[output]
Values printed by pr
intf

(C) 2007-2022 Hirohisa AMAN

5


Instruction coverage example: x=
1,y=0
Source code

Execution?

void foo(int x, int y){

---

int sum, n;

○

sum = 0;

○

for ( n = 0; n < x; n+
+ ){

Instruction
Coverage
(C0)

○

○

sum += n;
}

--○

if ( sum < y ){
printf("%dn", sum);
}
}
else{
else{
printf("%dn", y);
}
}

}
}

☓

← sum = 0 (that‘s why)

------○
---

[Note]
It is necessary to
define in advance

which instructions are
counted as executable.

------(C) 2007-2022 Hirohisa--AMAN

6


Example of 100% instruction cove 100% instruction cove instruction cove
rage
Source code

x=1, y=0

x=1, y=1

total

---

---

---

int sum, n;

○

○

○

sum = 0;

○

○

○

○

○

○

○

○

○

--○

--○

--○

☓

○

○

--------○

--------☓

--------○

--------- 7

---------

void foo(int x, int y){

for ( n = 0; n < x; n+
+ ){
sum += n;
}
if ( sum < y ){
printf("%dn", sum);
}
}
else{
else{
printf("%dn", y);
}
}

}
}

--------(C) 2007-2022 Hirohisa
AMAN


When drawing a flowchart
Test case ① x=1, y=0
START
sum = 0
F.
sum < y

n=0

nn++

F.

If 100% instruction cove some
instructions are
Not running

T.
Sum output

Y output

T.
sum += n
END
(C) 2007-2022 Hirohisa AMAN

8


When drawing a flowchart
Test case ② x=1, y=1
START
sum = 0
F.
sum < y

n=0

nn++

F.

In test case ①
did not run
cover instruction

T.
Sum output

Y output

T.
sum += n
END
(C) 2007-2022 Hirohisa AMAN

9


White box test method (2)
Branch coverage method
 If all conditional branches (if statement,
while statement, for statement), execut
e at least once when it becomes True a
nd when it becomes False
 Again, It would be nice to be able to cover
set of multiple test cases
 Percentage who experienced “T” and “F” of
all conditions is called branch coverage rate
 This is also known as C1
(C) 2007-2022 Hirohisa AMAN

10


Branch coverage example: x=1,
y=0
Source code

Exec?

T, F

---

---

int sum, n;

○

---

sum = 0;

○

---

void foo(int x, int y){

for ( n = 0; n < x; n+
+ ){

○

F

○

sum += n;

---

}

--○

if ( sum < y ){
printf("%dn", sum);

☓

--F
---

}

---

---

else{

--○

---

printf("%dn", y);
}

}

T

-----

---

---

---

(C) 2007-2022 Hirohisa AMAN

Branch
coverage rate
(C1)


Example of 100% instruction cove 100% instruction cove branch coverag
e rate
Source code

x=1, y=0

x=1, y=1

total

---

---

---

int sum, n;

---

---

---

sum = 0;

---

---

---

void foo(int x, int y){

for ( n = 0; n < x; n+
+ ){
sum += n;

T.

F.

---

}

--- F.

T.

F.
---

T.

F.
---

T. ---

T. --- F.

if ( sum < y ){
printf("%dn", sum);

---

---

---

}

---

---

---

else{

---

---

---

---

---

---

---

---

---

---

---

printf("%dn", y);
}
}

(C) 2007-2022 Hirohisa AMAN

12

---


When drawing a flowchart
Test case ① x=1, y=0
T and F of 100% instruction cove
both
experienc
e

START
F only
sum = 0
F.
sum < y

n=0

nn++

F.

T.
Sum output

Y output

Theref 100% instruction coveore,
Branch
coverage
rate
(C1) = 3/4

T.
sum += n
END
(C) 2007-2022 Hirohisa AMAN

13


When drawing a flowchart
Test case ② x = 1, y = 1
In test case ①
did not run
"T." side cover

START
sum = 0
F.

sum < y

n=0

nn++

F.

T.
Sum output

Y output

T.
sum += n
END
(C) 2007-2022 Hirohisa AMAN

14


Consider another example
 Consider a program that performs si
mple security checks on passwords
[Check items]
 Is password length more than 6 characte
rs?
 Is your password the same as your usern
ame?

(C) 2007-2022 Hirohisa AMAN

15


Example of 100% instruction cove 75% instruction cove branch coverage
test cases

START

pwd length
>= 6

①
③

F.

F.

No output

②

T.

Password
!= Username

②

(username, password)
① ("taro","")
② ("taro","f 100% instruction coveoobar")
③ ("taro","taro")

T.
OKoutput

END
(C) 2007-2022 Hirohisa AMAN

16


(Note) in the program
“}” does not count as an
instruction

[Exercise 1]
Calculate C0 and C1
 I ran two test ca
ses for the right
code
 (n, limit)
= (2, 1), (-2, 0)

 In this case,
Calculate C0 and
C1

if ( n < 0 ){
n *= -1;
}
sum = 0;
for ( i = 0; i < n; i+
+ ){
sum += i;
if ( sum > limit ){
sum = 0;
}
}
printf("%dn", sum);

(C) 2007-2022 Hirohisa AMAN

17


[Exercise 1] Answer: 100% for
① and ②
test cases

START
i=0
F.

n<0

① ②

i
T.

n *= -1

①

F.

T.

sum += i
①

Sum output
sum = 0

i++

(n, limit)
=
① (2, 1)
② (-2, 0)

F.
sum > limit
T.

②

①

sum = 0
END
(C) 2007-2022 Hirohisa AMAN

18


[Exercise 2]
Calculate branch coverage
int is_prime(int n){
int k;
is_prime(1);
if ( n < 2 ){
return 0;
is_prime(2);
}
for ( k = 2; k < n; k+
test and calculate t+ ){
he branch coverage
if ( n % k == 0 ){
(C1)
return 0;
}
　
}
(*Drawing a flowchart

return 1;
Not required)
}
(C) 2007-2022 Hirohisa AMAN
19

 Right function


Note for exercise 2
Source code

n=1
実行 　 条件
条件

int k;

n=1, n=2 for each
While checking whether or
not it is conditional executed

n=2
実行 　 条件
条件

total

---

---

---

---

---

---

---

---

---

---

---

---

---

---

---

---

---

---

---

---

---

if ( n < 2 ){
return 0;
}
for ( k = 2; k < n; k+
+ ){
if ( n % k == 0 ){
return 0;
}
}
return 1;

(C) 2007-2022 Hirohisa AMAN

20