## Tutorial 8

MSc Computing and Security

1. Consider the deceptively simple program below. Consider the three initial values of a, b and c to be inputs to the program. So the search space for test data generation is a vector of three values <a, b, c>. In this exercise, we are concerned with branch and target coverage. There are four targets: 1, 2, 3, 4 in the program for which we wish to generate test data.
int a, b, c

if (a>b)
c = 1 /* target 1 */ ;
else {
c = 2;

if (b==c)
/* target 2 */ ;
}

if (c==3)
do something; /* target 3 */ ;
if (b==42)
do something else; /* target 4*/ ;


a). Consider the following 3 input vectors and show what each output each produces.

i). <1, 2, 3>

The program will reach /* target 2 */ and end with <1, 2, 2>.

ii). <-1, -1, 4>

The program will end with <-1, -1, 2>.

iii). <24, 52, 1>

The program will end with <24, 52, 2>.

iv). None of these cover target 1. Which comes closest?

ii). comes closest as it gets closest to a > b.

b). Explain why it is not possible to cover target 3 in this program. Which input vector or vectors come closest to hitting target 3?

Because of the first if-else statement, c is always set to either 1 or 2. When a < b, then c = 2 and that is the closest the program gets to hitting target 3.

c). Which input vectors hit target 4?

<1, 42, 2>

d). Is it possible to achieve 100% branch coverage? If yes, give the test suite with 100% coverage. If not, explain why not and construct a test suite that covers all the reachable branches.

It is not possible to attain 100% branch coverage as target 3 is unreachable. The following test suite covers the other branches:

• $TC_1$ : <3, 2, 1> (target 1)
• $TC_2$ : <1, 2, 1> (target 2)
• $TC_3$ : <1, 42, 1> (target 4)

This could be done with 2 test cases, but then you are combining 2 targets into one test.

1. Assume you have a part of a program:
if ((a && b) || c) do X;
else do Y;


Construct a test suite that gives 100% condition coverage; 100% branch coverage; 100% MC/DC.

Test Suite:

• $TC_1$ : TTT
• $TC_2$ : TTF
• $TC_3$ : TFF
• $TC_4$ : FFF
• $TC_5$ : FFT
• $TC_6$ : FTT
• $TC_7$ : FTF
• $TC_8$ : TFT

$(a \&\& b) || c$

a:
TTF = T x
FTF = F x
b:
TTF = T
TFF = F x
c:
TFT = T x
TFF = F

1. Calculate the McCabe Number (cyclomatic complexity) for the following graphs:
• a). $e - n + c = 1 - 2 + 2 = 1$
• b). $e - n + c = 4 - 4 + 2 = 2$
• c). $e - n + c = 2 - 4 + 2 = 0$
• d). $e - n + 2 = 15 - 13 + 2 = 4$

a, b, c) are control flow graphs, whereas d is an arbitrary graph.