Tutorial 8

  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.
1int a, b, c
2
3if (a>b)
4    c = 1 /* target 1 */ ;
5else {
6    c = 2;
7
8    if (b==c)
9    /* target 2 */ ;
10}
11
12if (c==3)
13    do something; /* target 3 */ ;
14if (b==42)
15    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:

  • : <3, 2, 1> (target 1)
  • : <1, 2, 1> (target 2)
  • : <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:
1if ((a && b) || c) do X;
2else do Y;

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

Test Suite:

  • : TTT
  • : TTF
  • : TFF
  • : FFF
  • : FFT
  • : FTT
  • : FTF
  • : TFT

1a:
2  TTF = T x
3  FTF = F x
4b:
5  TTF = T
6  TFF = F x
7c:
8  TFT = T x
9  TFF = F
  1. Calculate the McCabe Number (cyclomatic complexity) for the following graphs:
  • a).
  • b).
  • c).
  • d).

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