Questions:
• 7. Use the following method printPrimes() for questions a–d.
基于Junit及Eclemma(jacoco)实现一个主路径覆盖的测试。
/*****************************************************************
* Finds and prints n prime integers
* Jeff Offutt, Spring 2003
*****************************************************************/
private static void printPrimes(int n)
{
int curPrime; // Value currently considered for primeness
int numPrimes; // Number of primes found so far;
boolean isPrime; // Is curPrime prime?
int[] primes = new int[MAXPRIMES]; // The list of prime numbers. // Initialize 2 into the list of primes.
primes[0] = 2;
numPrimes = 1;
curPrime = 2;
while(numPrimes < n)
{
curPrime++; // next number to consider...
isPrime = true;
for(int i = 0; i <= numPrimes-1; i++)
{ // for each previous prime.
if(isDivisible(primes[i], curPrime))
{ // Found a divisor, curPrime is not prime.
isPrime = false;
break; // out of loop through primes.
}
}
if(isPrime)
{ // save it!
primes[numPrimes] = curPrime;
numPrimes++;
}
} // End while // print all the primes out.
for(int i = 0; i <= numPrimes-1; i++)
{
System.out.println("Prime: " + primes[i]);
}
}// End printPrimes.
(a) Draw the control flow graph for the printPrimes() method.
(b) Consider test cases t1 = (n = 3) and t2 = (n = 5). Although these tour the same prime paths in printPrimes(), they do not necessarily find the same faults. Design a simple fault that t2 would be more likely to discover than t1 would.
(c) For printPrimes(), find a test case such that the corresponding test path visits the edge that connects the beginning of the while statement to the for statement without going through the body of the while loop.
(d) Enumerate the test requirements for node coverage, edge coverage, and prime path coverage for the path for printPrimes().
Answers:
(a)
(b) MAXPRIMES >= n
所以当MAXPRIMES = 4时，t2会发生数组越界，而t1不会。
(c) n = 1;
(d)点覆盖：TR = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
边覆盖：TR = {(1,2), (2,3), (2,11), (3,4), (4,5), (4,8), (5,6), (5,7), (6,8), (7,4), (8,9), (8,10), (9,10), (10,2), (11,12), (12,13), (12,15), (13,14), (14,12)}
主路径覆盖：
TR = {(1, 2, 3, 4, 5, 6, 8, 9, 10),(1, 2, 3, 4, 5, 6, 8, 10),(1, 2, 3, 4, 8, 9, 10),(1, 2, 3, 4, 8, 10),(1, 2, 3, 4, 5, 7),(1, 2, 11, 12, 13, 14),(1, 2, 11, 12, 15),
(2, 3, 4, 5, 6, 8, 9, 10, 2),(2, 3, 4, 5, 6, 8, 10, 2),(2, 3, 4, 8, 9, 10, 2),(2, 3, 4, 8, 10, 2),
(3, 4, 5, 6, 8, 9, 10, 2, 11, 12, 13, 14),(3, 4, 5, 6, 8, 9, 10, 2, 11, 12, 15),(3, 4, 5, 6, 8, 10, 2, 11, 12, 13, 14),(3, 4, 5, 6, 8, 10, 2, 11, 12, 15),(3, 4, 8, 9, 10, 2, 11, 12, 13, 14),(3, 4, 8, 9, 10, 2, 11, 12, 15),(3, 4, 8, 10, 2, 11, 12, 13, 14),(3, 4, 8, 10, 2, 11, 12, 15),
(4, 5, 7, 4),(4, 5, 6, 8, 9, 10, 2, 3, 4),(4, 5, 6, 8, 10, 2, 3, 4),
(5, 7, 4, 5),(5, 6, 8, 9, 10, 2, 3, 4, 5),(5, 6, 8, 10, 2, 3, 4, 5),(5, 7, 4, 8, 9, 10, 2, 3),(5, 7, 4, 8, 10, 2, 3),(5, 7, 4, 8, 9, 10, 2, 11, 12, 13, 14),(5, 7, 4, 8, 10, 2, 11, 12, 13, 14),(5, 7, 4, 8, 9, 10, 2, 11, 12, 15),(5, 7, 4, 8, 10, 2, 11, 12, 15),
(6, 8, 9, 10, 2, 3, 4, 5, 6),(6, 8, 10, 2, 3, 4, 5, 6),(6, 8, 9, 10, 2, 3, 4, 5, 7),(6, 8, 10, 2, 3, 4, 5, 7),
(7, 4, 5, 7),(7, 4, 5, 6, 8, 9, 10, 2, 3, 4),(7, 4, 5, 6, 8, 10, 2, 3, 4),(7, 4, 5, 6, 8, 9, 10, 2, 11, 12, 13, 14),(7, 4, 5, 6, 8, 10, 2, 11, 12, 13, 14),(7, 4, 5, 6, 8, 9, 10, 2, 11, 12, 15),(7, 4, 5, 6, 8, 10, 2, 11, 12, 15),
(8, 10, 2, 3, 4, 8),(8, 9, 10, 2, 3, 4, 8),(8, 10, 2, 3, 4, 5, 6, 8),(8, 9, 10, 2, 3, 4, 5, 6, 8),
(9, 10, 2, 3, 4, 5, 6, 8, 9),(9, 10, 2, 3, 4, 8, 9),
(10, 2, 3, 4, 5, 6, 8, 10),(10, 2, 3, 4, 5, 6, 8, 9, 10),(10, 2, 3, 4, 8, 10),(10, 2, 3, 4, 8, 9, 10),
(12, 13, 14, 12),
(13, 14, 12, 15),(13, 14, 12, 13),
(14, 12, 13, 14)}
(附加)基于Junit及Eclemma(jacoco)实现一个主路径覆盖的测试:
首先，假设MAXPRIMES=10并补全代码；
其次，确定测试用例并编写测试文件；
