软件测试第三次作业-覆盖测试

Source code of printPrimes(int n):

/******************************************************* 
     * Finds and prints n prime integers 
     * Jeff Offutt, Spring 2003 
     ******************************************************/ 
    public static String printPrimes (int n) 
    { 
        String list = new String();
        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 

a) CFG:

b)

Let MAXPRIMES = 4. Then if we use t1=(n=3) to test the program, we won’t find a fault. But if we use t2=(n=5) to test it, we will find a fault because of the overflow of the array(primes[]).

c)

t=(n=1)

d)

　　Node coverage={1,2,3,4,5,6,7,8,9,10,11,12,13}

　　Edge coverage={(1,2),(2,3),(2,4),(3,5),(4,11),(5,6),(5,9),(6,5),(6,8),(7,5),(8,9),(9,2),(9,10),

　　　　　　　　　　  (10,2),(11,12),(11,13),(12,11)}

 

     Prime path coverage={(1 2 4 11 13),

　　　　　　　　　　　　　 (1 2 4 11 12)

                                     (1 2 3 5 6 7)

 (1 2 3 5 6 8 9 10)

 (1 2 3 5 9 10)

 (2 3 5 6 8 9 2)

 (2 3 5 9 2)

 (2 3 5 9 10 2)

 (3 5 6 8 9 2 3)

 (3 5 6 8 9 10 2 3)

 (3 5 9 10 2 3)

 (3 5 9 2 3)

 (3 5 6 8 9 2 4 11 13)

 (3 5 6 8 9 2 4 11 12)

 (3 5 6 8 9 10 2 4 11 13)

 (3 5 6 8 9 10 2 4 11 12)

 (3 5 9 10 2 4 11 13)

 (3 5 9 10 2 4 11 12)

 (3 5 9 2 4 11 13)

 (3 5 9 2 4 11 12)

 (5 6 7 5)

 (5 6 8 9 2 3 5)

 (5 6 8 9 10 2 3 5)

 (5 9 10 2 3 5)

 (5 9 2 3 5)

 (5 6 8 9 2 4 11 13)

 (5 6 8 9 2 4 11 12)

 (5 6 8 9 10 2 4 11 13)

 (5 6 8 9 10 2 4 11 12)

 (5 9 10 2 4 11 13)

 (5 9 10 2 4 11 12)

 (5 9 2 4 11 13)

 (5 9 2 4 11 12)

 (6 7 5 6)

 (6 7 5 9 2 3)

 (6 7 5 9 10 2 3)

 (6 8 9 2 3 5 6)

 (6 8 9 10 2 3 5 6)

 (6 7 5 9 2 4 11 13)

 (6 7 5 9 2 4 11 12)

 (6 7 5 9 10 2 4 11 13)

 (6 7 5 9 10 2 4 11 12)

 (7 5 6 7)

 (7 5 6 8 9 2 3)

 (7 5 6 8 9 2 4 11 13)

 (7 5 6 8 9 2 4 11 12)

 (7 5 6 8 9 10 2 3)

 (7 5 6 8 9 10 2 4 11 13)

 (7 5 6 8 9 10 2 4 11 12)

 (8 9 2 3 5 6 7)

 (8 9 2 3 5 6 8)

 (8 9 10 2 3 5 6 7)

 (8 9 10 2 3 5 6 8)

 (9 2 3 5 6 8 9)

 (9 2 3 5 9)

 (9 10 2 3 5 6 8 9)

 (9 10 2 3 5 9)

 (10 2 3 5 6 8 9 10)

 (10 2 3 5 9 10)

 (11 12 11)

 (12 11 13)

 (12 11 12)

 }

 

• Prime path test:

　　　　

         T = (n = 5)

         Excepted = “2 3 5 7 11 “

         Output = “2 3 5 7 11 “

• Printscreen: