Q: Use the following method printPrimes() for questions a–d.

/******************************************************* 
     * Finds and prints n prime integers 
     * Jeff Offutt, Spring 2003 
     ******************************************************/ 
    public 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 (isDivisable(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 tourthe 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.

#define MAXPRIMES 4

Then, the volume of the array will be out of bound, and 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.

When n = 1; statement is 0 -> 1 -> 10, without going through the while loop.

(d) Enumerate the test requirements for node coverage, edge coverage, and prime path coverage for the graph for printPrimes().

node coverage:

{0, 1，2，3，4，5，6，7，8，9，10，11，12，13, 14}

edge coverage:

{(0, 1), (1, 2), (1, 10), (2, 3), (3, 4), (4, 5), (4, 8), (5, 6), (6, 4), (5, 7), (7, 8), (8, 1), (8, 9), (9, 1), (10, 11), (11, 12), (12, 13), (13, 11), (11, 14*)}

prime path coverage(PPC):

{

(0, 1, 2, 3, 4, 5, 6, 4, 5, 7, 8, 9, 1, 10, 11, 14),

(0, 1, 2, 3, 4, 5, 6, 4, 8, 9, 1, 10, 11, 14),

(0, 1, 2, 3, 4, 5, 7, 8, 9, 1, 10, 11, 14),

(0, 1, 2, 3, 4, 8, 1, 10, 11, 14),

(0, 1, 10, 11, 12, 13, 11, 14),}

二. 基于 Junit 及 Eclemma 实现一个PPC的测试

结果截图: (主路径覆盖测试详见 [ST2017] Lab1)

posted on 2017-03-10 11:58 E-C-Ares 阅读(181) 评论(0) 编辑收藏举报