标签:cti loop 实现 set sid error sign case run
Question1: Use the following method printPrimes() for questions a–d.
1 /******************************************************* 2 * Finds and prints n prime integers 3 * Jeff Offutt, Spring 2003 4 ******************************************************/ 5 public static void printPrimes (int n) 6 { 7 int curPrime; // Value currently considered for primeness 8 int numPrimes; // Number of primes found so far. 9 boolean isPrime; // Is curPrime prime? 10 int [] primes = new int [MAXPRIMES]; // The list of prime numbers. 11 12 // Initialize 2 into the list of primes. 13 primes [0] = 2; 14 numPrimes = 1; 15 curPrime = 2; 16 while (numPrimes < n) 17 { 18 curPrime++; // next number to consider ... 19 isPrime = true; 20 for (int i = 0; i <= numPrimes-1; i++) 21 { // for each previous prime. 22 if (curPrime%primes[i]==0) 23 { // Found a divisor, curPrime is not prime. 24 isPrime = false; 25 break; // out of loop through primes. 26 } 27 } 28 if (isPrime) 29 { // save it! 30 primes[numPrimes] = curPrime; 31 numPrimes++; 32 } 33 } // End while 34 35 // Print all the primes out. 36 for (int i = 0; i <= numPrimes-1; i++) 37 { 38 System.out.println ("Prime: " + primes[i]); 39 } 40 } // end printPrimes
(a) Draw the control flow graph for the printPrimes() method.
Answer:
(b) Consider test cases t1=(n=3) and t2=(n=5). Although these tour the same prime paths in ptinrPrimes(), they do not necessarily find the same faults. Design a simple fault that t2 would be more likely to discover than t1 would.
Answer:
When MAXPRIMES = 4, t2=(n=5) is overflow, but t1=(n=3) has not error.
(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 withtout going through the body of the while loop.
Answer:
When n = 1, the program will run through edge (2, 12) without running into while loop.
(d) Enumerate the test requirements for node coverage, edge coverage, amd prime path coverage for the graph fpr printPrime().
Answer:
1.Node Coverage
TR = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
2.Edge Coverage
TR = {(1,2),(2,3),(3,4),(4,5),(5,6),(6,8),(8,5),(6,7),(7,9),(5,9),(9,10),(9,11),(10,11),(11,2),(2,12),(12,13),(13,14),(14,15),(15,13),(13,16)};
3.Prime Path Coverage
TR={(1,2,3,4,5,6,8),(1,2,3,4,5,6,7,9,10,11),(1,2,3,4,5,6,7,9,11),(1,2,3,4,5,9,11),(1,2,3,4,5,9,10,11),(5,6,8,5),(6,8,5,6),(8,5,6,8),(8,5,6,7,9,11),(8,5,6,7,9,10,11),(1,2,12,13,16),(1,2,12,13,14,15),(13,14,15,13),(14,15,13,14),(15,13,14,15),(14,15,13,16),(15,13,16)};
Quetion2:基于Junit及Eclemma(jacoco)实现一个主路径覆盖的测试
测试代码:
package st_hw3; import static org.junit.Assert.*; import org.junit.Before; import org.junit.Test; public class PrimesTest { Primes prime; @Before public void setUp() throws Exception { prime = new Primes(); } @Test public void test() { prime.printPrimes(5); } }
测试结果:
EclEmma覆盖测试:
标签:cti loop 实现 set sid error sign case run
原文地址:http://www.cnblogs.com/marscarl/p/6545425.html