homework3
1 /** 2 * Finds and prints n prime integers 3 * Jeff Offutt, Spring 2003 4 */ 5 6 7 private static void printPrimes(int n) { 8 int curPrime; //Value currently considered for primeness 9 int numPrimes; // Number of primes found so far; 10 boolean isPrime; //Is curPrime prime? 11 int[] primes = new int[MAXPRIMES];// The list of primes. 12 13 // Initialize 2 into the list of primes. 14 primes[0] = 2; 15 numPrimes = 1; 16 curPrime = 2; 17 while(numPrimes < n) { 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(isDvisible(primes[i],curPrime)) { 23 //Found a divisor, curPrime is not prime. 24 isPrime = false; 25 break; 26 } 27 } 28 if(isPrime) { 29 // save it! 30 primes[numPrimes] = curPrime; 31 numPrimes++; 32 33 } 34 }// End while 35 36 // print all the primes out 37 for(int i = 0; i < numPrimes; i++) { 38 System.out.println("Prime: " + primes[i] ); 39 40 } 41 42 }// End printPrimes.
(a) Draw the control flow graph for the printPrime() method.
(b) Consider test cases ti = (n = 3) and t2 = ( n = 5). Although these tour the same prime paths in printPrime(), they don't necessarily find
the same faults. Design a simple fault that t2 would be more likely to discover than t1 would.
MAXPRIMES=4即可,此时t2会遇到数组越界问题
(c) For printPrime(), 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.
n=1
(d) Enumerate the test requirements for node coverage, edge coverage,and prime path coverage for the path for printPrimes().
点覆盖:
{1,2,3,4,5,6,7,8,9,10,11,12,13,14}
边覆盖:
{(1,2),(2,3),(2,10),(3,4),(4,5),(4,8),(5,6),(5,6),(5,7),(6,4),(7,8),(8,2),(8,9),(9,2),(10,11),(11,12),(11,14),(12,13),(13,11)}
主路径覆盖:
{(1,2,3,4,5,7,8,9), (1,2,3,4,8,9), (1,2,3,4,5,6), (1,2,10,11,12,13), (1,2,10,11,14), (2,3,4,5,7,8,2), (2,3,4,5,7,8,9,2), (2,3,4,8,2), (2,3,4,8,9,2), (3,4,5,7,8,9,2,10,11,14), (3,4,5,7,8,9,2,10,11,12,13), (3,4,5,7,8,9,10,11,14), (3,4,5,7,8,9,10,11,12,13), (3,4,5,7,8,2,10,11,14), (3,4,5,7,8,2,10,11,12,13), (3,4,8,2,10,11,14), (3,4,8,2,10,11,12,13), (3,4,5,7,8,9,2,3), (3,4,5,7,8,2,3), (3,4,8,9,2,3), (3,4,8,2,3), (4,5,7,8,2,3,4), (4,5,7,8,9,2,3,4), (4,5,6,4), (4,8,2,3,4), (5,6,4,5), (5,7,8,9,2,3,4,5), (5,7,8,2,3,4,5), (6,4,5,6), (7,8,9,2,3,4,5,7), (7,8,4,5,7), (8,9,2,3,4,5,7,8), (8,2,3,4,5,7,8), (9,2,3,4,5,7,8,9), (9,2,3,4,8,9), (11,12,13,11), (12,13,11,12), (13,11,12,13)}
测试代码:
package printPrimes; public class PrintPrimes { private static int MAXPRIMES = 100; 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 primes. // 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(isDvisible(primes[i],curPrime)) { //Found a divisor, curPrime is not prime. isPrime = false; break; } } if(isPrime) { // save it! primes[numPrimes] = curPrime; numPrimes++; } }// End while // print all the primes out for(int i = 0; i < numPrimes; i++) { System.out.println("Prime: " + primes[i] ); } } private static boolean isDvisible(int i, int curPrime) { // TODO Auto-generated method stub if(curPrime%i==0) return false; return true; } }
package printPrimes; import static org.junit.Assert.*; import org.junit.Before; import org.junit.Test; public class PrintPrimesTest { @Test public void test() { PrintPrimes p = new PrintPrimes(); p.printPrimes(5); } }
测试结果: