Software testing hw3 -- printPrimes()

/******************************************************* 
     * 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 (curPrime%primes[i]==0) 
                { // 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 ow graph for the printPrimes() method

控制流图

(b) Considertestcasest1=(n=3)andt2=(n=5).Although these tourthe same prime paths in printPrimes(), they do not necessarily find the same faults.Designasimplefaultthat t2 would bemorelikelytodiscover than t1 would.

答:将MAXPRIMES设置为4时,t2会发生数组越界错误,但t1不会发生错误。

 

(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.

 

答:当n=1时,会通过numPrimes>=直接从上图的2节点跳转到12节点,不经过while循环体内部。

 

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

 

答:

点覆盖:{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16}

边覆盖:{(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)}

主路径覆盖:{(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)}

 

 

测试

import static org.junit.Assert.*;

import org.junit.Before;
import org.junit.Test;

public class PrintPrimesTest {
    
    private PrintPrimes pptest;

    @Before
    public void setUp() throws Exception {
        pptest = new PrintPrimes();
    }

    @Test
    public void testPrintPrimes() {
        pptest.printPrimes(5);
        
    }
}
package stHW3;
public class printPrime {
    public String printPrimes (int n) 
    { 
        final int MAXPRIMES=100;
        int curPrime; // Value currently considered for primeness 
        int numPrimes; // Number of primes found so far. 
        boolean isPrime; // Is curPrime prime? 
        String str = "";
        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 (curPrime%primes[i]==0) 
                { // 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++) 
        { 
            str += primes[i]+" "; 
        }
        return str;
    } // end printPrimes
}

 

posted @ 2018-03-26 23:07  季鑫达  阅读(180)  评论(0编辑  收藏  举报