软件测试：Homework 3

Homework 3

 

 

 

Questions：

 

private 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; 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] );

        }
        
    }// 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 tour the 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.

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.

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

 

Answers:

 

一．the answers of questions a-d

a）printPrimes()方法的控制流图如下：

      

 

b）MAXPRIMES = 4，此时 test case t2 会发生数组越界，而test case t1 不会，所以t2 比t1更容易发现错误。

 

c ）test case t3 = (n = 1)；

 

d）

    (1) Node coverage

         TR = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}

         Test Paths: [1,2,3,4,5,6,7,5,6,8,9,10,2,11,12,13,14,15]

 

    (2) Edge coverage

         TR = {(1,2) , (2,3) , (2,11) , (3,4) , (4,5) , (5,6) , (5,9) , (6,7) , (6,8) , (7,5) , (8,9) , (9,2) , (9,10) , (10,2) , (11,12) , (12,13) , (12,15) , (13,14) , (14,12)}

         Test Paths: [1, 2, 3, 4, 5, 6, 7, 5, 6, 8, 9, 10, 2, 11, 12, 13, 14, 12, 15 ]

                             [1, 2, 3, 4, 5, 9, 2, 11, 12, 15 ]

 

    (3) Prime Path coverage:

         TR = {[5,6,7,5] , [6,7,5,6] , [7,5,6,7] , [12,13,14,12] , [13,14,12,13] , [13,14,12,15] , [14,12,13,14] ,

                   [1,2,11,12,15] , 

                   [2,3,4,5,6,7] , [2,3,4,5,9,2] , [3,4,5,9,2,3] , [4,5,9,2,3,4] , [5,9,2,3,4,5] , [9,2,3,4,5,9] ,

                   [1,2,3,4,5,6,7] , [1,2,3,4,5,9,10] , [1,2,11,12,13,14,15] , [2,3,4,5,9,10,2] , [3,4,5,9,10,2,3] , [4,5,9,10,2,3,4] , [5,9,10,2,3,4,5] , [9,10,2,3,4,5,9] , [10,2,3,4,5,9,10] ,

                   [1,2,11,12,13,14,12,15] , [2,3,4,5,6,8,9,2] , [3,4,5,6,8,9,2,3] , [3,4,5,9,2,11,12,15] , [4,5,6,8,9,2,3,4] , [5,6,8,9,2,3,4,5] , [6,7,5,9,2,11,12,15] , [6,8,9,2,3,4,5,6] ,

                   [8,9,2,3,4,5,6,8] , [9,2,3,4,5,6,8,9] ,

                   [1,2,3,4,5,6,8,9,10] , [2,3,4,5,6,8,9,10,2] , [3,4,5,6,8,9,10,2,3] , [3,4,5,9,10,2,11,12,15] , [4,5,6,8,9,10,2,3,4] , [5,6,8,9,10,2,3,4,5] , [6,7,5,9,10,2,11,12,15] ,

                   [6,8,9,10,2,3,4,5,6] , [8,9,10,2,3,4,5,6,8] , [9,10,2,3,4,5,6,8,9] , [10,2,3,4,5,6,8,9,10] ,

                   [3,4,5,6,8,9,10,11,12,15] ,

                   [3,4,5,6,8,9,10,2,11,12,15] , [3,4,5,9,2,11,12,13,14,12,15] , [6,7,5,9,2,11,12,13,14,12,15] ,

                   [3,4,5,9,10,2,11,12,13,14,12,15] , [6,7,5,9,10,2,11,12,13,14,12,15] ,

                   [3,4,5,6,8,9,2,11,12,13,14,12,15] ,

                   [3,4,5,6,8,9,10,2,11,12,13,14,12,15]}

         Test Paths: [ 1, 2, 11, 12, 15 ]

                            [ 1, 2, 11, 12, 13, 14, 15 ]

                            [ 1, 2, 11, 12, 13, 14, 12, 15 ]

                            [ 1, 2, 3, 4, 5, 6, 7, 5, 6, 7, 5, 6, 8, 9, 2, 3, 4, 5, 9, 10, 2, 3, 4, 5, 9, 2, 3, 4, 5, 9, 2, 11, 12, 13, 14, 12, 15 ]

                            [ 1, 2, 3, 4, 5, 9, 10, 2, 3, 4, 5, 6, 8, 9, 10, 2, 3, 4, 5, 9, 2, 11, 12, 15 ]

                            [ 1, 2, 3, 4, 5, 6, 8, 9, 2, 3, 4, 5, 6, 8, 9, 10, 2, 3, 4, 5, 9, 10, 2, 11, 12, 15 ]

                            [1, 2, 3, 4, 5, 6, 8, 9, 10, 2, 3, 4, 5, 6, 8, 9, 10, 2, 3, 4, 5, 6, 7, 5, 9, 2, 11, 12, 13, 14, 12, 15 ]

     

二．基于Junit及Eclemma（jacoco）实现一个主路径覆盖的测试。   

        测试类：

        Triangle.java 

package jUnitTest;

public class Triangle {
    public String check(int a, int b, int c){        
        String result = "";
        if(a == b && b == c){
            result = "该三角形为等边三角形";
        }else if(a == b || a == c || b ==c){
            result = "该三角形为等腰三角形";
        }else if(a + b > c && a + c > b && b + c > a){
            result = "该三角形为不等边三角形";
        }else{
            result = "这不是一个三角形";
        }

        return result;     
    }

}

 

        测试代码及用例：

        testTriangle.java

package jUnitTest;

import static org.junit.Assert.*;

import org.junit.Test;

public class testTriangle{
    Triangle triangle = new Triangle();
    String equilateral = "该三角形为等边三角形";
    String isosceles = "该三角形为等腰三角形";
    String scalene = "该三角形为不等边三角形";
    String nonATriangle = "这不是一个三角形";
    String result;

    @Test
    public void testEquilateral(){
        result = triangle.check(3, 3, 3);
        assertEquals(equilateral, result);
    }

    @Test
    public void testIsosceles(){
        result = triangle.check(3, 1, 3);
        assertEquals(isosceles, result);
    }

    @Test
    public void testScalene(){
        result = triangle.check(2, 3, 4);
        assertEquals(scalene, result);
    }

    @Test
    public void testNonTri(){
        result = triangle.check(3, 2, 5);
        assertEquals(nonATriangle, result);
    }

}

 

        测试结果：