码迷,mamicode.com
首页 > 其他好文 > 详细

hm3

时间:2018-03-27 01:58:09      阅读:210      评论:0      收藏:0      [点我收藏+]

标签:void   alt   without   print   val   current   cts   init   state   

作业题目:

/**
* Finds and prints n prime integers
* Jeff Offutt, Spring 2003
*/


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 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.
(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.
(d) Enumerate the test requirements for node coverage, edge coverage,and prime path coverage for the path for printPrimes().
(e) List test paths that achieve node coverage but not edge coverage ont the graph.
(f) List test paths that achieve edge coverage but not prime path coverage on the gra

----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

(a)

(b)令 MAXPRIMES = 4 ,t1 不能检查出错误,t2 发生数组越界,使得 t2 比 t1 更容易发现

(c) t3 = ( n = 1 )

(d)

点覆盖:{1,2,3,4,5,6,7,8,9,10,11,12}

边覆盖:{1,2},{2,3},{3,4},{4,5},{4,7},{5,4},{5,6},{6,7},{7,2},{7,8},{8,2},{2,9},{9,10},{10,11}

    {11,10},{10,12}

主路径覆盖:{1,2,3,4,5,6,7,8},{1,2,9,10,11},{1,2,9,10,12},{2,3,4,5,6,7,8,2},{2,3,4,7,2},{2,3,4,5,6,7,2},{4,5,4},{5,4,5},{3,4,5,6,7,8,2,9,10,12},{3,4,5,6,7,8,2,9,10,11},{10,11,10},{11,10,11},{11,10,12}

(e)

点覆盖测试路径:{1,2,3,4,5,6,7,8,2,9,10,11,10,12}

(f)

边覆盖测试路径:{1,2,3,4,5,4,5,6,7,8,2,9,10,11,10,12},{1,2,3,4,7,2,9,10,12};

hm3

标签:void   alt   without   print   val   current   cts   init   state   

原文地址:https://www.cnblogs.com/qg123456/p/8654708.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!