題目

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.

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? ? // 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

? ? for(int i = 0; i < numPrimes; i++) {// print all the primes out

? ? ? ? System.out.println("Prime: " + primes[i] );

? ? }

}

1.Draw the control flow graph for the printPrime() method

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2.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.

while(numPrimes < n) 改成 while(numPrimes < 4) 這樣當(dāng)n為4和5的時(shí)候結(jié)果是一樣的,所以能發(fā)現(xiàn)錯(cuò)誤,但是前者不會(huì)

3.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.

printPrimes(1); 不進(jìn)入循環(huán),不符合循環(huán)條件

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

node coverage:[1,2,3,4,5,6,7,8,2,10,11,12],[1,2,3,4,5,9,4,7,2,10,12]

edge coverage:[1,2,3,4,5,6,7,8,2,10,11,12],[1,2,3,4,5,9,4,7,2,10,12]

prime path coverage:[1,2,3,4,5,9,4,5,9,4.5,6,7,8,2,3,4,5,6,7,8,2,3,4,5,6,7,8,2,3,4,5,6,7,2,3,4,5,6,7,2,3,4,7,2,3,4,7,2,10,11,10,11,10,12]

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轉(zhuǎn)載于:https://www.cnblogs.com/sukebei/p/8641459.html

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