Question

Code Analysis (1)

public int countPositive(int[] x) {
//Effects: If x == null throw NullPointerException
//Else return the number of positive (non-zero) elements in x
int count = 0;
for (int i = 0; i < ; i++) {
if (x[i] >= 0) {
count++;
}
}
return count;
}
// Test Input: x = [-4, 2, 0, 2]
// Expected output = 2

If possible, identify a test case that does not execute the fault:

A.) Test input: x = [1, 2, 3]; Expected output = 3

B.) Test input: x = [0, 0, 0]; Expected output = 0

C.) Test input: x = []; Expected output = 0

D.) No test case possible

Answer 1

The code counts the number of positive elements in an array, but has a fault in the condition used to check if an element is positive or not.

Step-by-step explanation:

The subject of this question is Computers and Technology. The purpose of the code is to count the number of positive (non-zero) elements in an array. The provided code has a fault in the condition used to check if an element is positive or not.

For the given test input x = [-4, 2, 0, 2], the expected output is 2. However, the condition x[i] >= 0 is incorrect, as it also includes zero. To fix this, you should change the condition to x[i] > 0.

Based on the available test cases, we can see that a test case that does not execute the fault is Test input: x = [0, 0, 0]; Expected output = 0. In this case, the code will correctly count zero elements as non-positive and the output will be 0.