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Consider the partially-filled array named a. What does the following loop do? (cin is a Scanner object)int[] a = {1, 3, 7, 0, 0, 0};int size = 3, capacity = 6;int value = cin.nextInt();while (value > 0){a[size] = value;size++;value = cin.nextInt();}1. May crashe at runtime because it can input more elements than the array can hold2. Reads up to 3 values and inserts them in the array in the correct position.3. Reads one value and places it in the remaining first unused space endlessly.4. Reads up to 3 values and places them in the array in the unused space.

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Answer:

Option 1: May crash at runtime because it can input more elements than the array can hold

Explanation:

Given the code as follows:

  1.        int[] a = {1, 3, 7, 0, 0, 0};
  2.        int size = 3, capacity = 6;
  3.        int value = cin.nextInt();
  4.        while (value > 0)
  5.        {
  6.            a[size] = value;
  7.            size++;
  8.            value = cin.nextInt();
  9.        }

From the code above, we know the a is an array with six elements (Line 1). Since the array has been initialized with six elements, the capacity of the array cannot be altered in later stage.

However, a while loop is created to keep prompting for user input an integer and overwrite the value in the array started from index 3 (Line 4- 9). In every round of loop, the index is incremented by 1 (Line 7). If the user input for variable value is always above zero, the while loop will persist.  This may reach a point where the index value is out of bound and crash the program. Please note the maximum index value for the array is supposedly be 5.  

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