Write a function for a function \"count\" that counts the number of odd and even

Question

Write a function for a function "count" that counts the number of odd and even numbers in an integer array. Note that the function has to return two results. Note that the four formal parameters have been left blank.

Hint: There should be four parameters to the function with the following names:

int input [size]; --->integer array of some size (specific size does not matter)
int size; --->the size of the array
int odd; --->number of odd numbers
int even; --->number of even numbers

Skeleton Code:

void count ( , , , )
{








}

Explanation / Answer

The following is C / C++ code. a % 2 is the remainder of dividing a by 2. If it's equal to zero, the number is even. If not, it's odd. You should pass even and odd, the counters, as pointers, so the function that calls count could get the answer.

void count (int n, int *array, int *even, int *odd)
{
    *even = 0;
    *odd = 0;
    int i;
    for (i = 0; i < n; i++)
    {
        // The number is even.
        if (array[i] % 2 == 0)
            *even += 1;
        // The number is odd.
        else
            *odd += 1;
    };
};

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