Average of odd cells? - c++

Question

For an exercise I am doing, I am supposed to find out the average of the items contained in odd numbered cells of an array and some other things. Finding the average of the odd numbered cells in the only thing I'm having a problem with. Here is my code, what am I doing wrong? The final function is the odd numbered cells average function. Thanks.
#include <iostream>
#include <iomanip>
using namespace std;
const int SIZE = 1000;
int randoms[SIZE];
int sum2 = 0;
int top = 0;
int maximum = 0;
int highest(int randoms[]);
int findsum(int randoms[]);
int average(int randoms[], int sum);
int oddavg(int randoms[]);
int main()
{
int sum = 0;
int top = 0;
int avg = 0;
int oddaverage = 0;
for (int i = 0; i < SIZE; i++)
{
randoms[i] = (rand() % 5000 + 1);
cout << randoms[i] << setw(10) << " ";
}
cout << endl << endl;
cout << "The sum of the values in the array is ";
sum = findsum(randoms);
cout << sum << endl;
cout << endl << endl;
cout << "The highest value in the array is ";
top = highest(randoms);
cout << top << endl;
cout << endl << endl;
cout << "The average of all of the numbers in the array is ";
avg = average(randoms, sum);
cout << avg << endl;
cout << endl << endl;
cout << "The average of all of the numbers in the odd cells is ";
oddaverage = oddavg(randoms);
}
int findsum(int randoms[])
{
for (int i = 0; i < SIZE; i++)
{
sum2 += randoms[i];
}
return sum2;
}
int highest(int randoms[])
{
for (int i = 0; i < SIZE; i++)
{
if (randoms[i] > maximum)
{
maximum = randoms[i];
}
}
top = maximum;
return top;
}
int average(int randoms[], int sum)
{
int avg = 0;
for (int i = 0; i < SIZE; i++)
{
avg = (sum / SIZE);
}
return avg;
}
int oddavg(int randoms[])
{
int avg = 0;
int sum = 0;
for (int i = 0; i < SIZE; i++)
{
if (randoms[i] / 2 == 1)
{
sum += randoms[i];
}
}
avg = sum / SIZE;
return avg;
}

Answer 1

Doing the odd/even test (using modulo as many have suggested) in this case is totally redundant, since the loop doesn't do anything else.
You can just use a stride of 2 and start at the first odd element:
for (int i = 1; i < SIZE; i += 2)
{
sum += randoms[i];
}
Then it's just a matter of dividing out by half of SIZE. If that number is even, then there are SIZE/2 odd numbers. If it's odd, then there are lbound(SIZE/2)+1 odd numbers. Fortunately, you can take advantage of integer truncation and just do:
double avg = double(sum) / double((SIZE+1) / 2);
And you don't even have to worry about divide-by-zero =)

Answer 2

should be if (randoms[i] % 2 == 1)
Also you need to count the number of odd numbers.

Answer 3

randoms[i] / 2 == 1, this will be true only when a cell value is 2 or 3, this is certainly not what you need to do.
If you need to sum values of cells with odds index then it should be if (i % 2 == 1). If instead you are looking for odd values (in any cell) it should be if (randoms[i] % 2 == 1).
Mind that % is the modulo operation which returns the integer remainder of the integer division.
And since you are calculating an average you should divide by the found amount of elements, not by the total.

Answer 4

int oddavg(int randoms[])
{
int cnt = 0;
int avg = 0;
for (int i = 0; i < SIZE; i++)
{
if ( randoms[i] % 2 != 0 )
{
++cnt;
avg += randoms[i];
}
}
return ( cnt == 0 ? avg : avg / cnt );
}