Checking and replacing a duplicate in a single for loop - c++

Question

So I have set up my code to be this (note I have my user input for the array and display of the results in another function and a beginner)
int main( )
{
using namespace std;
int a[8];
populateArray(a, 8);
cout<<"The array is: ";
displayArray(a, 8);
cout<<"\nThis array becomes: "<<sameElement(a, 8);
int sameElement(int a[], int size)
{
using namespace std;
for(int i = 0; i < size; i ++)
{
if( a[i] == a[i +1])
{
a[i + 1] = 0;
return i;
}
else
{
return a[i];
}
}
}
Been trying to see if a user puts a duplicate number(that works fine and great). That same array that the user input I want to replace that duplicate to a zero. Sadly I get either a single 0, 1, or 2 and it replaces the entire array. not sure how to return it back to my main as it also says that I have nothing returning from my function despite invoking it in my function and trying to find the right way to return my function. I have declared my function on top and they both match with the parameter as well. So still don't see why I keep getting an issue of returning my function to main?
Example:
The array: 1 2 2 3 4 5 5 7
Array becomes 1 0 0 3 4 0 0 7

Answer 1

I suppose you want to replace duplicate entries of an array in sameElement function. You're getting single integer because sameElement function returns single integer.
Instead of returning value in for loop, let the function run through. Also note that a[i+1] will go over limit of array a, so you have to change ending condition to size - 1.
void sameElement(int a[], int size)
{
for(int i = 0; i < size - 1; i++)
{
if( a[i] == a[i +1])
{
a[i] = 0;
if(i + 1 == size - 1)
{
a[i + 1] = 0;
}
}
}
}
Because the function no longer returns value, you must display it on its own.
cout << "\nThis array becomes: ";
sameElement(a, 8);
displayArray(a, 8);
Lastly, I don't know if this is intentional but sameElement function will only find duplicate entries if they are next to each other: it won't turn 1 2 1 2 1 2 1 2 to zeroes.