Reference parameters in C++

Question

What happens when we define one reference parameter in function definition and send two parameters from function call
Example
~~~~~~~
void location (int &x, int y=4)
{
y+=2;
x+=y;
}
int main ()
{
int px=10, py=2;
location (px);
location (px,py); // my doubt is this function call.
cout<<px<<py;
}

Answer 1

(py) value replaces the 2nd parameter's default value and nothing happens to (py) value because it's not being sent to the function by reference.
output:
14 2

Answer 2

Your parameter will just take the value of py. Since the py is passed by value, no changes will be made to py.

Answer 3

yup instead of 4, 2 is passed to the function

Answer 4

then in second function call py value is replaced?!!

Answer 5

yes thanks I understood the logic now thank u

Answer 6

it doesnt give that output : 14 2