Why does the second variable not change in the field? - c++

Question

I'm testing some programms for my lectures. I'm creating classes and use a paramterlist to initalize a field but the second variable doesn't change.
#include <iostream>
using namespace std;
class Punkt {
int x;
int y;
public:
Punkt(int a = 0, int b = 0)
{
x = a;
y = b;
}
void printXY()
{
cout << "x= " << x << " y= " << y << endl;
}
};
int main() {
Punkt pFeld[] = { (1, 1), (2, 2), (3, 3) };
for (int i = 0; i < 3; i++)
pFeld[i].printXY();
cin.get();
};
No error messages. Expected result was that x and y change, while actual result is that only x changes and y stays 0.

Answer 1

This
(1, 1)
is an expression with the comma operator.
In fact this initialization
Punkt pFeld[] = { (1, 1), (2, 2), (3, 3) };
is equivalent to
Punkt pFeld[] = { 1, 2, 3 };
So the constructor with the second default argument equal to 0 is called three times.
Use instead
{ 1, 1 }
Here is your updated code
#include <iostream>
using namespace std;
class Punkt {
int x;
int y;
public:
Punkt(int a = 0, int b = 0)
{
x = a;
y = b;
}
void printXY()
{
cout << "x= " << x << " y= " << y << endl;
}
};
int main() {
Punkt pFeld[] = { {1, 1}, {2, 2}, {3, 3} };
for (int i = 0; i < 3; i++)
pFeld[i].printXY();
cin.get();
}
Its output is
x= 1 y= 1
x= 2 y= 2
x= 3 y= 3
Pay attention to that the semicolon after the function main is redundant.

Answer 2

Passing (1, 1) to the constructor of Punkt, the comma operator will return the 2nd operand as the result (the 1st operand is discarded), so you're only passing one int with value 1 to the constructor. That's why y is always initialized as 0.
What you want should be
Punkt pFeld[] = { {1, 1}, {2, 2}, {3, 3} }; // list initialization since C++11
or
Punkt pFeld[] = { Punkt(1, 1), Punkt(2, 2), Punkt(3, 3) };