Why?

[randysimes]

I was wondering why the output of the following code is:
x = 1
y = 2
z = 3
x = 2 (understand here up)
y = 2 (Why isn't it 3)
z = 5 (Why isn't it 4)

int F(int& x, int y)
{
x = x+1;
y = y +1;
return x + y;
}

int main()
{
int x, y, z;
x = 1, y = 2, z = 3;
std::cout << "x=" << x << '\n'
              << "y=" << y << '\n'
              << "z=" << z << '\n';
z = F(x,y);
std::cout << "x=" << x << '\n'
              << "y=" << y << '\n'
              << "z=" << z << '\n';
return 0;
}

[Banfa]

It has do do with the differing types of the parameters to F

int F(int& x, int y);

If you know and understand the difference between the declarations of the parameters int& x and int y then you should be able to easily explain your results.

So do you know what that & in the declaration means and what that in turn implies for program execution?

[randysimes]

I can see that you can change the value of the & variable, that it.

[Banfa]

That is the effect but the question you need to answer is what is the syntactic meaning of the & in this construct? Why does it cause the value of x outside the function to be modified.

[randysimes]

Is it the addess of the variable or is it passed by reference? I couldn't find much about & online.

[Frinavale]

You'd be interested in looking up the topic of pointers and passing parameters by reference.

[Banfa]

Try reading this