A little question about &

[isnork]

Hi,
I have a little doubt with this:

Whatever y(...);
Whatever& x = y;

With that code, x will act as a pointer to y??

Thanks in advance!

[quirkyusername]

not quite, x is a reference to a Whatever object

http://cplusplus.com/doc/tutorial/pointers/

[Disch]

Basically yeah.

Another way to think of it is that x "is" y. Just another name for the same object.

[isnork]

Oh! Ok!!

Thanks a lot!

[TheNoobie]

Example..

#include <iostream>
void MyFunction(int, int &);

int main()
{
	int number, value;
	std::cout << "Please enter a number: ";
	std::cin >> number;
	std::cout << "How much would you like to multiply it by: ";
	std::cin >> value;
	
	MyFunction(value, number);
	
	std::cout << "Your multiplied number is: " << number;
	return 0;
}

void MyFunction(int value, int &rNumber)
{
	rNumber *= value;
}

This will allow you to change a parameter in a function outside of it's scope.

[isnork]

I've got another doubt:

Having something like this:

class A {
  private:
     Whatever& x;
  public:
      A(const A& y);
};

I want to create another copy of x from the copy constructor, as if I typed:
x = new Whatever( y.GetX() )

However, this won't compile. Is there a way to do this? I think x should be Whatever* instead of Whatever& .
Is that so?

Thanks!

[king214]

I want to create another copy of x from the copy constructor, as if I typed:
x = new Whatever( y.GetX() )

What exactly do you want x to refer to?

I think x should be Whatever* instead of Whatever& .
Is that so?

Well pointers and references are almost the same thing although in some situations references are just a little cleaner syntax-wise, but anyhing you can do with a reference you can do with a pointer. Although thats not always true the other way around...

[isnork]

What exactly do you want x to refer to?

I want a local copy of class member x, so I can modify its values without changing the ones of y::x[i].

So, when typing [i]x = new Whatever( y.GetX() ); x is pointing to a new copy of y::x.

What I'm asking if its possible to achieve the same as above using Whatever& x instead of
Whatever* x directly.

I tried to explain it as clearly as I could, so my apologies if its still unclear :D

Thanks!

[LB]

A reference is assigned to when you initialize it, the first time. And it can never be changed again. Ever. For instance:

int x = 5;
int y = 7;
int& z = x; //z is now the same variable as x
z = y; //x and z now equal y, but neither is the same variable as y 

[isnork]

I think that I get it now. Thanks to everyone!