Passing by Reference

[AnthonyF]

I wrote this function:

ostream& operator<<(ostream& out, LoginData& a)
{
	out << a.getLogin();
	return out;
}

and I originally wrote the first line of the function as
out << a->getLogin();

because I thought I was passing a pointer to a LoginData object. My compiler told me that I was passing a class and that I should use "." instead of "->" so I changed it and it compiles fine and seems to work correctly. So I then tested what would happen if I wrote ostream& operator<<(ostream& out, LoginData a) and it compiles and seems to work the same so now I am extremely confused about what I am doing and why both LoginData and LoginData& compile and work.

[Athar]

A reference is an alias for an object. If you don't pass by reference (that is, by value), a copy of the object you're passing is made.

[AnthonyF]

But when I pass by reference, aren't I passing a pointer to the object instead of the object itself?

[Zhuge]

Sort of. It's like a pointer that is always dereferenced. As Athar said, if you don't pass a reference, a separate copy of the object is made for the function to use.

[jsmith]

passing by reference allows you to access the data as if it were by value, even though behind the scenes the compiler may be passing a pointer rather than the actual value.