Thread: Passing an address to a function

Lets say we have a variable we want to change through a function.

What is the best way to do this, is one method better than another?

Code:

int number;

void ChangeNumber(int &number)
{

    // change number

}

Is there a better way to do that.

You should prefer using references unless it's possible to pass a "nothing" object, in which case you have no choice but to use pointers because there's no such thing as a null reference.

Just to make sure I understand.

Code:

void ChangeNumber(int *number)
{

    // change number

}

I am at a point in programming where I have just recently started to learn the language. I have used C++ for about 2-3 years but I just toyed with it. I am just now getting into it and learning how/why it works.

So the & operator gets the address-of and * creates a reference.

Now, what is it actually doing. I understand that the & operator passes the memory location of the variable. But what does * do?

>So the & operator gets the address-of and * creates a reference.
No, not really. Operators in C++ are overloaded to do different things based on the context. When used in a declaration, the ampersand means that you are creating a reference:

Code:

int i = 5;
int& ri = i; // Any changes to ri reflect i

When used as a value, the ampersand takes the address of an object:

Code:

int i = 5;

cout<< &i <<endl; // Prints i's address

It also does other things, such as bitwise AND, but we'll ignore that for now.

The asterisk in a declaration creates a pointer (ie. a variable that holds an address):

Code:

int i = 5;
int *pi = &i; // pi "points to" i

When used in an expression, the asterisk means to dereference a pointer so that you are looking at the contents of the address rather than the address itself:

Code:

cout<< *pi <<endl; // Prints 5 because pi points to i

Thank you both for the info.