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:

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int number;

void ChangeNumber(int &number)
{

    // change number

}

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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:

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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?

& is the address-of when used with a variable
& means its a reference when used with parameters and variable declartions
* means its a pointer when used with parameters and variable declartions
* means to deference a pointer when used with a variable

Code:

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int a = 10;
int *b; // B is a pointer
b = &a;  // &a is the address of a
int &c = a; // &c means c is a reference to a
*b = 10; // go to the location b points to and change its value to 10

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>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:

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

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

Code:

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int i = 5;

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

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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:

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

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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:

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

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Thank you both for the info.