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Like for instance in a function like so:

void Monster::Attack(Player& player)

this function would be in the .cpp file for the Monster class, and using the header for the Player class.

would the new object player have access to all the goodies that are embedded into the Player class. Also why couldn't:

void Monster::Attack(Player player)

be done?

Jcodin
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  • I'm not sure what you're asking but a reference is used when you want to modify the state of a class without having to dereference pointers.. Monster's attack function should take a reference or pointer if it is modifying the player state.. If you do `Attack(Player player)` then the attack function gets a "copy" of that player. – Brandon Apr 12 '14 at 05:36
  • Possible duplicate of http://stackoverflow.com/questions/16747807/in-c-how-do-references-and-pointers-in-parameters-actually-work-what-are-thei – user35443 Apr 12 '14 at 05:36
  • The semantics of storage qualifiers has been written about time and time again. I recommend a good google search for how they work. It'll serve you more good than getting the same answer parroted back at you from somewhere across the tubes. – CodaFi Apr 12 '14 at 05:37
  • That's someone helpful actually, so using a reference will modify the data and not have to return anything instead? – Jcodin Apr 12 '14 at 05:38
  • In addition to what CantChooseUsernames said, a reference is used whenever you want to pass an object around without making a copy of it. It is basically nothing more than a glorified non-nullable pointer, nothing more. The compiler emits the same machine code whether a reference or a pointer is used, it just validates that a reference can never be NULL, that's all. – Remy Lebeau Apr 12 '14 at 05:41

3 Answers3

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You can choose the appropriate function once you understand the pros and cons.

With

void Monster::Attack(Player& player);

You are able to avoid the cost of making a copy of the input argument. You are able to modify the input object. That can be a good thing or a bad thing depending on your requirements.

With

void Monster::Attack(Player player);

You are incurring the cost of making a copy of the input argument. You can modify player in the function but that won't change the object in the calling function. Once again, this can be a good thing or a bad thing depending on your requirements.

R Sahu
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  • so using a copy, basically does like so. Makes a replica of the object and class and function (or whatever its being copied into), but with references your passing the original in to be modified – Jcodin Apr 12 '14 at 20:21
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I believe references are again pointers internally. The syntax is just a little different than that of pointers, but under the hood the same things happen. You are making changes to the original object.

The problem with the second method declaration is that it passes the player parameter not using a pointer or reference, but as an actually object. That means pass-by-value, which means a copy is made. And then you make changes to the copy and what happens? The original remains untouched, which is probably not what you want.

dandan78
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  • So if theres things that need to change and not look back use references or pointers, something dynamic that flows with the game. But if you want to run with just copies the initial values in the class will not change just temporarily while the function is called – Jcodin Apr 12 '14 at 05:52
  • Or another way of putting it your changing the blueprint instead the house? – Jcodin Apr 12 '14 at 05:54
  • Not really, no. Changing the blueprint would be akin to changing the class definition. To expand on your house analogy, say you have a method that operates on a house and changes the color of the roof and doesn't pass by reference/pointer. It would take the original house, makes a copy of it (create a new house with the same attributes as the original), changes the roof color to say green *on the copy*, then the copy gets destroyed when the method returns and what you're left with is the original, whose roof is still the same red color. – dandan78 Apr 12 '14 at 06:01
  • so when you want to make an actually change to the object itself. When that happens does that not go into the class of the object and change the value of whatever data member equaled to? Or just for that object. So i'm trying to understand it in terms of memory. So every object stores at a new address but takes up the same amount of space as the class itself. And every object made takes up a different address and so on and so forth – Jcodin Apr 12 '14 at 06:35
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My advice would be to understand references first, separate from the context of function parameters.

Here's some sample code:

int a = 5;
int &b = a;

Now, a and b are both labels for the same variable. This code is actually identical to:

int b = 5;
int &a = b;

Anywhere you subsequently use a, you can use b instead; no exceptions to this rule. The key point is that you can have multiple identifiers that identify the same variable.

Moving onto the function context. If you have:

int a;
foo(a);

and 

foo(int &b)
{

then main's a and foo's b are both labels for the same variable. Doing an operation on b in foo is exactly the same as doing the operation on a in main. They are both the same variable, just with different names. Similar to how Prince William and the Duke of Cambridge are the same guy.

M.M
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  • So my question would be, does every object take up a place in the memory equal to size of the class, so when you reference it you actually get the address of the object to modify data members. Vs just a copy that just modifies it for the time of the function – Jcodin Apr 12 '14 at 20:18
  • Yes. When you write a variable name, it's said to *designate* a storage area. Then it depends on the context whether or not you retrieve the value from that area, or operate with its address. – M.M Apr 13 '14 at 00:26