在 C++ 中，对象和指向对象的指针有什么区别？

Question

In java and objective-c, a variable representing an object is generally a pointer to that object. However, it seems that in C++, it's common to have non-pointer types hold objects. What's the difference between the two?

If I pass a struct as an argument to a function, I believe I'm passing by value, meaning I'm actually creating a new struct in memory, and changes to that struct inside the function its passed to won't affect the "source" struct outside the function. However, if I pass a pointer to a struct, there's still only one original struct, and changes to the struct referenced by the pointer will be visible to any code which is aware of this struct. Do I have that right?

So, is there any difference with objects? When I pass a non-pointer object to a function, does the entire object get copied?

Answer 1

It's exactly as you said.

When you pass an object by value, its copy constructor is invoked to produce a new instance of such object that will be used inside the function. The changes done to such new object won't be reflected to the original one ¹ .

As with structures, the default copy constructor just does a shallow copy of the original object - ie, its fields are copied ² to the new instance; in many cases this is not desirable (eg if the object wraps a pointer/another resource), so there are classes which redefine the copy constructor or disable it completely. Objects of these last classes can only be passed by pointer or reference.

Passing objects by value can be costly if they are bigger than a pointer (in size) or in general if their copy constructor isn't "cheap". On the other hand, in comparison to pointers, the pass-by-value yields the usual advantages of not having to specify the pointer ownership, letting the callee do whatever it wants with the object, etc.

Notice that passing an object by value kills the polymorphism. This because a function receiving an object by value receives a statically typed object, with a precise size and type, so any attempt to pass an object of a derived class will result in object slicing (the copy constructor for the base class is called, that by default just copies the fields that are available in the base class).

This is the reason why often the preferred method of passing objects is by const reference. This yields several advantages:

• no copies involved; the object that the callee will see will be exactly the one specified at the moment of the call;
• no changes to the original object can be made, thanks to the const qualifier;
• if however the callee needs to change a copy of the object, it can still construct a copy by itself from the reference;
• no awkward pointer syntax;
• polymorphism preserved, since behind the scenes we're actually passing a pointer;
• no big doubts about object ownership: the general rule about references is that they are owned by the caller.

---

1. As far as the "raw fields" of the object are concerned; naturally if the original object and the copy continue to share a pointer/handle to the same resource some modifications to one may affect the other.

2. Primitive types (and in general PODs) are copied bitwise, while the copy constructor is called for non-POD types.

Answer 2

The difference mostly has to do with where in memory an object is allocated. For instance:

int main() {
    MyObject x;   //allocates space for an instance of MyObject on the stack
    MyObject* y;  //allocates space for a pointer on the stack
    MyObject* z = new MyObject();  //allocates space for a pointer on the 
                                   //stack and an object instance in the heap and
                                   //sets the pointer to point to the new instance
    MyObject* a = &x;  //allocates space for a pointer on the stack and 
                       //makes it point to 'x'
    ...
}

int someFunc(MyObject byValue, MyObject* byReference) {
   //the 'byValue' parameter will be passed by creating a copy of the 
   //entire source object on the stack (can be quite expensive for 
   //complex object types)

   //the 'byReference' parameter will be passed by creating a 
   //copy of the source pointer on the stack and setting it to 
   //point to the source object in memory
}

Answer 3

In C++, a variable is the variable that it is representing. It is the actual object in memory, at the actual location.

However, you can choose to make such a variable represent a pointer instead, in which case it'll say "Hey, I'm me, I am pointing over there, The object you want isn't here. it's THERE, Yes, there. Go on, get there!".

Unless you're explicitly using C++'s "reference type", which I suspect you're not, then ALL arguments you pass are by value.

Answer 4

The answers to the questions in your second and third paragraph are both "yes". More specifically, if you pass an object to a function by value, the function will receive a copy of that object (created by the copy constructor).

Answer 5

When you pass an object to a function by value, it's copied by means of its class's copy constructor. If you haven't defined a copy constructor, there'll be a default one supplied by the compiler (unless you take special steps to avoid that), which is equivalent to copying the members by hand.

The preferred thing to do is often actually to pass a const reference rather than either a pointer or the object itself.

(You might want to be aware that actually a struct is just a class whose members are public by default; in particular, structs can have user-defined copy constructors too. A struct is not necessarily just plain inert data.)

Answer 6

You have that right.

Indeed, that's how it works. A pointer stores the memory address of a variable.

When you pass a pointer (to an object) for a function as a parameter, it means that function will have access to that object through it's memory address instead of a new object being created on the stack.

Check this thread for more info on pointers.

Answer 7

• A pointer stores an memory address.
• A variable stores a value.

Ex:

Player * a; // Allocated an int value to stored an address, you will able to access to value at that address by a -> but you have to allocated it and free when it done

Player a; // Allocated a block of memory that size equal = Player size.