C++ Returning Pointers/References

Question

I have a fairly good understanding of the dereferencing operator, the address of operator, and pointers in general.

I however get confused when I see stuff such as this:

int* returnA() {
    int *j = &a;
    return j;
}

int* returnB() {
    return &b;
}

int& returnC() {
    return c;
}

int& returnC2() {
    int *d = &c;
    return *d;
}

1. In returnA() I'm asking to return a pointer; just to clarify this works because j is a pointer?
2. In returnB() I'm asking to return a pointer; since a pointer points to an address, the reason why returnB() works is because I'm returning &b?
3. In returnC() I'm asking for an address of int to be returned. When I return c is the & operator automatically "appended" c?
4. In returnC2() I'm asking again for an address of int to be returned. Does *d work because pointers point to an address?

Assume a, b, c are initialized as integers.

Can someone validate if I am correct with all four of my questions?

Answer 1

In returnA() I'm asking to return a pointer; just to clarify this works because j is a pointer?

Yes, int *j = &a initializes j to point to a. Then you return the value of j, that is the address of a.

In returnB() I'm asking to return a pointer; since a pointer points to an address, the reason why returnB() works is because I'm returning &b?

Yes. Here the same thing happens as above, just in a single step. &b gives the address of b.

In returnC() I'm asking for an address of int to be returned. When I return c is the & operator automatically appended?

No, it is a reference to an int which is returned. A reference is not an address the same way as a pointer is - it is just an alternative name for a variable. Therefore you don't need to apply the & operator to get a reference of a variable.

In returnC2() I'm asking again for an address of int to be returned. Does *d work because pointers point to an address?

Again, it is a reference to an int which is returned. *d refers to the original variable c (whatever that may be), pointed to by c. And this can implicitly be turned into a reference, just as in returnC.

Pointers do not in general point to an address (although they can - e.g. int** is a pointer to pointer to int). Pointers are an address of something. When you declare the pointer like something*, that something is the thing your pointer points to. So in my above example, int** declares a pointer to an int*, which happens to be a pointer itself.

Answer 2

In returnC() and returnC2() you are not asking to return the address.

Both these functions return references to objects.
A reference is not the address of anything it is an alternative name of something (this may mean the compiler may (or may not depending on situation) use an address to represent the object (alternatively it may also know to keep it in register)).

All you know that a reference points at a specific object.
While a reference itself is not an object just an alternative name.

Answer 3

All of your examples produce undefined run-time behavior. You are returning pointers or references to items that disappear after execution leaves the function.

Let me clarify:

int * returnA()
{
  static int a;  // The static keyword keeps the variable from disappearing. 
  int * j = 0;   // Declare a pointer to an int and initialize to location 0.
  j = &a;        // j now points to a.
  return j;      // return the location of the static variable (evil).
}

In your function, the variable j is assigned to point to a's temporary location. Upon exit of your function the variable a disappears, but it's former location is returned via j. Since a no longer exists at the location pointed to by j, undefined behavior will happen with accessing *j.

Variables inside functions should not be modified via reference or pointer by other code. It can happen although it produces undefined behavior.

Being pedantic, the pointers returned should be declared as pointing to constant data. The references returned should be const:

const char * Hello()
{
  static const char text[] = "Hello";
  return text;
}

The above function returns a pointer to constant data. Other code can access (read) the static data but cannot be modified.

const unsigned int& Counter()
{
  static unsigned int value = 0;
  value = value + 1;
  return value;
}

In the above function, the value is initialized to zero on the first entry. All next executions of this function cause value to be incremented by one. The function returns a reference to a constant value. This means that other functions can use the value (from afar) as if it was a variable (without having to dereference a pointer).

In my thinking, a pointer is used for an optional parameter or object. A reference is passed when the object must exist. Inside the function, a referenced parameter means that the value exists, however a pointer must be checked for null before dereferencing it. Also, with a reference, there is more guarantee that the target object is valid. A pointer could point to an invalid address (not null) and cause undefined behavior.

Answer 4

Semantically, references do act as addresses. However, syntactically, they are the compiler's job, not yours, and you can treat a reference as if it is the original object it points to, including binding other references to it and having them refer to the original object too. Say goodbye to pointer arithmetic in this case.

The downside of that is that you can't modify what they refer to - they are bound at construct time.

Answer 5

Although Peter answered your question, one thing that's clearly confusing you is the symbols * and &. The tough part about getting your head around these is that they both have two different meanings that have to do with indirection (even excluding the third meanings of * for multiplication and & for bitwise-and).

• *, when used as part of a type indicates that the type is a pointer: int is a type, so int* is a pointer-to-int type, and int** is a pointer-to-pointer-to-int type.

• & when used as part of a type indicates that the type is a reference. int is a type, so int& is a reference-to-int (there is no such thing as reference-to-reference). References and pointers are used for similar things, but they are quite different and not interchangable. A reference is best thought of as an alias, or alternate name, for an existing variable. If x is an int, then you can simply assign int& y = x to create a new name y for x. Afterwords, x and y can be used interchangeably to refer to the same integer. The two main implications of this are that references cannot be NULL (since there must be an original variable to reference), and that you don't need to use any special operator to get at the original value (because it's just an alternate name, not a pointer). References can also not be reassigned.

• * when used as a unary operator performs an operation called dereference (which has nothing to do with reference types!). This operation is only meaningful on pointers. When you dereference a pointer, you get back what it points to. So, if p is a pointer-to-int, *p is the int being pointed to.

• & when used as a unary operator performs an operation called address-of. That's pretty self-explanatory; if x is a variable, then &x is the address of x. The address of a variable can be assigned to a pointer to the type of that variable. So, if x is an int, then &x can be assigned to a pointer of type int*, and that pointer will point to x. E.g. if you assign int* p = &x, then *p can be used to retrieve the value of x.

So remember, the type suffix & is for references, and has nothing to do with the unary operatory &, which has to do with getting addresses for use with pointers. The two uses are completely unrelated. And * as a type suffix declares a pointer, while * as a unary operator performs an action on pointers.