In C++, what does & mean after a function's return type?

Question

In a C++ function like this:

int& getNumber();

what does the & mean? Is it different from:

int getNumb         


        

Answer 1

Yes, it's different.

The & means you return a reference. Otherwise it will return a copy (well, sometimes the compiler optimizes it, but that's not the problem here).

An example is vector. The operator[] returns an &. This allows us to do:

my_vector[2] = 42;

That wouldn't work with a copy.

Answer 2

It means that it is a reference type. What's a reference?

Wikipedia:

In the C++ programming language, a reference is a simple reference datatype that is less powerful but safer than the pointer type inherited from C. The name C++ reference may cause confusion, as in computer science a reference is a general concept datatype, with pointers and C++ references being specific reference datatype implementations. The declaration of the form:

Type & Name

where is a type and is an identifier whose type is reference to .

Examples:

1. int A = 5;
2. int& rA = A;
3. extern int& rB;
4. int& foo ();
5. void bar (int& rP);
6. class MyClass { int& m_b; /* ... */ };
7. int funcX() { return 42 ; }; int (&xFunc)() = funcX;

Here, rA and rB are of type "reference to int", foo() is a function that returns a reference to int, bar() is a function with a reference parameter, which is reference to int, MyClass is a class with a member which is reference to int, funcX() is a function that returns an int, xFunc() is an alias for funcX.

Rest of the explanation is here

Answer 3

It's a reference, which is exactly like a pointer except you don't have to use a pointer-dereference operator (* or ->) with it, the pointer dereferencing is implied.

Especially note that all the lifetime concerns (such as don't return a stack variable by address) still need to be addressed just as if a pointer was used.

Answer 4

It's different.

int g_test = 0;

int& getNumberReference()
{
     return g_test;
}

int getNumberValue()
{
     return g_test;
}

int main()
{
    int& n = getNumberReference();
    int m = getNumberValue();
    n = 10;
    cout << g_test << endl; // prints 10
    g_test = 0;
    m = 10;
    cout << g_test << endl; // prints 0
    return 0;
}

the getNumberReference() returns a reference, under the hood it's like a pointer that points to an integer variable. Any change applyed to the reference applies to the returned variable.

The getNumberReference() is also a left-value, therefore it can be used like this:

getNumberReference() = 10;

Answer 5

Yes, the int& version returns a reference to an int. The int version returns an int by value.

See the section on references in the C++ FAQ

Answer 6

"&" means reference, in this case "reference to an int".