The "&" denotes a reference instead of a pointer to an object (In your case a constant reference).
The advantage of having a function such as
foo(string const& myname)
over
foo(string const* myname)
is that in the former case you are guaranteed that myname is non-null, since C++ does not allow NULL references. Since you are passing by reference, the object is not copied, just like if you were passing a pointer.
Your second example:
const string &GetMethodName() { ... }
Would allow you to return a constant reference to, for example, a member variable. This is useful if you do not wish a copy to be returned, and again be guaranteed that the value returned is non-null. As an example, the following allows you direct, read-only access:
class A
{
public:
int bar() const {return someValue;}
//Big, expensive to copy class
}
class B
{
public:
A const& getA() { return mA;}
private:
A mA;
}
void someFunction()
{
B b = B();
//Access A, ability to call const functions on A
//No need to check for null, since reference is guaranteed to be valid.
int value = b.getA().bar();
}
You have to of course be careful to not return invalid references.
Compilers will happily compile the following (depending on your warning level and how you treat warnings)
int const& foo()
{
int a;
//This is very bad, returning reference to something on the stack. This will
//crash at runtime.
return a;
}
Basically, it is your responsibility to ensure that whatever you are returning a reference to is actually valid.
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;
Best Answer
The
&
operator has three meanings in C++.2 & 1 == 3
int x = 3; int* ptr = &x;
int x = 3; int& ref = x;
Here you have a reference type modifier. Your function
class1 &class1::instance()
is a member function of typeclass1
calledinstance
, that returns a reference-to-class1
. You can see this more clearly if you writeclass1& class1::instance()
(which is equivalent to your compiler).