What does '&' do in a C++ declaration?

Question

I am a C guy and I'm trying to understand some C++ code. I have the following function declaration:

int foo(const string &myname) {   cout << "called foo for: " << myname << endl;   return 0; } 

How does the function signature differ from the equivalent C:

int foo(const char *myname) 

Is there a difference between using string *myname vs string &myname? What is the difference between & in C++ and * in C to indicate pointers?

Similarly:

const string &GetMethodName() { ... } 

What is the & doing here? Is there some website that explains how & is used differently in C vs C++?

Answer 1

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.

Answer 2

Here, & is not used as an operator. As part of function or variable declarations, & denotes a reference. The C++ FAQ Lite has a pretty nifty chapter on references.