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

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++?

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.

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.

string * and string& differ in a couple of ways. First of all, the pointer points to the address location of the data. The reference points to the data. If you had the following function:

int foo(string *param1);

You would have to check in the function declaration to make sure that param1 pointed to a valid location. Comparatively:

int foo(string &param1);

Here, it is the caller's responsibility to make sure the pointed to data is valid. You can't pass a "NULL" value, for example, int he second function above.

With regards to your second question, about the method return values being a reference, consider the following three functions:

string &foo();
string *foo();
string foo();

In the first case, you would be returning a reference to the data. If your function declaration looked like this:

string &foo()
{
    string localString = "Hello!";
    return localString;
}

You would probably get some compiler errors, since you are returning a reference to a string that was initialized in the stack for that function. On the function return, that data location is no longer valid. Typically, you would want to return a reference to a class member or something like that.

The second function above returns a pointer in actual memory, so it would stay the same. You would have to check for NULL-pointers, though.

Finally, in the third case, the data returned would be copied into the return value for the caller. So if your function was like this:

string foo()
{
    string localString = "Hello!";
    return localString;
}

You'd be okay, since the string "Hello" would be copied into the return value for that function, accessible in the caller's memory space.