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 << \"ca         


        

Answer 1

Your function declares a constant reference to a string:

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

A reference has some special properties, which make it a safer alternative to pointers in many ways:

• it can never be NULL
• it must always be initialised
• it cannot be changed to refer to a different variable once set
• it can be used in exactly the same way as the variable to which it refers (which means you do not need to deference it like a pointer)

How does the function signature differ from the equivalent C:

int foo(const char *myname)

There are several differences, since the first refers directly to an object, while const char* must be dereferenced to point to the data.

Is there a difference between using string *myname vs string &myname?

The main difference when dealing with parameters is that you do not need to dereference &myname. A simpler example is:

int add_ptr(int *x, int* y)
{
    return *x + *y;
}
int add_ref(int &x, int &y)
{
    return x + y;
}

which do exactly the same thing. The only difference in this case is that you do not need to dereference x and y as they refer directly to the variables passed in.

const string &GetMethodName() { ... }

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

This returns a constant reference to a string. So the caller gets to access the returned variable directly, but only in a read-only sense. This is sometimes used to return string data members without allocating extra memory.

There are some subtleties with references - have a look at the C++ FAQ on References for some more details.