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?


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) 


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
  int bar() const {return someValue;}
  //Big, expensive to copy class

class B
 A const& getA() { return mA;}
 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.

Source: StackOverflow.com

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