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

#include
using namespace std;
int add(int &number);

int main ()
{
            int number;
            int result;
            number=5;
            cout << "The value of the variable number before calling the function : " << number << endl;
            result=add(&number);
            cout << "The value of the variable number after the function is returned : " << number << endl;
            cout << "The value of result : " << result << endl;
            return(0);
}

int add(int &p)
{
            *p=*p+100;
            return(*p);
}

This is invalid code on several counts. Running it through g++ gives:

crap.cpp: In function ‘int main()’:
crap.cpp:11: error: invalid initialization of non-const reference of type ‘int&’ from a temporary of type ‘int*’
crap.cpp:3: error: in passing argument 1 of ‘int add(int&)’
crap.cpp: In function ‘int add(int&)’:
crap.cpp:19: error: invalid type argument of ‘unary *’
crap.cpp:19: error: invalid type argument of ‘unary *’
crap.cpp:20: error: invalid type argument of ‘unary *’

A valid version of the code reads:

#include
using namespace std;
int add(int &number);

int main ()
{
            int number;
            int result;
            number=5;
            cout << "The value of the variable number before calling the function : " << number << endl;
            result=add(number);
            cout << "The value of the variable number after the function is returned : " << number << endl;
            cout << "The value of result : " << result << endl;
            return(0);
}

int add(int &p)
{
            p=p+100;
            return p;
}

What is happening here is that you are passing a variable "as is" to your function. This is roughly equivalent to:

int add(int *p)
{
      *p=*p+100;
      return *p;
}

However, passing a reference to a function ensures that you cannot do things like pointer arithmetic with the reference. For example:

int add(int &p)
{
            *p=*p+100;
            return p;
}

is invalid.

If you must use a pointer to a reference, that has to be done explicitly:

int add(int &p)
{
                    int* i = &p;
            i=i+100L;
            return *i;
}

Which on a test run gives (as expected) junk output:

The value of the variable number before calling the function : 5
The value of the variable number after the function is returned : 5
The value of result : 1399090792